Chapter 9 Areas Of Parallelograms And Triangles (Additional Questions)

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Solution:

According to the given definition, two figures are on the same base if they have a common side.

The two figures are $\triangle ABC$ and parallelogram $ABCD$.

The sides of $\triangle ABC$ are AB, BC, and AC.

The sides of parallelogram $ABCD$ are AB, BC, CD, and DA.

For $\triangle ABC$ and parallelogram $ABCD$ to be on the same base BC, the side BC must be common to both figures.

From the list of sides, BC is a side of both $\triangle ABC$ and parallelogram $ABCD$.

Therefore, the common side is BC.

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The correct option is (C) Side BC.

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Solution:

The figures are $\triangle ABC$ and parallelogram $ABCD$.

They are on the same base BC, which means BC lies on one parallel line.

For $\triangle ABC$ to be between the same parallels as the base BC, the opposite vertex A must lie on the other parallel line.

For parallelogram $ABCD$ to be between the same parallels as the base BC, the opposite side AD must lie on the other parallel line.

For both figures to be between the same two parallel lines, the line containing BC must be parallel to the line containing both vertex A and the side AD.

This means the line containing AD must be parallel to the line containing BC.

Thus, AD must be parallel to BC.

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The correct option is (A) AD is parallel to BC.

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Solution:

This question is based on a fundamental theorem related to the areas of parallelograms.

The theorem states that two parallelograms on the same base (or on equal bases) and between the same parallels are equal in area.

Given that the two parallelograms are on the same base and between the same parallels, their areas must be equal.

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The correct option is (B) Areas are equal.

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Solution:

The area of a parallelogram is calculated by multiplying its base by its corresponding height.

The formula is:

Area of parallelogram = Base $\times$ Height

Comparing this with the given options:

(A) $\frac{1}{2} \times \text{base} \times \text{height}$ is the formula for the area of a triangle.

(B) $\text{base} \times \text{height}$ matches the formula for the area of a parallelogram.

(C) $(\text{side})^2$ is the formula for the area of a square.

(D) Length $\times$ Width is the formula for the area of a rectangle (a special type of parallelogram, but "base $\times$ height" is the general formula). While length can be a base and width can be a height for a rectangle, the general formula for any parallelogram requires the perpendicular height to the chosen base.

Thus, the correct formula is base $\times$ height.

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The correct option is (B) $\text{base} \times \text{height}$.

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Solution:

This question relates to a fundamental geometric theorem concerning the areas of triangles and parallelograms.

The theorem states that if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.

Given that the triangle and the parallelogram are on the same base and between the same parallels, their areas follow this relationship.

Area of triangle = $\frac{1}{2} \times$ Area of parallelogram

Therefore, the area of the triangle is half of the area of the parallelogram.

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The correct option is (C) Half of.

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Solution:

This question asks about the relationship between the areas of two triangles that share a common base (or have equal bases) and are situated between the same pair of parallel lines.

According to a theorem in geometry, triangles standing on the same base (or equal bases) and lying between the same parallel lines have equal areas.

Let the two triangles be $\triangle ABC$ and $\triangle DBC$, with common base BC, and let BC be parallel to a line containing A and D. Since they are on the same base BC and between the same parallels (BC and the line through A and D), their areas are equal.

Area($\triangle ABC$) = $\frac{1}{2} \times \text{Base} \times \text{Height}$

Area($\triangle DBC$) = $\frac{1}{2} \times \text{Base} \times \text{Height}$

Since the base (BC) is the same for both triangles, and the perpendicular distance between the parallel lines is the height for both triangles, their areas must be equal.

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The correct option is (B) Equal.

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Solution:

We are given $\triangle ABC$ where D is the midpoint of side BC.

This means that AD is a median of $\triangle ABC$.

The base of $\triangle ABD$ is BD and the base of $\triangle ACD$ is CD.

Since D is the midpoint of BC, we have BD = CD.

Let h be the perpendicular height from vertex A to the base BC. This height is common for both $\triangle ABD$ (with base BD) and $\triangle ACD$ (with base CD) as their bases lie on the same line segment BC and they share the common vertex A.

The area of $\triangle ABD$ is given by:

Area$(\triangle ABD) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BD \times h$

The area of $\triangle ACD$ is given by:

Area$(\triangle ACD) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times CD \times h$

Since BD = CD, we can substitute BD for CD in the second equation:

Area$(\triangle ACD) = \frac{1}{2} \times BD \times h$

Comparing the areas:

Area$(\triangle ABD) = \frac{1}{2} \times BD \times h$

Area$(\triangle ACD) = \frac{1}{2} \times BD \times h$

Thus, Area$(\triangle ABD) =$ Area$(\triangle ACD)$.

This illustrates the property that a median of a triangle divides it into two triangles of equal areas.

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The correct option is (C) Area$(\triangle ABD) =$ Area$(\triangle ACD)$.

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Solution:

The formula for the area of a triangle is:

Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height}$

Given:

Base = $10 \text{ cm}$

Height = $5 \text{ cm}$

Substituting the given values into the formula:

Area = $\frac{1}{2} \times 10 \text{ cm} \times 5 \text{ cm}$

Area = $\frac{1}{2} \times 50 \text{ cm}^2$

Area = $25 \text{ cm}^2$

Thus, the area of the triangle is $25 \text{ cm}^2$.

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The correct option is (B) $25 \text{ cm}^2$.

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Solution:

The formula for the area of a parallelogram is:

Area = Base $\times$ Height

Given:

Area = $60 \text{ cm}^2$

Base = $12 \text{ cm}$

Let the height be $h$ cm.

Substitute the given values into the formula:

$60 \text{ cm}^2 = 12 \text{ cm} \times h$

To find the height $h$, divide the area by the base:

$h = \frac{60 \text{ cm}^2}{12 \text{ cm}}$

$h = 5 \text{ cm}$

Thus, the corresponding height of the parallelogram is 5 cm.

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The correct option is (A) 5 cm.

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Solution:

Let's analyze each option based on the definitions of figures on the same base and between the same parallels.

Recall the definitions:

Two figures are on the same base if they have a common side.

Two figures are between the same parallels if their common base lies on one parallel line and the vertex (or opposite side) of each figure lies on the other parallel line.

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(A) A rectangle and a parallelogram with the same length and equal area.

Let the rectangle have sides $l$ and $w$. Its area is $lw$.

Let the parallelogram have sides $a$ and $b$ and the angle between them be $\theta$. Its area is $ab \sin\theta$.

The condition "same length" means that a specific length associated with the rectangle is equal to a specific length associated with the parallelogram. This could refer to a side length, a diagonal length, or even a height. It does not explicitly state that they share a common side ("same base") or even have equal bases. The condition "equal area" is given: $lw = ab \sin\theta$.

If we assume "same length" means a corresponding side, say $l=a$, then $lw = ab \sin\theta \implies lw = lb \sin\theta \implies w = b \sin\theta$. If $l$ (or $a$) is considered the base for both figures, then $w$ is the height of the rectangle and $b \sin\theta$ is the height of the parallelogram. In this case, their heights corresponding to this side are equal. While they *can* be placed on the same base of length $l$ and between parallels $w$ (or $b \sin\theta$) apart, the condition itself does not *necessarily* mean they are currently arranged this way, nor does it guarantee they share the *same base segment*.

If "same length" refers to something else (like a diagonal), then equal areas give even less direct information about bases and heights related to a common pair of parallel lines.

Therefore, this condition does NOT necessarily mean they are on the same base *and* between the same parallels.

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(B) Two parallelograms with equal area and equal base.

Let the parallelograms have areas $A_1, A_2$, bases $b_1, b_2$, and heights $h_1, h_2$.

Given: $A_1 = A_2$ and $b_1 = b_2$.

Area of parallelogram = base $\times$ height, so $A_1 = b_1 h_1$ and $A_2 = b_2 h_2$.

Since $A_1 = A_2$ and $b_1 = b_2$, we have $b_1 h_1 = b_1 h_2$. As $b_1 \neq 0$, this implies $h_1 = h_2$.

The two parallelograms have equal bases (lengths) and equal heights. Figures with equal bases and equal heights can always be placed such that their bases lie on the same line and their opposite sides lie on a parallel line. Thus, they can be placed between the same parallels with equal (collinear) bases on one line. While they are not necessarily on the *same base segment* (a common side), they are necessarily related in a way that allows them to be between the same parallels if their equal bases are made collinear.

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(C) A triangle and a parallelogram on the same base and with vertices on the same line parallel to the base.

This is precisely the definition of a triangle and a parallelogram being on the same base and between the same parallels. The common base is on one parallel line, and the opposite vertex of the triangle and the opposite side of the parallelogram are on the other parallel line.

This condition is necessarily that they are on the same base and between the same parallels.

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(D) Two triangles with equal area and equal base.

Let the triangles have areas $A_1, A_2$, bases $b_1, b_2$, and heights $h_1, h_2$.

Given: $A_1 = A_2$ and $b_1 = b_2$.

Area of triangle = $\frac{1}{2} \times$ base $\times$ height, so $A_1 = \frac{1}{2} b_1 h_1$ and $A_2 = \frac{1}{2} b_2 h_2$.

Since $A_1 = A_2$ and $b_1 = b_2$, we have $\frac{1}{2} b_1 h_1 = \frac{1}{2} b_1 h_2$. As $b_1 \neq 0$, this implies $h_1 = h_2$.

The two triangles have equal bases (lengths) and equal heights. Triangles with equal bases and equal heights can always be placed such that their bases lie on the same line and their third vertices lie on a parallel line. Thus, they can be placed between the same parallels with equal (collinear) bases on one line. Similar to option (B), while they are not necessarily on the *same base segment*, they are necessarily related in a way that allows them to be between the same parallels if their equal bases are made collinear.

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Comparing the options, option (C) explicitly describes figures on the same base and between the same parallels. Options (B) and (D) provide conditions (equal area, equal base length) that imply equal heights, a key characteristic of figures between the same parallels (when on equal bases). Option (A), with the condition "same length" (which is not specified as a base) and equal area, is the least specific and does not necessarily force the figures into the configuration of being on the same base *segment* and between the same parallels.

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The correct option is (A) A rectangle and a parallelogram with the same length and equal area.

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Solution:

We are given two triangles, $\triangle ABC$ and $\triangle DBC$, which are on the same base BC.

We are also given that Area$(\triangle ABC) =$ Area$(\triangle DBC)$.

The formula for the area of a triangle is given by:

Area $= \frac{1}{2} \times \text{base} \times \text{height}$

For $\triangle ABC$, the base is BC and let the corresponding height from vertex A to BC be $h_A$.

Area$(\triangle ABC) = \frac{1}{2} \times BC \times h_A$

For $\triangle DBC$, the base is BC and let the corresponding height from vertex D to BC be $h_D$.

Area$(\triangle DBC) = \frac{1}{2} \times BC \times h_D$

Since Area$(\triangle ABC) =$ Area$(\triangle DBC)$, we have:

$\frac{1}{2} \times BC \times h_A = \frac{1}{2} \times BC \times h_D$

As BC is the base and is non-zero, we can cancel $\frac{1}{2} \times BC$ from both sides of the equation:

$h_A = h_D$

This implies that the perpendicular distance from vertex A to the line containing BC is equal to the perpendicular distance from vertex D to the line containing BC.

When two points are equidistant from a given line, they lie on a line that is parallel to the given line.

Therefore, the line segment joining A and D must be parallel to the line containing BC.

Thus, AD is parallel to BC.

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The correct option is (B) AD is parallel to BC.

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Solution:

The formula for the area of a rectangle is:

Area of rectangle = Length $\times$ Width

Given:

Length = $8 \text{ cm}$

Width = $5 \text{ cm}$

Substituting the given values into the formula:

Area = $8 \text{ cm} \times 5 \text{ cm}$

Area = $40 \text{ cm}^2$

Thus, the area of the rectangle is $40 \text{ cm}^2$.

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The correct option is (A) $40 \text{ cm}^2$.

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Solution:

Let's match the properties in Column A with the corresponding figure types in Column B:

(i) Area = $\text{base} \times \text{height}$ is the formula for the area of a parallelogram. Therefore, (i) matches with (b).

(ii) A median of a triangle divides the triangle into two triangles of equal area. Therefore, (ii) matches with (a).

(iii) The area of a rhombus can be calculated as half the product of its diagonals. Therefore, (iii) matches with (c).

(iv) Area = $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$ is the formula for the area of a trapezium. Therefore, (iv) matches with (d).

The correct matching is:

(i) - (b)

(ii) - (a)

(iii) - (c)

(iv) - (d)

Comparing this with the given options, we find that option (C) corresponds to this matching.

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The correct option is (C) (i)-(b), (ii)-(a), (iii)-(c), (iv)-(d).

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Solution:

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): A median of a triangle divides it into two triangles of equal areas.

Consider $\triangle ABC$ and let AD be a median, where D is the midpoint of BC.

The median AD divides $\triangle ABC$ into two triangles, $\triangle ABD$ and $\triangle ACD$.

The bases of these two triangles on the line BC are BD and CD, respectively.

Since D is the midpoint of BC, BD = CD.

The height of $\triangle ABD$ with respect to base BD is the perpendicular distance from A to the line BC.

The height of $\triangle ACD$ with respect to base CD is also the perpendicular distance from A to the line BC.

Let $h$ be the height from A to BC.

Area$(\triangle ABD) = \frac{1}{2} \times BD \times h$

Area$(\triangle ACD) = \frac{1}{2} \times CD \times h$

Since BD = CD, it follows that Area$(\triangle ABD) =$ Area$(\triangle ACD)$.

Thus, Assertion (A) is True.

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Reason (R): The two triangles formed by a median have the same base and are between the same parallels.

The two triangles, $\triangle ABD$ and $\triangle ACD$, have bases BD and CD respectively. Since D is the midpoint of BC, BD = CD. So, they have equal bases, which can be interpreted as "same base length".

The bases BD and CD lie on the same line, BC.

The vertex opposite to these bases is A, which is common to both triangles.

Triangles with bases on the same line and having a common vertex opposite to the base are considered to be between the line containing the bases and a line through the common vertex parallel to the base line.

In this case, $\triangle ABD$ and $\triangle ACD$ have their bases on the line BC and their common vertex A on a line parallel to BC (any line is parallel to itself, but more significantly, the height from A to BC is constant for both). Thus, they are between the same parallels (the line containing BC and the line through A parallel to BC).

So, Reason (R) is also True.

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Furthermore, Reason (R) correctly explains why Assertion (A) is true. The theorem states that triangles on the same base (or equal bases) and between the same parallels have equal areas. Reason (R) states exactly these conditions for the two triangles formed by a median (interpreting "same base" as equal base length and "between the same parallels" as sharing the same height from the opposite vertex to the collinear bases).

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Therefore, both A and R are true and R is the correct explanation of A.

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The correct option is (A) Both A and R are true and R is the correct explanation of A.

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Solution:

Let's examine the Assertion (A) and Reason (R).

Assertion (A): Parallelograms on the same base have equal areas.

The area of a parallelogram is given by the formula Area = base $\times$ height.

If two parallelograms are on the same base, their base length is the same. However, their areas will be equal only if their corresponding heights are also equal.

Consider two parallelograms on the same base but with different heights. Their areas will be different.

Therefore, Assertion (A) is False.

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Reason (R): This is true only if they are also between the same parallels.

The statement "This" in Reason (R) refers to the assertion, i.e., "Parallelograms on the same base have equal areas".

Reason (R) states that parallelograms on the same base have equal areas *only if* they are also between the same parallels.

If two parallelograms are on the same base, their bases lie on the same line. If they are also between the same parallels, their opposite sides lie on another line parallel to the base line. The perpendicular distance between these parallel lines is the height for both parallelograms corresponding to that base.

Since the base is the same and the height is the same (because they are between the same parallels), their areas will be equal (Area = base $\times$ height).

Conversely, if parallelograms on the same base have equal areas, and Area = base $\times$ height, then base $\times$ height$_1$ = base $\times$ height$_2$. Since the base is the same and non-zero, this implies height$_1$ = height$_2$. Having the same height (perpendicular distance from the opposite side/vertices to the base) means they must lie between the same pair of parallel lines.

Thus, the statement "Parallelograms on the same base have equal areas" is indeed true *if and only if* they are between the same parallels. Reason (R) correctly states that being between the same parallels is a necessary condition for this equality of area when on the same base.

Therefore, Reason (R) is True.

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Assertion (A) is false and Reason (R) is true.

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The correct option is (D) A is false but R is true.

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Given:

A parallelogram ABCD.

Diagonal AC is drawn, forming two triangles $\triangle ABC$ and $\triangle ADC$.

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To Find:

The relationship between Area$(\triangle ABC)$ and Area$(\triangle ADC)$.

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Solution:

In parallelogram ABCD, AC is a diagonal.

We know that a diagonal of a parallelogram divides it into two congruent triangles.

Let's prove the congruence of $\triangle ABC$ and $\triangle ADC$.

In $\triangle ABC$ and $\triangle ADC$:

AB = CD (Opposite sides of a parallelogram are equal)

BC = DA (Opposite sides of a parallelogram are equal)

AC = AC (Common side)

Therefore, $\triangle ABC \cong \triangle CDA$ by SSS congruence rule.

Since congruent triangles have equal areas, we have:

Area$(\triangle ABC) =$ Area$(\triangle CDA)$

Note that $\triangle CDA$ is the same triangle as $\triangle ADC$, just named differently.

So, Area$(\triangle ABC) =$ Area$(\triangle ADC)$.

Alternatively, we can consider $\triangle ABC$ and $\triangle ADC$ are on the same base AC (if we reorient the parallelogram) and between the same parallels AD and BC extended or AB and DC extended.

Another approach: Consider base BC for $\triangle ABC$ and base AD for $\triangle ADC$. In a parallelogram, BC = AD and BC is parallel to AD. Let h be the perpendicular distance between the parallel lines BC and AD. This height h is the height for $\triangle ABC$ with base BC and also the height for $\triangle ADC$ with base AD.

Area$(\triangle ABC) = \frac{1}{2} \times BC \times h$

Area$(\triangle ADC) = \frac{1}{2} \times AD \times h$

Since BC = AD (opposite sides of a parallelogram),

Area$(\triangle ABC) = \frac{1}{2} \times AD \times h =$ Area$(\triangle ADC)$.

Both methods confirm that the areas are equal.

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The correct option is (C) Area$(\triangle ABC) =$ Area$(\triangle ADC)$.

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Given:

Area of $\triangle ABC = 30 \text{ cm}^2$.

AD is a median of $\triangle ABC$.

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To Find:

The area of $\triangle ABD$.

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Solution:

We know that a median of a triangle divides it into two triangles of equal area.

In $\triangle ABC$, AD is a median to the side BC. This means D is the midpoint of BC.

The median AD divides $\triangle ABC$ into two triangles: $\triangle ABD$ and $\triangle ACD$.

According to the property of a median:

Area$(\triangle ABD) =$ Area$(\triangle ACD)$

The sum of the areas of these two triangles is equal to the area of the original triangle $\triangle ABC$:

Area$(\triangle ABC) =$ Area$(\triangle ABD) +$ Area$(\triangle ACD)$

Since Area$(\triangle ABD) =$ Area$(\triangle ACD)$, we can write:

Area$(\triangle ABC) =$ Area$(\triangle ABD) +$ Area$(\triangle ABD)$

Area$(\triangle ABC) = 2 \times$ Area$(\triangle ABD)$

We are given Area$(\triangle ABC) = 30 \text{ cm}^2$. Substituting this value:

$30 \text{ cm}^2 = 2 \times$ Area$(\triangle ABD)$

To find the Area$(\triangle ABD)$, divide the area of $\triangle ABC$ by 2:

Area$(\triangle ABD) = \frac{30 \text{ cm}^2}{2}$

Area$(\triangle ABD) = 15 \text{ cm}^2$

Thus, the area of $\triangle ABD$ is $15 \text{ cm}^2$.

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The correct option is (B) $15 \text{ cm}^2$.

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Solution:

Let the common base of the parallelogram and the rectangle be $b$.

Let the area of the parallelogram be $A_p$ and its corresponding height be $h_p$.

Area of parallelogram, $A_p = b \times h_p$.

Let the area of the rectangle be $A_r$ and its height (width) be $h_r$. Since the rectangle is on the same base $b$, its length is $b$.

Area of rectangle, $A_r = b \times h_r$.

We are given that the areas are equal, so $A_p = A_r$.

$b \times h_p = b \times h_r$

Since $b$ is the base length and must be greater than 0, we can divide both sides by $b$:

$h_p = h_r$

This means the height of the parallelogram corresponding to the base $b$ is equal to the height of the rectangle corresponding to the base $b$.

When two figures are on the same base and have the same corresponding height, they lie between the same pair of parallel lines. The common base lies on one parallel line, and the vertices/sides opposite the base lie on the other parallel line.

Let's examine the options:

(A) The parallelogram is a rectangle: This is not necessarily true. A parallelogram is a rectangle only if its angles are 90 degrees. Equal base and height do not guarantee this.

(B) The rectangle is a square: This requires the length and width of the rectangle to be equal ($b = h_r$), which is not implied by the given conditions.

(C) They are between the same parallels: Since they are on the same base and have equal heights, they must be located between the same two parallel lines.

(D) The heights are different: This is false, as we concluded that $h_p = h_r$.

Therefore, if a parallelogram and a rectangle are on the same base and have equal areas, they must be between the same parallels.

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The correct option is (C) They are between the same parallels.

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Given:

Area of triangle $= 48 \text{ cm}^2$

Base of triangle $= 12 \text{ cm}$

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To Find:

The corresponding height of the triangle.

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Solution:

The formula for the area of a triangle is:

Area $= \frac{1}{2} \times \text{base} \times \text{height}$

Let the height be $h$ cm.

Substitute the given values into the formula:

$48 = \frac{1}{2} \times 12 \times h$

$48 = 6 \times h$

To find the height $h$, divide the area by 6:

$h = \frac{48}{6}$

$h = 8$

Thus, the corresponding height of the triangle is 8 cm.

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The correct option is (B) 8 cm.

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Solution:

Let the two triangles be $\triangle_1$ and $\triangle_2$. We are given that Area$(\triangle_1) =$ Area$(\triangle_2)$.

Let the base and height of $\triangle_1$ be $b_1$ and $h_1$, respectively. Let the base and height of $\triangle_2$ be $b_2$ and $h_2$, respectively.

Area$(\triangle_1) = \frac{1}{2} b_1 h_1$

Area$(\triangle_2) = \frac{1}{2} b_2 h_2$

Since Area$(\triangle_1) =$ Area$(\triangle_2)$, we have $\frac{1}{2} b_1 h_1 = \frac{1}{2} b_2 h_2$, which simplifies to $b_1 h_1 = b_2 h_2$.

The question asks which condition, when added to Area$(\triangle_1) =$ Area$(\triangle_2)$, is NOT sufficient to conclude that the triangles are on the same base and between the same parallels.

Recall that "on the same base" means having a common side segment.

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Let's examine each option:

(A) They have the same base.

If the triangles have the same base (let's call it $b$, so $b_1 = b_2 = b$), and their areas are equal ($b_1 h_1 = b_2 h_2 \implies b h_1 = b h_2$). Since $b \neq 0$, this implies $h_1 = h_2$. If two triangles share the same base and have equal heights, their vertices opposite the base must lie on a line parallel to the base. Thus, they are on the same base and between the same parallels. This condition is sufficient.

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(B) They have equal bases and equal corresponding heights.

If they have equal bases ($b_1 = b_2$) and equal corresponding heights ($h_1 = h_2$), their areas are equal ($A_1 = \frac{1}{2} b_1 h_1 = \frac{1}{2} b_2 h_2 = A_2$), which is consistent with the given information. However, having equal bases (length) does not mean they share a common side segment ("same base"). For example, two triangles with base length 5 cm and height 4 cm can have their bases on different lines. While triangles with equal bases and equal heights *can* be placed between the same parallels (by placing their equal bases collinearly), this condition itself does not guarantee they are on the *same base* (common segment). Therefore, this condition is NOT sufficient to conclude they are on the same base and between the same parallels.

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(C) They have equal bases and vertices on the same side of the common base.

The phrase "common base" implies they share the same base segment. So they are on the same base. If they are on the same base and have equal areas, their heights must be equal ($h_1 = h_2$). If they are on the same base, have equal heights, and their vertices are on the same side of the base, they are between the same parallels. This condition is sufficient.

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(D) They have equal areas and equal corresponding heights.

We are already given that they have equal areas ($b_1 h_1 = b_2 h_2$). If they also have equal corresponding heights ($h_1 = h_2$, where $h_1, h_2 \neq 0$), then substituting $h_1$ for $h_2$ in the area equation gives $b_1 h_1 = b_2 h_1$, which implies $b_1 = b_2$. So, this condition is equivalent to stating that the triangles have equal bases (length) and equal corresponding heights. As discussed in option (B), having equal bases (length) and equal heights does not guarantee they share a common side segment ("same base"). While they can be placed between the same parallels, they are not necessarily on the *same base*. Therefore, this condition is NOT sufficient to conclude they are on the same base and between the same parallels.

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Both options (B) and (D) describe the same scenario ($b_1=b_2$ and $h_1=h_2$) and are not sufficient to conclude the triangles are on the same base (common segment). However, standard geometric theorems distinguish between "same base" (common segment) and "equal bases" (same length). Conditions involving "equal bases" and "equal heights" or "equal areas" imply that the figures are on *equal bases* and between the same parallels, but not necessarily on the *same base*. Option (D) presents the given condition (equal areas) along with another condition (equal heights) which together lead to having equal bases (length) and equal heights, the properties that allow placement on equal bases between the same parallels, but not necessarily on the same base.

Thus, option (D) is not sufficient to conclude they are on the same base and between the same parallels, specifically because it doesn't guarantee they share a common side segment.

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The correct option is (D) They have equal areas and equal corresponding heights.

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Given:

Parallelogram PQRS.

Base PQ = $8 \text{ cm}$.

Height corresponding to base PQ = $5 \text{ cm}$.

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To Find:

The area of the parallelogram.

---

Solution:

The formula for the area of a parallelogram is:

Area = Base $\times$ Height

Given Base = $8 \text{ cm}$ and Height = $5 \text{ cm}$.

Substituting these values into the formula:

Area = $8 \text{ cm} \times 5 \text{ cm}$

Area = $40 \text{ cm}^2$

Thus, the area of the parallelogram is $40 \text{ cm}^2$.

---

The correct option is (B) $40 \text{ cm}^2$.

---

Given:

In $\triangle ABC$, AD is a median.

E is any point on the median AD.

---

To Find:

The relationship of Area$(\triangle ABE)$ with other areas.

---

Solution:

Since AD is the median of $\triangle ABC$, D is the midpoint of BC.

This implies that the median AD divides $\triangle ABC$ into two triangles of equal area:

Area$(\triangle ABD) =$ Area$(\triangle ACD)$ ... (i)

Now consider $\triangle EBC$. Since D is the midpoint of BC, ED is the median of $\triangle EBC$.

Therefore, the median ED divides $\triangle EBC$ into two triangles of equal area:

Area$(\triangle EBD) =$ Area$(\triangle ECD)$ ... (ii)

Now, let's look at $\triangle ABD$ and $\triangle ACD$.

Area$(\triangle ABD) =$ Area$(\triangle ABE) +$ Area$(\triangle EBD)$

Area$(\triangle ACD) =$ Area$(\triangle ACE) +$ Area$(\triangle ECD)$

From (i), we know Area$(\triangle ABD) =$ Area$(\triangle ACD)$.

So, Area$(\triangle ABE) +$ Area$(\triangle EBD) =$ Area$(\triangle ACE) +$ Area$(\triangle ECD)$.

From (ii), we know Area$(\triangle EBD) =$ Area$(\triangle ECD)$.

Substituting this into the equation above:

Area$(\triangle ABE) +$ Area$(\triangle EBD) =$ Area$(\triangle ACE) +$ Area$(\triangle EBD)$

Subtracting Area$(\triangle EBD)$ from both sides (since Area$(\triangle EBD)$ = Area$(\triangle ECD)$):

Area$(\triangle ABE) =$ Area$(\triangle ACE)$

This shows that the line segment AD (and any point on it) connects vertex A to the midpoint D of the opposite side BC, and the triangles formed by joining E (on AD) to B and C have equal areas ($\triangle ABE$ and $\triangle ACE$).

---

The correct option is (A) Area$(\triangle ACE)$.

---

Given:

Two parallelograms are on the same base.

Area of the first parallelogram $= 50 \text{ cm}^2$.

Area of the second parallelogram $= 70 \text{ cm}^2$.

---

Analysis:

Let the common base of the two parallelograms be $b$.

The area of a parallelogram is given by the formula: Area = base $\times$ height.

For the first parallelogram, Area$_1 = b \times h_1 = 50$, where $h_1$ is the height corresponding to the base $b$.

For the second parallelogram, Area$_2 = b \times h_2 = 70$, where $h_2$ is the height corresponding to the base $b$.

We have $b \times h_1 = 50$ and $b \times h_2 = 70$.

Since $50 \neq 70$, it must be true that $b \times h_1 \neq b \times h_2$.

As the base $b$ is the same for both parallelograms ($b \neq 0$), the heights must be different. That is, $h_1 \neq h_2$.

Specifically, $h_1 = \frac{50}{b}$ and $h_2 = \frac{70}{b}$. Since $50 \neq 70$, $h_1 \neq h_2$.

Two figures on the same base are between the same parallels if and only if their corresponding heights are equal.

Since the heights $h_1$ and $h_2$ are different, the two parallelograms cannot be between the same parallels.

---

Let's evaluate the options based on this analysis:

(A) They are between the same parallels: If they were between the same parallels, their heights would be equal, which would imply their areas are equal (since the base is the same). But their areas are different, so this is not possible.

(B) They are not between the same parallels: If they are on the same base but not between the same parallels, their heights must be different. Different heights with the same base lead to different areas. This matches the given information ($50 \text{ cm}^2$ and $70 \text{ cm}^2$). This is possible.

(C) Their bases are different: The question explicitly states they are "on the same base", so this option is incorrect.

(D) Their heights are equal: As shown above, if the bases are the same and areas are different, the heights must be different. So this option is incorrect.

Therefore, the given situation is possible if and only if the two parallelograms are not between the same parallels.

---

The correct option is (B) They are not between the same parallels.

---

Solution:

Let's examine each statement:

(A) Triangles on the same base and between the same parallels have equal areas.

This is a fundamental theorem in geometry. If two triangles share the same base and their vertices opposite the base lie on a line parallel to the base, then the height corresponding to the base is the same for both triangles. Since Area $= \frac{1}{2} \times \text{base} \times \text{height}$, and both the base and height are equal, their areas are equal.

This statement is TRUE.

---

(B) The area of a triangle is half the product of its base and height.

This is the standard formula for calculating the area of a triangle.

Area $= \frac{1}{2} \times \text{base} \times \text{height}$

This statement is TRUE.

---

(C) A diagonal of a parallelogram divides it into two triangles of equal area.

Consider a parallelogram ABCD and its diagonal AC. This diagonal divides the parallelogram into $\triangle ABC$ and $\triangle ADC$. In a parallelogram, opposite sides are equal (AB=CD, BC=AD), and the diagonal is common (AC=AC). By SSS congruence rule, $\triangle ABC \cong \triangle CDA$. Since congruent figures have equal areas, Area$(\triangle ABC) =$ Area$(\triangle ADC)$.

This statement is TRUE.

---

(D) All figures with equal areas are congruent.

Congruent figures have the same size and shape. Equal areas mean the figures enclose the same amount of space. For example, a rectangle with dimensions $2 \times 6$ cm has an area of $12 \text{ cm}^2$. A rectangle with dimensions $3 \times 4$ cm also has an area of $12 \text{ cm}^2$. However, these two rectangles are not congruent as their shapes are different.

This statement is FALSE.

---

The statements that are true are (A), (B), and (C).

---

The correct options are (A), (B), and (C).

---

Given:

An equilateral triangle, let's call it $\triangle ABC$, with Area($\triangle ABC$) = $A$.

D, E, and F are the mid-points of the sides BC, CA, and AB respectively.

---

To Find:

The area of the triangle formed by joining the mid-points, i.e., Area($\triangle DEF$).

---

Solution:

In $\triangle ABC$, D, E, and F are the mid-points of the sides BC, CA, and AB respectively.

According to the Midpoint Theorem:

1. The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half the length of the third side.

Applying the Midpoint Theorem to $\triangle ABC$:

FE || BC and $FE = \frac{1}{2} BC$

DE || AB and $DE = \frac{1}{2} AB$

DF || AC and $DF = \frac{1}{2} AC$

Since $\triangle ABC$ is an equilateral triangle, AB = BC = AC. Let the side length be $s$. So, AB = BC = AC = $s$.

Then, $FE = \frac{1}{2} s$, $DE = \frac{1}{2} s$, $DF = \frac{1}{2} s$.

This shows that $\triangle DEF$ is also an equilateral triangle with side length $\frac{1}{2}s$.

Now, consider the four triangles formed: $\triangle AFE$, $\triangle BFD$, $\triangle CDE$, and $\triangle DEF$.

We have AF = FB = $\frac{1}{2}s$, BD = DC = $\frac{1}{2}s$, CE = EA = $\frac{1}{2}s$.

Also, as shown above, FE = DE = DF = $\frac{1}{2}s$.

Let's compare the sides of the four triangles:

In $\triangle AFE$: AF = $\frac{1}{2}s$, AE = $\frac{1}{2}s$, FE = $\frac{1}{2}s$. So, $\triangle AFE$ is equilateral.

In $\triangle BFD$: BF = $\frac{1}{2}s$, BD = $\frac{1}{2}s$, FD = $\frac{1}{2}s$. So, $\triangle BFD$ is equilateral.

In $\triangle CDE$: CD = $\frac{1}{2}s$, CE = $\frac{1}{2}s$, DE = $\frac{1}{2}s$. So, $\triangle CDE$ is equilateral.

In $\triangle DEF$: DE = $\frac{1}{2}s$, EF = $\frac{1}{2}s$, FD = $\frac{1}{2}s$. So, $\triangle DEF$ is equilateral.

All four triangles $\triangle AFE$, $\triangle BFD$, $\triangle CDE$, and $\triangle DEF$ are equilateral triangles with the same side length $\frac{1}{2}s$.

Therefore, these four triangles are congruent to each other (by SSS congruence rule).

Since congruent triangles have equal areas, the areas of these four triangles are equal.

Area($\triangle AFE$) = Area($\triangle BFD$) = Area($\triangle CDE$) = Area($\triangle DEF$).

The large triangle $\triangle ABC$ is composed of these four triangles:

Area($\triangle ABC$) = Area($\triangle AFE$) + Area($\triangle BFD$) + Area($\triangle CDE$) + Area($\triangle DEF$).

Given Area($\triangle ABC$) = $A$.

$A =$ Area($\triangle DEF$) + Area($\triangle DEF$) + Area($\triangle DEF$) + Area($\triangle DEF$)

$A = 4 \times$ Area($\triangle DEF$)

Therefore, Area($\triangle DEF$) = $\frac{A}{4}$.

The area of the triangle formed by joining the mid-points of the sides of an equilateral triangle is one-fourth the area of the original triangle.

---

The correct option is (C) $A/4$.

---

Given:

Two triangles have equal areas.

The two triangles have equal bases.

---

Analysis:

Let the two triangles be $\triangle_1$ and $\triangle_2$.

Area$(\triangle_1) =$ Area$(\triangle_2)$

Base$_1 =$ Base$_2$ (Let's call this common base length $b$).

The area of a triangle is given by Area $= \frac{1}{2} \times \text{base} \times \text{height}$.

For $\triangle_1$: Area$_1 = \frac{1}{2} \times \text{Base}_1 \times \text{Height}_1 = \frac{1}{2} \times b \times \text{Height}_1$.

For $\triangle_2$: Area$_2 = \frac{1}{2} \times \text{Base}_2 \times \text{Height}_2 = \frac{1}{2} \times b \times \text{Height}_2$.

Since Area$_1 =$ Area$_2$, we have:

$\frac{1}{2} \times b \times \text{Height}_1 = \frac{1}{2} \times b \times \text{Height}_2$

Assuming $b \neq 0$ (a triangle must have a non-zero base) and dividing both sides by $\frac{1}{2} \times b$, we get:

Height$_1 =$ Height$_2$

So, the two triangles have equal bases (lengths) and equal corresponding heights.

If two triangles have equal bases (length) and equal corresponding heights, they can be placed such that their bases lie on the same line and their vertices opposite to the bases lie on a line parallel to the first line. This means they are between the same parallels.

---

Let's check the options:

(A) Congruent: Having equal bases and equal heights does not guarantee congruence. For example, a triangle with sides 3, 4, 5 (a right triangle) and a triangle with sides 3, some value, some value can have the same base and height but different shapes (and thus not congruent).

(B) Equilateral: This is not necessarily true. The conditions only relate area, base, and height, not all side lengths and angles.

(C) Between the same parallels: As shown by the equality of heights when on equal bases (or on the same base), the vertices opposite the base are equidistant from the line containing the base, which means they lie on a parallel line.

(D) Right-angled: This is not necessarily true. The conditions do not impose any restriction on the angles of the triangles.

Therefore, if two triangles have equal areas and equal bases, they must be between the same parallels.

---

The correct option is (C) Between the same parallels.

---

Given:

Parallelogram ABCD.

Area of parallelogram ABCD $= 80 \text{ cm}^2$.

---

To Find:

The area of $\triangle ABD$.

---

Solution:

In parallelogram ABCD, BD is a diagonal.

A property of parallelograms states that a diagonal divides the parallelogram into two congruent triangles.

The diagonal BD divides parallelogram ABCD into two triangles: $\triangle ABD$ and $\triangle CDB$.

We can show that $\triangle ABD \cong \triangle CDB$ using SSS or SAS congruence criteria:

• AB = CD (Opposite sides of a parallelogram)
• AD = CB (Opposite sides of a parallelogram)
• BD is common to both triangles

By SSS congruence rule, $\triangle ABD \cong \triangle CDB$.

Since congruent triangles have equal areas:

Area$(\triangle ABD) =$ Area$(\triangle CDB)$

The area of the parallelogram is the sum of the areas of the two triangles:

Area(ABCD) = Area$(\triangle ABD) +$ Area$(\triangle CDB)$

Substitute Area$(\triangle CDB)$ with Area$(\triangle ABD)$:

Area(ABCD) = Area$(\triangle ABD) +$ Area$(\triangle ABD)$

Area(ABCD) = $2 \times$ Area$(\triangle ABD)$

We are given Area(ABCD) $= 80 \text{ cm}^2$.

$80 \text{ cm}^2 = 2 \times$ Area$(\triangle ABD)$

To find the area of $\triangle ABD$, divide the area of the parallelogram by 2:

Area$(\triangle ABD) = \frac{80 \text{ cm}^2}{2}$

Area$(\triangle ABD) = 40 \text{ cm}^2$

Thus, the area of $\triangle ABD$ is $40 \text{ cm}^2$.

---

The correct option is (B) $40 \text{ cm}^2$.

---

Given:

A triangle and a parallelogram have the same base.

Area of triangle = Area of parallelogram.

---

To Find:

The relation between their corresponding heights.

---

Solution:

Let the common base of the triangle and the parallelogram be $b$.

Let the height of the triangle corresponding to the base $b$ be $h_t$.

Let the height of the parallelogram corresponding to the base $b$ be $h_p$.

The area of the triangle is given by: Area$_t = \frac{1}{2} \times \text{base} \times \text{height}_t = \frac{1}{2} b h_t$.

The area of the parallelogram is given by: Area$_p = \text{base} \times \text{height}_p = b h_p$.

We are given that Area$_t =$ Area$_p$.

So, $\frac{1}{2} b h_t = b h_p$

Assuming $b \neq 0$ (a figure must have a non-zero base) and dividing both sides by $b$, we get:

$\frac{1}{2} h_t = h_p$

To find the relation between $h_t$ and $h_p$, we can rearrange the equation:

$h_t = 2 h_p$

This means the height of the triangle is double the height of the parallelogram when they have the same area and the same base.

---

Let's check the options:

(A) Height of triangle is double the height of parallelogram: $h_t = 2 h_p$. This matches our result.

(B) Height of parallelogram is double the height of triangle: $h_p = 2 h_t$. This is incorrect.

(C) Heights are equal: $h_t = h_p$. This would imply $\frac{1}{2} b h_t = b h_t$, which means $\frac{1}{2} = 1$ (if $b, h_t \neq 0$), which is false. So heights are not equal.

(D) Height of triangle is half the height of parallelogram: $h_t = \frac{1}{2} h_p$. This implies $2 h_t = h_p$, which is the same as option (B), incorrect.

Therefore, the height of the triangle is double the height of the parallelogram.

---

The correct option is (A) Height of triangle is double the height of parallelogram.

---

Given:

In $\triangle ABC$, D is the midpoint of BC.

E is the midpoint of AD.

Area($\triangle ABC$) $= 1200 \text{ sq metres}$.

---

To Find:

The area of $\triangle BEC$.

---

Solution:

Since D is the midpoint of BC, AD is a median of $\triangle ABC$.

A median of a triangle divides it into two triangles of equal area.

Therefore, Area($\triangle ABD$) $=$ Area($\triangle ACD$) $=$ $\frac{1}{2}$ Area($\triangle ABC$).

Now consider $\triangle ABD$. E is the midpoint of AD.

BE is a line segment joining vertex B to the midpoint E of the side AD.

Triangles with the same base (or equal bases) and the same height have equal areas.

Consider $\triangle ABE$ and $\triangle EBD$. They share the same vertex B.

Their bases AE and ED lie on the same line segment AD.

Since E is the midpoint of AD, AE = ED.

The height of $\triangle ABE$ from B to AD is the perpendicular distance from B to the line AD.

The height of $\triangle EBD$ from B to AD is also the perpendicular distance from B to the line AD.

Since they have equal bases (AE=ED) and the same height (from B to AD), their areas are equal.

Similarly, consider $\triangle ACD$. E is the midpoint of AD.

CE is a line segment joining vertex C to the midpoint E of the side AD.

Consider $\triangle ACE$ and $\triangle ECD$. They share the same vertex C.

Their bases AE and ED lie on the same line segment AD.

Since E is the midpoint of AD, AE = ED.

The height of $\triangle ACE$ from C to AD is the perpendicular distance from C to the line AD.

The height of $\triangle ECD$ from C to AD is also the perpendicular distance from C to the line AD.

Since they have equal bases (AE=ED) and the same height (from C to AD), their areas are equal.

Now, the area of $\triangle ABD$ is the sum of Area($\triangle ABE$) and Area($\triangle EBD$).

Area($\triangle ABD$) $=$ Area($\triangle ABE$) + Area($\triangle EBD$).

From (iii), Area($\triangle ABE$) $=$ Area($\triangle EBD$). So,

Area($\triangle ABD$) $=$ Area($\triangle EBD$) + Area($\triangle EBD$) $=$ $2 \times$ Area($\triangle EBD$).

From (i), Area($\triangle ABD$) $= 600 \text{ sq metres}$.

$600 = 2 \times$ Area($\triangle EBD$)

Area($\triangle EBD$) $=$ $\frac{600}{2} = 300 \text{ sq metres}$.

Similarly, the area of $\triangle ACD$ is the sum of Area($\triangle ACE$) and Area($\triangle ECD$).

Area($\triangle ACD$) $=$ Area($\triangle ACE$) + Area($\triangle ECD$).

From (iv), Area($\triangle ACE$) $=$ Area($\triangle ECD$). So,

Area($\triangle ACD$) $=$ Area($\triangle ECD$) + Area($\triangle ECD$) $=$ $2 \times$ Area($\triangle ECD$).

From (ii), Area($\triangle ACD$) $= 600 \text{ sq metres}$.

$600 = 2 \times$ Area($\triangle ECD$)

Area($\triangle ECD$) $=$ $\frac{600}{2} = 300 \text{ sq metres}$.

The area of $\triangle BEC$ is the sum of Area($\triangle EBD$) and Area($\triangle ECD$).

Area($\triangle BEC$) $=$ Area($\triangle EBD$) + Area($\triangle ECD$).

Substitute the values from (v) and (vi):

Area($\triangle BEC$) $= 300 \text{ sq metres} + 300 \text{ sq metres}$

Area($\triangle BEC$) $= 600 \text{ sq metres}$.

---

Alternatively, from (v) and (vi):

Area($\triangle EBD$) = $\frac{1}{4}$ Area($\triangle ABC$)

Area($\triangle ECD$) = $\frac{1}{4}$ Area($\triangle ABC$)

Area($\triangle BEC$) $=$ Area($\triangle EBD$) + Area($\triangle ECD$) $=$ $\frac{1}{4}$ Area($\triangle ABC$) + $\frac{1}{4}$ Area($\triangle ABC$) $=$ $\frac{2}{4}$ Area($\triangle ABC$) $=$ $\frac{1}{2}$ Area($\triangle ABC$).

Area($\triangle BEC$) $=$ $\frac{1}{2} \times 1200 \text{ sq metres} = 600 \text{ sq metres}$.

---

The correct option is (A) $600 \text{ sq metres}$.

---

Solution:

The term "between the same parallels" means that the base of each figure lies on one parallel line, and the vertex (or the side opposite the base) of each figure lies on the other parallel line.

The height of a figure (such as a triangle or a parallelogram) corresponding to a given base is the perpendicular distance from the opposite vertex (or the line containing the opposite side) to the line containing the base.

If two figures are between the same pair of parallel lines, their bases are on one line and their opposite vertices/sides are on the other line.

The perpendicular distance between any two parallel lines is constant.

Therefore, for any figure placed between the same two parallel lines in this configuration, its height (corresponding to the base on one parallel) will be equal to the perpendicular distance between the parallel lines.

---

Let the two parallel lines be $l_1$ and $l_2$. Let the distance between them be $h$.

If two triangles are between $l_1$ and $l_2$, their bases are on $l_1$ and their third vertices are on $l_2$. Their heights are both equal to $h$.

If two parallelograms are between $l_1$ and $l_2$, their bases are on $l_1$ and their opposite sides are on $l_2$. Their heights are both equal to $h$.

If a triangle and a parallelogram are between $l_1$ and $l_2$, the triangle's base is on $l_1$ and its opposite vertex is on $l_2$ (height $h$), and the parallelogram's base is on $l_1$ and its opposite side is on $l_2$ (height $h$). Their heights are both equal to $h$.

In all cases, if the figures are between the same parallels, their corresponding heights must be the same.

---

Considering the options:

(A) Two triangles between the same parallels have the same height.

(B) Two parallelograms between the same parallels have the same height.

(C) A triangle and a parallelogram between the same parallels have the same height.

All these statements are true because being between the same parallels implies they share the same perpendicular distance between the lines, which is their height.

Therefore, all the scenarios listed in options (A), (B), and (C) imply that the figures must have the same height.

---

The correct option is (D) All of the above.

---

Given:

Area of rhombus = $96 \text{ cm}^2$.

Length of one diagonal ($d_1$) = $16 \text{ cm}$.

---

To Find:

The length of the other diagonal ($d_2$).

---

Solution:

The formula for the area of a rhombus is:

Area $= \frac{1}{2} \times \text{product of diagonals}$

Area $= \frac{1}{2} \times d_1 \times d_2$

Substitute the given values into the formula:

$96 \text{ cm}^2 = \frac{1}{2} \times 16 \text{ cm} \times d_2$

Simplify the right side of the equation:

$96 \text{ cm}^2 = 8 \text{ cm} \times d_2$

To find the length of the other diagonal ($d_2$), divide the area by 8 cm:

$d_2 = \frac{96 \text{ cm}^2}{8 \text{ cm}}$

$d_2 = 12 \text{ cm}$

Thus, the length of the other diagonal is $12 \text{ cm}$.

---

The correct option is (B) 12 cm.

---

Given:

A triangle and a parallelogram are on the same base.

The triangle and the parallelogram are between the same parallels.

Base = $10 \text{ cm}$

Height = $6 \text{ cm}$

---

To Find:

The ratio of the area of the triangle to the area of the parallelogram.

---

Solution:

Let the common base be $b$. According to the problem, $b = 10 \text{ cm}$.

Since the triangle and the parallelogram are between the same parallels, their corresponding heights are equal. Let the common height be $h$. According to the problem, $h = 6 \text{ cm}$.

The area of the triangle is given by:

Area$_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height}$

Area$_{\text{triangle}} = \frac{1}{2} \times b \times h$

The area of the parallelogram is given by:

Area$_{\text{parallelogram}} = \text{base} \times \text{height}$

Area$_{\text{parallelogram}} = b \times h$

The ratio of the area of the triangle to the area of the parallelogram is:

Ratio = $\frac{\text{Area}_{\text{triangle}}}{\text{Area}_{\text{parallelogram}}} = \frac{\frac{1}{2} \times b \times h}{b \times h}$

Since $b \neq 0$ and $h \neq 0$, we can cancel $b \times h$ from the numerator and the denominator.

Ratio = $\frac{1}{2}$

The ratio is $1:2$.

Note that the specific values of the base (10 cm) and height (6 cm) were not needed to find the ratio, only the fact that they are common for both figures when placed on the same base and between the same parallels.

---

The correct option is (B) $1:2$.

---

Solution:

Let's analyze each statement:

(A) Area is always positive.

For a non-degenerate triangle (a triangle that actually exists and is not just a straight line or a point), the area is a measure of the two-dimensional space it occupies. This measure is a positive value. The area is zero only for a degenerate triangle (where the vertices are collinear). Assuming we are dealing with actual triangles, the area is positive.

This statement is TRUE (for non-degenerate triangles).

(B) Area depends on the chosen base and its corresponding height.

The formula for the area of a triangle is Area $= \frac{1}{2} \times \text{base} \times \text{height}$. This formula directly shows that the area is calculated using a base and its corresponding height. While a triangle has three bases (its sides) and three corresponding heights, the product of any base and its corresponding height is constant, leading to the same area value regardless of which base-height pair is chosen. So, the area is defined by any base and its corresponding height.

This statement is TRUE.

(C) Congruent triangles have equal areas.

Congruent figures are identical in shape and size. If two triangles are congruent, all their corresponding sides and angles are equal. Consequently, they occupy the same amount of area.

This statement is TRUE.

(D) Triangles with equal areas are always congruent.

This statement is the converse of statement (C). Having equal areas does not necessarily mean that two figures are congruent. For example, consider a triangle with base $b_1=4$ and height $h_1=6$. Its area is $\frac{1}{2} \times 4 \times 6 = 12$. Now consider a triangle with base $b_2=3$ and height $h_2=8$. Its area is $\frac{1}{2} \times 3 \times 8 = 12$. Both triangles have an area of 12, but they can have different side lengths and angles, meaning they are not necessarily congruent.

This statement is FALSE.

---

The statement that is FALSE is (D).

---

The correct option is (D) Triangles with equal areas are always congruent.

---

Given:

Area of $\triangle ABC = 50 \text{ cm}^2$.

D is the midpoint of AB.

E is the midpoint of AC.

---

To Find:

The area of $\triangle ADE$.

---

Solution:

In $\triangle ABC$, D and E are the midpoints of sides AB and AC respectively.

According to the Midpoint Theorem, the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.

Therefore, DE is parallel to BC, and $DE = \frac{1}{2} BC$.

Now consider $\triangle ADE$ and $\triangle ABC$.

We have:

$\angle DAE = \angle BAC$ (Common angle)

$\frac{AD}{AB} = \frac{\text{length of AD}}{\text{length of AB}}$

So, $\frac{AD}{AB} = \frac{1}{2}$.

Similarly, $\frac{AE}{AC} = \frac{\text{length of AE}}{\text{length of AC}}$

So, $\frac{AE}{AC} = \frac{1}{2}$.

Since $\angle DAE = \angle BAC$ and $\frac{AD}{AB} = \frac{AE}{AC} = \frac{1}{2}$, by the SAS similarity criterion, $\triangle ADE \sim \triangle ABC$.

For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

The ratio of corresponding sides is $\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC} = \frac{1}{2}$.

So, $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = \left(\frac{DE}{BC}\right)^2 = \left(\frac{\frac{1}{2} BC}{BC}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.

Thus, Area$(\triangle ADE) = \frac{1}{4} \times$ Area$(\triangle ABC)$.

Substitute the given area of $\triangle ABC$:

Area$(\triangle ADE) = \frac{1}{4} \times 50 \text{ cm}^2$

Area$(\triangle ADE) = \frac{\cancel{50}^{25}}{\cancel{4}_{2}} \text{ cm}^2$

Area$(\triangle ADE) = \frac{25}{2} \text{ cm}^2$

Area$(\triangle ADE) = 12.5 \text{ cm}^2$.

Therefore, the area of $\triangle ADE$ is $12.5 \text{ cm}^2$.

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The correct option is (B) $12.5 \text{ cm}^2$.

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Solution:

We are given a parallelogram ABCD and specifically considering the triangle BCD within this figure.

Two figures are said to be on the same base if they have a common side.

Let's list the sides of the parallelogram ABCD: AB, BC, CD, DA.

Let's list the sides of the triangle BCD: BC, CD, DB.

We need to find a side that is common to both the parallelogram ABCD and the triangle BCD.

• The side AB is a side of the parallelogram, but not a side of $\triangle BCD$.
• The side BC is a side of the parallelogram and also a side of $\triangle BCD$. Thus, BC is a common side.
• The side CD is a side of the parallelogram and also a side of $\triangle BCD$. Thus, CD is a common side.
• The side DA is a side of the parallelogram, but not a side of $\triangle BCD$.

The question asks which option correctly identifies figures on the same base, specifically considering the parallelogram ABCD and the triangle BCD.

Let's look at the options provided:

(A) Parallelogram ABCD and $\triangle ABC$ on base AB. This option considers $\triangle ABC$ instead of $\triangle BCD$.

(B) Parallelogram ABCD and $\triangle BCD$ on base BC. This option correctly identifies both figures and states that BC is the common base. As found above, BC is indeed a common side.

(C) Parallelogram ABCD and $\triangle ACD$ on base CD. This option considers $\triangle ACD$ instead of $\triangle BCD$.

(D) Parallelogram ABCD and $\triangle ABD$ on base AD. This option considers $\triangle ABD$ instead of $\triangle BCD$.

Based on the problem statement which focuses on the parallelogram ABCD and triangle BCD, the only option that correctly identifies a common base for *these specific figures* is option (B).

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The correct option is (B) Parallelogram ABCD and $\triangle BCD$ on base BC.

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Solution:

A physical quantity is classified as either a scalar quantity or a vector quantity.

A scalar quantity is a quantity that has only magnitude and no direction.

A vector quantity is a quantity that has both magnitude and direction.

Area, as defined in the question ("a measure of the region enclosed by a plane figure"), refers to the size or extent of the two-dimensional space covered by the figure. This measure is represented by a numerical value and a unit (e.g., $10 \text{ cm}^2$, $5 \text{ m}^2$).

While in more advanced contexts (like physics or vector calculus), an "area vector" can be defined (which has a magnitude equal to the area and a direction perpendicular to the surface), the basic measure of the area of a plane figure is purely its size.

The size (magnitude) of the area does not have an inherent direction in the plane itself.

Therefore, area, in this context, is considered a scalar quantity.

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The correct option is (B) Scalar.

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Given:

Two parallelograms are on the same base.

The two parallelograms have equal areas.

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To Find:

The relation between their corresponding heights.

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Solution:

Let the common base of the two parallelograms be $b$.

Let the area of the first parallelogram be $A_1$ and its corresponding height be $h_1$.

Let the area of the second parallelogram be $A_2$ and its corresponding height be $h_2$.

The formula for the area of a parallelogram is: Area = base $\times$ height.

For the first parallelogram: $A_1 = b \times h_1$.

For the second parallelogram: $A_2 = b \times h_2$.

We are given that the areas are equal: $A_1 = A_2$.

So, $b \times h_1 = b \times h_2$.

Since $b$ is the base length of a parallelogram, $b \neq 0$. We can divide both sides of the equation by $b$:

$\frac{b \times h_1}{b} = \frac{b \times h_2}{b}$

$h_1 = h_2$

Thus, if two parallelograms are on the same base and have equal areas, their corresponding heights must be equal.

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Let's check the options:

(A) Heights must be equal: This matches our conclusion.

(B) Heights must be different: This contradicts our conclusion.

(C) One height is double the other: This is incorrect.

(D) One height is half the other: This is incorrect.

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The correct option is (A) Heights must be equal.

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Given:

The vertices of a triangle are $(0,0)$, $(a,0)$, and $(0,b)$.

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To Find:

The area of the triangle.

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Solution:

Let the vertices of the triangle be O$(0,0)$, P$(a,0)$, and Q$(0,b)$.

These points lie on the coordinate axes.

• The point O$(0,0)$ is the origin.
• The point P$(a,0)$ lies on the x-axis.
• The point Q$(0,b)$ lies on the y-axis.

The side OP lies along the x-axis, and its length is the distance between $(0,0)$ and $(a,0)$, which is $|a|$. Let's assume $a > 0$ for simplicity, so the length is $a$. This side can be considered the base of the triangle.

The side OQ lies along the y-axis, and its length is the distance between $(0,0)$ and $(0,b)$, which is $|b|$. Let's assume $b > 0$ for simplicity, so the length is $b$. This side is perpendicular to the base OP (since the axes are perpendicular) and passes through the vertex Q opposite to the base OP. Thus, this side can be considered the height of the triangle corresponding to the base OP.

The triangle OPQ is a right-angled triangle with the right angle at the origin O.

The base of the triangle is $b = a$.

The corresponding height of the triangle is $h = b$.

The formula for the area of a triangle is:

Area $= \frac{1}{2} \times \text{base} \times \text{height}$

Substitute the base $a$ and height $b$ into the formula:

Area $= \frac{1}{2} \times a \times b$

Area $= \frac{1}{2}ab$

If $a$ or $b$ are negative, the lengths of the base and height are $|a|$ and $|b|$ respectively. The area would be $\frac{1}{2} |a| |b| = \frac{1}{2} |ab|$. The option $\frac{1}{2}ab$ implies $a$ and $b$ might represent dimensions or the absolute value of the product is taken.

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Alternatively, using the determinant formula for the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:

Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Let $(x_1, y_1) = (0,0)$, $(x_2, y_2) = (a,0)$, $(x_3, y_3) = (0,b)$.

Area $= \frac{1}{2} |0(0 - b) + a(b - 0) + 0(0 - 0)|$

Area $= \frac{1}{2} |0 + ab + 0|$

Area $= \frac{1}{2} |ab|$

Assuming $a, b > 0$ or taking the magnitude, Area $= \frac{1}{2}ab$.

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Comparing with the given options, the area is $\frac{1}{2}ab$.

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The correct option is (C) $\frac{1}{2}ab$.

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Given:

The wall is in the shape of a parallelogram.

Base of the parallelogram = $5 \text{ m}$.

Height of the parallelogram = $3 \text{ m}$.

Cost of painting per square metre = $\textsf{₹} 20$.

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To Find:

The total cost of painting the wall.

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Solution:

First, we need to find the area of the wall (parallelogram).

The formula for the area of a parallelogram is:

Area $= \text{Base} \times \text{Height}$

Substitute the given values:

Area $= 5 \text{ m} \times 3 \text{ m}$

Area $= 15 \text{ m}^2$

Now, we need to calculate the total cost of painting. The cost is $\textsf{₹} 20$ per square metre.

Total Cost $= \text{Area} \times \text{Cost per square metre}$

Total Cost $= 15 \text{ m}^2 \times \textsf{₹} 20/\text{m}^2$

Total Cost $= 15 \times 20$ $\textsf{₹}$

Total Cost $= 300$ $\textsf{₹}$

Thus, the total cost of painting the wall is $\textsf{₹} 300$.

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The correct option is (C) $\textsf{₹} 300$.

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Solution:

Let's examine each statement:

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(A) A trapezium is always a parallelogram.

A trapezium (or trapezoid) is a quadrilateral with at least one pair of parallel sides. A parallelogram is a quadrilateral with both pairs of opposite sides parallel. A trapezium is a parallelogram only if it has two pairs of parallel sides. However, a trapezium can have only one pair of parallel sides. Thus, this statement is false.

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(B) A rectangle is always a rhombus.

A rectangle is a parallelogram with four right angles. A rhombus is a parallelogram with four equal sides. A rectangle is a rhombus only if all its sides are equal (i.e., it is a square). A general rectangle does not have all sides equal. Thus, this statement is false.

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(C) A triangle has exactly one median.

A median of a triangle is a line segment from a vertex to the midpoint of the opposite side. Since a triangle has three vertices, it has three medians (one from each vertex). Thus, this statement is false.

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(D) Area of a triangle is positive.

The area of a non-degenerate triangle (a triangle whose vertices are not collinear) is a positive measure of the two-dimensional space it occupies. The area is zero only for a degenerate triangle. In standard geometry problems, the term "triangle" refers to a non-degenerate one unless otherwise specified. Thus, the area of a triangle is considered positive.

This statement is true for non-degenerate triangles, which are typically assumed.

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Comparing the statements, the only statement that is true is (D).

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The correct option is (D) Area of a triangle is positive.

The area of a plane figure or a shape is the measure of the amount of surface enclosed by its boundary. It is a two-dimensional quantity that represents the space occupied by the shape on a flat surface.

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The standard units used for measuring area are derived from the standard units of length by squaring them. Since area is a two-dimensional measure (length $\times$ width), the units are in square units.

Some common standard units for measuring area include:

• Square metres ($m^2$)
• Square centimetres ($cm^2$)
• Square kilometres ($km^2$)
• Square inches ($in^2$)
• Square feet ($ft^2$)
• Square yards ($yd^2$)
• Acres (often used for land area)
• Hectares (1 hectare = $10,000$ $m^2$)

The choice of unit depends on the size of the area being measured; for example, $cm^2$ is used for small areas, $m^2$ for medium areas, and $km^2$ or hectares for large areas like land or countries.

The formula for the area of a parallelogram with base $b$ and corresponding height $h$ is:

Area of parallelogram $= \text{base} \times \text{height}$

$A = b \times h$

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Given:

Base of the parallelogram, $b = 12$ cm

Corresponding height, $h = 7$ cm

To Find:

Area of the parallelogram.

Solution:

Using the formula for the area of a parallelogram:

Area $= b \times h$

Area $= 12$ cm $\times 7$ cm

Area $= 84$ $cm^2$

Therefore, the area of the parallelogram is $84$ $cm^2$.

The formula for the area of a triangle with base $b$ and corresponding height $h$ is:

Area of triangle $= \frac{1}{2} \times \text{base} \times \text{height}$

$A = \frac{1}{2} \times b \times h$

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Given:

Base of the triangle, $b = 9$ meters

Corresponding height, $h = 6$ meters

To Find:

Area of the triangle.

Solution:

Using the formula for the area of a triangle:

Area $= \frac{1}{2} \times b \times h$

Area $= \frac{1}{2} \times 9$ m $\times 6$ m

Area $= \frac{1}{2} \times 54$ $m^2$

Area $= 27$ $m^2$

Therefore, the area of the triangle is $27$ $m^2$.

In $\triangle ABC$, since D is the midpoint of BC, the line segment AD is a median of the triangle.

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We can say that the areas of $\triangle ABD$ and $\triangle ACD$ are equal.

Area($\triangle ABD$) = Area($\triangle ACD$)

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The property that supports this answer is:

A median of a triangle divides it into two triangles of equal areas.

This is because $\triangle ABD$ and $\triangle ACD$ have equal bases (BD = DC, since D is the midpoint) and they share the same height (the perpendicular distance from vertex A to the base BC).

There is a property relating the areas of a parallelogram and a triangle when they are on the same base and between the same parallels.

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The property states that the area of a triangle is half the area of a parallelogram if both are on the same base and between the same parallel lines.

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Given:

Area of the parallelogram = $72$ sq cm

The parallelogram and the triangle are on the same base and between the same parallels.

To Find:

Area of the triangle.

Solution:

Let Area($\triangle$) be the area of the triangle and Area($\square$) be the area of the parallelogram.

According to the property mentioned above:

Area($\triangle$) = $\frac{1}{2} \times$ Area($\square$)

We are given that Area($\square$) = $72$ $cm^2$.

So, Area($\triangle$) = $\frac{1}{2} \times 72$ $cm^2$

Area($\triangle$) = $36$ $cm^2$

Therefore, the area of the triangle is $36$ sq cm.

There is a geometric property related to parallelograms on the same base and between the same parallels.

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The property states that two parallelograms on the same base and between the same parallel lines have equal areas.

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Given:

Parallelogram ABCD and Parallelogram PQRS are on the same base AB.

They are between the same parallel lines AB and DQ.

Area of parallelogram ABCD = $45$ sq meters.

To Find:

Area of parallelogram PQRS.

Solution:

Since parallelograms ABCD and PQRS are on the same base AB and between the same parallels AB and DQ, according to the property mentioned above, their areas must be equal.

Area(parallelogram PQRS) = Area(parallelogram ABCD)

Area(parallelogram PQRS) = $45$ $m^2$

Therefore, the area of parallelogram PQRS is $45$ sq meters.

We are given two triangles $\triangle ABC$ and $\triangle DBC$ on the same base BC, and their areas are equal. Also, vertices A and D lie on the same side of the base BC.

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Let $h_1$ be the height of $\triangle ABC$ from A to BC, and $h_2$ be the height of $\triangle DBC$ from D to BC.

Area($\triangle ABC$) $= \frac{1}{2} \times BC \times h_1$

Area($\triangle DBC$) $= \frac{1}{2} \times BC \times h_2$

We are given that Area($\triangle ABC$) $=$ Area($\triangle DBC$).

Since BC is the common base and $BC \neq 0$, we can cancel $\frac{1}{2} \times BC$ from both sides of the equation (i).

This gives us:

$h_1 = h_2$

This means that the perpendicular distance from A to BC is equal to the perpendicular distance from D to BC.

Since A and D are on the same side of BC, the line segment AD must be parallel to BC.

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Therefore, the line segment AD is parallel to the base BC.

This is based on the property that two triangles on the same base (or equal bases) and having equal areas lie between the same parallels.

The formula for the area of a triangle is:

Area $= \frac{1}{2} \times \text{base} \times \text{height}$

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Given:

Area of the triangular field = $150$ $m^2$

Height of the triangular field = $10$ meters

To Find:

The length of the base of the triangular field.

Solution:

Let the base of the triangle be $b$ meters.

Using the formula for the area of a triangle:

Area $= \frac{1}{2} \times \text{base} \times \text{height}$

$150 = \frac{1}{2} \times b \times 10$

$150 = 5 \times b$

To find $b$, we can divide both sides of the equation by 5:

$b = \frac{150}{5}$

$b = 30$

The length of the base is 30 meters.

Therefore, the length of the base of the triangular field is $30$ meters.

The formula for the area of a parallelogram is:

Area $= \text{base} \times \text{height}$

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Given:

Area of the parallelogram = $260$ $cm^2$

Base of the parallelogram = $20$ cm

To Find:

The corresponding height of the parallelogram.

Solution:

Let the height of the parallelogram be $h$ cm.

Using the formula for the area of a parallelogram:

Area $= \text{base} \times \text{height}$

$260 = 20 \times h$

To find $h$, we can divide both sides of the equation by 20:

$h = \frac{260}{20}$

$h = \frac{26}{2}$

$h = 13$

The corresponding height is 13 cm.

Therefore, the corresponding height of the parallelogram is $13$ cm.

We are asked to answer the rephrased Question 10.

The rephrased question involves two triangles on the same base and with their vertices opposite to the base lying on a line parallel to the base. There is a specific theorem related to this scenario.

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The theorem states that two triangles on the same base (or equal bases) and between the same parallel lines have equal areas.

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Given:

Triangles $\triangle PQR$ and $\triangle SQR$ share the same base QR.

The vertices P and S lie on a line parallel to QR.

Area$(\triangle PQR) = 40$ sq cm.

To Find:

Area$(\triangle SQR)$.

Solution:

Since $\triangle PQR$ and $\triangle SQR$ are on the same base QR and between the same parallel lines (QR and the line containing P and S), their areas must be equal according to the theorem stated above.

Area$(\triangle SQR)$ = Area$(\triangle PQR)$

Given that Area$(\triangle PQR) = 40$ $cm^2$.

Area$(\triangle SQR) = 40$ $cm^2$.

Therefore, the area of $\triangle SQR$ is $40$ sq cm.

The figure is formed by combining a rectangle and a triangle. To find the total area, we need to calculate the area of each shape separately and then add them together.

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The formula for the area of a rectangle is:

Area of rectangle = Length $\times$ Width

The formula for the area of a triangle is:

Area of triangle = $\frac{1}{2} \times$ Base $\times$ Height

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Given:

For the rectangle:

Length $= 10$ cm

Width $= 5$ cm

For the triangle:

Base $= 10$ cm (This base is joined to one side of the rectangle)

Height $= 4$ cm

To Find:

The total area of the figure.

Solution:

First, calculate the area of the rectangle:

Area of rectangle $= \text{Length} \times \text{Width}$

Area of rectangle $= 10$ cm $\times 5$ cm

Area of rectangle $= 50$ $cm^2$

Next, calculate the area of the triangle:

Area of triangle $= \frac{1}{2} \times \text{Base} \times \text{Height}$

Area of triangle $= \frac{1}{2} \times 10$ cm $\times 4$ cm

Area of triangle $= \frac{1}{2} \times 40$ $cm^2$

Area of triangle $= 20$ $cm^2$

Now, find the total area of the figure by adding the area of the rectangle and the area of the triangle:

Total Area = Area of rectangle + Area of triangle

Total Area $= 50$ $cm^2$ + $20$ $cm^2$

Total Area $= 70$ $cm^2$

Therefore, the total area of the figure is $70$ sq cm.

We are asked to answer the rephrased Question 12.

The rephrased question provides the base and the corresponding height of the parallelogram, which is sufficient to calculate its area.

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The formula for the area of a parallelogram is:

Area $= \text{base} \times \text{corresponding height}$

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Given:

Base of the parallelogram, $b = 18$ cm

Corresponding height, $h = 7$ cm

To Find:

The area of the parallelogram.

Solution:

Using the formula for the area of a parallelogram:

Area $= b \times h$

Area $= 18$ cm $\times 7$ cm

Area $= 126$ $cm^2$

We can perform the multiplication:

$\begin{array}{cc}& & 1 & 8 \\ \times & & & 7 \\ \hline & 1 & 2 & 6 \\ \hline \end{array}$

Therefore, the area of the parallelogram is $126$ sq cm.

The area of a rhombus can be calculated using the lengths of its diagonals.

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The formula for the area of a rhombus is:

Area $= \frac{1}{2} \times d_1 \times d_2$

where $d_1$ and $d_2$ are the lengths of the diagonals.

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Given:

Length of diagonal 1, $d_1 = 10$ cm

Length of diagonal 2, $d_2 = 16$ cm

To Find:

The area of the rhombus.

Solution:

Using the formula for the area of a rhombus:

Area $= \frac{1}{2} \times d_1 \times d_2$

Area $= \frac{1}{2} \times 10$ cm $\times 16$ cm

Area $= \frac{1}{2} \times 160$ $cm^2$

Area $= 80$ $cm^2$

Alternatively, we can cancel the terms:

Area $= \frac{1}{\cancel{2}_1} \times \cancel{10}^5 \times 16$ $cm^2$

Area $= 5 \times 16$ $cm^2$

Area $= 80$ $cm^2$

Therefore, the area of the rhombus is $80$ sq cm.

The area of a trapezium is given by the formula:

Area $= \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$

where 'height' is the perpendicular distance between the parallel sides.

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Given:

Length of one parallel side, $a = 8$ cm

Length of the other parallel side, $b = 12$ cm

Area of the trapezium = $100$ $cm^2$

To Find:

The perpendicular distance between the parallel sides (height), $h$.

Solution:

Let the height of the trapezium be $h$ cm.

Using the formula for the area of a trapezium:

Area $= \frac{1}{2} \times (a + b) \times h$

Substitute the given values into the formula:

$100 = \frac{1}{2} \times (8 + 12) \times h$

$100 = \frac{1}{2} \times (20) \times h$

$100 = 10 \times h$

To find $h$, divide both sides of the equation by 10:

$h = \frac{100}{10}$

$h = 10$

The perpendicular distance between the parallel sides is 10 cm.

Therefore, the height of the trapezium is $10$ cm.

Yes, two triangles can have the same area but different perimeters.

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The area of a triangle depends on its base and corresponding height, while the perimeter depends on the sum of the lengths of all three sides. It is possible to have different combinations of side lengths that result in the same area but different sums of side lengths.

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Example:

Consider two right-angled triangles:

Triangle 1:

Base $= 6$ cm

Height $= 4$ cm

Area of Triangle 1 $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \text{ cm} \times 4 \text{ cm} = \frac{1}{2} \times 24 \text{ cm}^2 = 12$ $cm^2$.

Using the Pythagorean theorem, the hypotenuse is $\sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}$ cm.

Perimeter of Triangle 1 $= 6 + 4 + 2\sqrt{13} = 10 + 2\sqrt{13}$ cm.

($2\sqrt{13} \approx 2 \times 3.605 = 7.21$. So Perimeter of Triangle 1 $\approx 10 + 7.21 = 17.21$ cm)

Triangle 2:

Base $= 8$ cm

Height $= 3$ cm

Area of Triangle 2 $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \text{ cm} \times 3 \text{ cm} = \frac{1}{2} \times 24 \text{ cm}^2 = 12$ $cm^2$.

Using the Pythagorean theorem, the hypotenuse is $\sqrt{8^2 + 3^2} = \sqrt{64 + 9} = \sqrt{73}$ cm.

Perimeter of Triangle 2 $= 8 + 3 + \sqrt{73} = 11 + \sqrt{73}$ cm.

($\sqrt{73} \approx 8.544$. So Perimeter of Triangle 2 $\approx 11 + 8.544 = 19.544$ cm)

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Both triangles have an area of $12$ $cm^2$, but their perimeters ($10 + 2\sqrt{13}$ cm and $11 + \sqrt{73}$ cm) are different.

Yes, two parallelograms can have the same area but different perimeters.

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The area of a parallelogram depends on its base and corresponding height, while the perimeter depends on the lengths of its adjacent sides. It is possible to have different combinations of base, height, and side lengths that result in the same area but different perimeters.

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Example:

Consider two rectangles, which are special cases of parallelograms:

Parallelogram 1 (Rectangle 1):

Length ($l_1$) $= 12$ cm

Width ($w_1$) $= 5$ cm

Base ($b_1$) $= l_1 = 12$ cm

Height ($h_1$) $= w_1 = 5$ cm

Area of Parallelogram 1 $= b_1 \times h_1 = 12 \text{ cm} \times 5 \text{ cm} = 60$ $cm^2$.

The adjacent sides are 12 cm and 5 cm.

Perimeter of Parallelogram 1 $= 2 \times (l_1 + w_1) = 2 \times (12 \text{ cm} + 5 \text{ cm}) = 2 \times 17 \text{ cm} = 34$ cm.

Parallelogram 2 (Rectangle 2):

Length ($l_2$) $= 10$ cm

Width ($w_2$) $= 6$ cm

Base ($b_2$) $= l_2 = 10$ cm

Height ($h_2$) $= w_2 = 6$ cm

Area of Parallelogram 2 $= b_2 \times h_2 = 10 \text{ cm} \times 6 \text{ cm} = 60$ $cm^2$.

The adjacent sides are 10 cm and 6 cm.

Perimeter of Parallelogram 2 $= 2 \times (l_2 + w_2) = 2 \times (10 \text{ cm} + 6 \text{ cm}) = 2 \times 16 \text{ cm} = 32$ cm.

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Both parallelograms (rectangles in this case) have an area of $60$ $cm^2$, but their perimeters ($34$ cm and $32$ cm) are different.

In a parallelogram ABCD, the diagonal AC divides the parallelogram into two triangles, $\triangle ABC$ and $\triangle ADC$.

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We know that in a parallelogram, opposite sides are equal. So, AB = CD and BC = AD.

Consider the two triangles $\triangle ABC$ and $\triangle ADC$:

• AB = CD (Opposite sides of a parallelogram)
• BC = AD (Opposite sides of a parallelogram)
• AC = AC (Common side)

By the SSS (Side-Side-Side) congruence criterion, $\triangle ABC \cong \triangle ADC$.

Congruent triangles have equal areas.

Therefore, Area($\triangle ADC$) = Area($\triangle ABC$).

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Given:

Area$(\triangle ABC) = 30$ sq units.

To Find:

Area$(\triangle ADC)$.

Justification:

In parallelogram ABCD, the diagonal AC divides the parallelogram into two congruent triangles, $\triangle ABC$ and $\triangle ADC$. This can be shown by SSS congruence (AB = CD, BC = AD, AC is common).

Since the triangles are congruent, their areas are equal.

Area($\triangle ADC$) = Area($\triangle ABC$)

Area($\triangle ADC$) = $30$ sq units.

Therefore, the area of $\triangle ADC$ is $30$ sq units.

To find the cost of ploughing the field, we first need to calculate the area of the triangular field. Then, we multiply the area by the given rate per square meter.

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The formula for the area of a triangle is:

Area $= \frac{1}{2} \times \text{base} \times \text{height}$

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Given:

Base of the triangular field, $b = 25$ meters

Height of the triangular field, $h = 12$ meters

Rate of ploughing $= \textsf{₹}50$ per square meter

To Find:

The total cost of ploughing the field.

Solution:

First, calculate the area of the triangular field:

Area $= \frac{1}{2} \times b \times h$

Area $= \frac{1}{2} \times 25$ m $\times 12$ m

Area $= \frac{1}{2} \times 300$ $m^2$

Area $= 150$ $m^2$

Now, calculate the total cost of ploughing:

Total Cost = Area $\times$ Rate per square meter

Total Cost $= 150$ $m^2$ $\times \textsf{₹}50$ / $m^2$

Total Cost $= \textsf{₹}(150 \times 50)$

Total Cost $= \textsf{₹}7500$

We can perform the multiplication:

$\begin{array}{cc}& & 1 & 5 & 0 \\ \times & & & 5 & 0 \\ \hline && 0 & 0 & 0 \\ & 7 & 5 & 0 & \times \\ \hline 7 & 5 & 0 & 0 \\ \hline \end{array}$

Therefore, the cost of ploughing the field is $\textsf{₹}7500$.

In the given figure, PQRS is a parallelogram, and T is a point on the side QR. We are considering the triangle $\triangle PST$.

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Observe that the triangle $\triangle PST$ and the parallelogram PQRS share the same base PS.

Also, the vertex T of the triangle $\triangle PST$ lies on the line segment QR. Since PQRS is a parallelogram, the side QR is parallel to the side PS.

Thus, $\triangle PST$ and parallelogram PQRS are on the same base PS and are between the same parallel lines PS and QR.

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There is a theorem that states: If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.

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Applying this theorem to $\triangle PST$ and parallelogram PQRS:

Area$(\triangle PST) = \frac{1}{2} \times$ Area$(PQRS)$

This relationship holds true regardless of the position of point T on the side QR.

Therefore, the area of triangle PST is half the area of parallelogram PQRS.

Two figures are said to be "on the same base" if they share a common side (or base) of equal length.

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Two figures are said to be "between the same parallels" if the vertices opposite to the common base lie on a line parallel to the base.

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When we say that two figures are "on the same base and between the same parallels", it means that:

1. They share a common side (the base).

2. The line segment connecting the vertices of the two figures opposite to the common base is parallel to the common base.

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Diagrammatic Illustration:

Consider a parallelogram ABCD and a triangle ABE.

In the figure (conceptual diagram):

The parallelogram is ABCD.

The triangle is ABE.

They share the common base AB.

The vertex opposite to the base AB in the parallelogram is represented by the line segment DC (since DC is parallel to AB and its length is equal to AB in a parallelogram). The vertices D and C lie on the line parallel to AB.

The vertex opposite to the base AB in the triangle is E.

If point E lies anywhere on the line containing DC (or the line parallel to AB passing through D and C), then $\triangle ABE$ and parallelogram ABCD are on the same base AB and between the same parallels AB and the line containing DC and E.

In this specific illustration, the condition "between the same parallels" means that vertex E of $\triangle ABE$ lies on the line DC, which is parallel to AB.

So, Parallelogram ABCD and $\triangle ABE$ are on the same base AB and between the same parallels AB and DC.

The formula for the area of a parallelogram is:

Area $= \text{base} \times \text{height}$

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Given:

Area of the parallelogram = $288$ $cm^2$

The height of the parallelogram is half its base ($h = \frac{1}{2}b$).

To Find:

The length of the base and the height of the parallelogram.

Solution:

Let the base of the parallelogram be $b$ cm and the height be $h$ cm.

According to the problem, $h = \frac{1}{2}b$.

The area of the parallelogram is given by:

Area $= b \times h$

Substitute the given area and the relationship between $b$ and $h$ into the formula:

$288 = b \times \left(\frac{1}{2}b\right)$

$288 = \frac{1}{2}b^2$

Multiply both sides by 2 to isolate $b^2$:

$2 \times 288 = b^2$}

$576 = b^2$

To find $b$, take the square root of both sides:

$b = \sqrt{576}$

Since length must be positive, we take the positive square root.

$b = 24$

The base of the parallelogram is 24 cm.

Now, find the height using the relationship $h = \frac{1}{2}b$:

$h = \frac{1}{2} \times 24$

$h = 12$

The height of the parallelogram is 12 cm.

Therefore, the base of the parallelogram is $24$ cm and the height is $12$ cm.

We are given the vertices of a triangle at the points (0,0), (5,0), and (0,8). We can find the area of this triangle using either a geometric approach based on plotting the points or using the coordinate formula for the area of a triangle.

---

Given:

Vertices of the triangle: A(0,0), B(5,0), C(0,8).

To Find:

The area of the triangle.

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Solution (Geometric Method):

Let the vertices be A(0,0), B(5,0), and C(0,8).

Point A (0,0) is the origin.

Point B (5,0) lies on the x-axis.

Point C (0,8) lies on the y-axis.

The side AB lies along the x-axis, and its length is the distance between (0,0) and (5,0).

Base AB $= \sqrt{(5-0)^2 + (0-0)^2} = \sqrt{5^2 + 0^2} = \sqrt{25} = 5$ units.

The side AC lies along the y-axis, and its length is the distance between (0,0) and (0,8).

Height AC $= \sqrt{(0-0)^2 + (8-0)^2} = \sqrt{0^2 + 8^2} = \sqrt{64} = 8$ units.

Since AB is on the x-axis and AC is on the y-axis, and the axes are perpendicular, $\triangle ABC$ is a right-angled triangle with the right angle at the origin A(0,0).

In a right-angled triangle, the legs can serve as the base and height.

Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height}$

Area $= \frac{1}{2} \times \text{Length of AB} \times \text{Length of AC}$

Area $= \frac{1}{2} \times 5 \times 8$

Area $= \frac{1}{2} \times 40$

Area $= 20$ square units.

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Alternate Solution (Using Coordinate Formula):

Let the vertices be $(x_1, y_1) = (0,0)$, $(x_2, y_2) = (5,0)$, and $(x_3, y_3) = (0,8)$.

The formula for the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is:

Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Substitute the coordinates into the formula:

Area $= \frac{1}{2} |0(0 - 8) + 5(8 - 0) + 0(0 - 0)|$

Area $= \frac{1}{2} |0(-8) + 5(8) + 0(0)|$

Area $= \frac{1}{2} |0 + 40 + 0|$

Area $= \frac{1}{2} |40|$

Area $= \frac{1}{2} \times 40$

Area $= 20$ square units.

Both methods give the same result.

The area of the triangle is $20$ square units.

Theorem: Parallelograms on the same base and between the same parallels are equal in area.

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Given:

Two parallelograms, ABCD and EBCF, are on the same base BC.

They are between the same parallel lines, BC and AF (where line AF contains points A, D, E, F).

---

To Prove:

Area(parallelogram ABCD) = Area(parallelogram EBCF)

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Diagram:

Draw two parallel lines, say $l_1$ and $l_2$. Let BC be a line segment on $l_1$. Draw points A, D, E, F on line $l_2$ such that A, D are the vertices opposite to BC for parallelogram ABCD, and E, F are the vertices opposite to BC for parallelogram EBCF. Assume the parallelograms overlap such that the vertices on line $l_2$ are in the order A, E, D, F.

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Proof:

Consider the two parallelograms ABCD and EBCF, which are on the same base BC and between the same parallel lines BC and AF.

Since ABCD is a parallelogram, AB is parallel to DC, and AD is parallel to BC. Also, AD = BC.

Since EBCF is a parallelogram, EB is parallel to FC, and EF is parallel to BC. Also, EF = BC.

Since AD || BC and EF || BC, and A, D, E, F lie on the line AF which is parallel to BC, the line segments AD and EF lie on the same line AF, and their lengths are equal to BC. Thus, AD = EF.

Consider the $\triangle ABE$ and $\triangle DCF$.

1. AB = DC (Opposite sides of parallelogram ABCD)

2. BE = CF (Opposite sides of parallelogram EBCF)

3. AE = DF

Since A, E, D, F are collinear and AD = EF, and E is between A and D (from diagram configuration), we have:

AD = AE + ED

EF = ED + DF

Since AD = EF, we have AE + ED = ED + DF.

Subtracting ED from both sides gives AE = DF.

By the SSS (Side-Side-Side) congruence criterion, $\triangle ABE \cong \triangle DCF$.

Congruent triangles have equal areas.

Therefore, Area($\triangle ABE$) = Area($\triangle DCF$).

Now consider the area of parallelogram ABCD. It can be seen as the union of $\triangle ABE$ and quadrilateral EBCD. The intersection is the line segment BE, which has zero area.

Area(parallelogram ABCD) = Area(quadrilateral EBCD) + Area($\triangle ABE$)

Similarly, the area of parallelogram EBCF can be seen as the union of $\triangle DCF$ and quadrilateral EBCD. The intersection is the line segment CF, which has zero area.

Area(parallelogram EBCF) = Area(quadrilateral EBCD) + Area($\triangle DCF$)

Since Area($\triangle ABE$) = Area($\triangle DCF$), we can substitute Area($\triangle ABE$) for Area($\triangle DCF$) in the second equation:

Area(parallelogram EBCF) = Area(quadrilateral EBCD) + Area($\triangle ABE$)

Comparing the expressions for Area(parallelogram ABCD) and Area(parallelogram EBCF):

Area(parallelogram ABCD) = Area(quadrilateral EBCD) + Area($\triangle ABE$)

Area(parallelogram EBCF) = Area(quadrilateral EBCD) + Area($\triangle ABE$)

Thus, Area(parallelogram ABCD) = Area(parallelogram EBCF).

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Note: A similar proof using congruent triangles and area decomposition can be constructed for the case where the parallelograms do not overlap, by considering different congruent triangles formed at the outer ends of the combined figure and the trapezium formed by the two parallelograms and the line connecting the bases.

Theorem: The area of a triangle is half the area of a parallelogram on the same base and between the same parallels.

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Given:

A parallelogram ABCD and a triangle ABE are on the same base AB.

They are between the same parallel lines AB and DC (where vertex E lies on the line DC).

---

To Prove:

Area$(\triangle ABE) = \frac{1}{2} \times$ Area(parallelogram ABCD)

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Diagram:

Draw two parallel lines. Let the lower line contain the points A and B, forming the base AB. Let the upper line contain the points D and C, such that ADCB forms a parallelogram. Let E be a point on the line segment DC. Draw the triangle ABE.

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Construction:

Draw a line through B parallel to AE, which meets the line DC at point F.

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Proof:

By construction, AB || DF and AE || BF. Thus, ABFE is a parallelogram.

The parallelogram ABFE and the parallelogram ABCD are on the same base AB and between the same parallel lines AB and DC.

According to the theorem about the areas of parallelograms on the same base and between the same parallels (as proved in the previous question):

Area(parallelogram ABFE) = Area(parallelogram ABCD)

Now, consider the parallelogram ABFE. The diagonal BE divides it into two triangles, $\triangle ABE$ and $\triangle FEB$.

We know that a diagonal of a parallelogram divides it into two congruent triangles. Therefore, $\triangle ABE \cong \triangle FEB$.

Congruent triangles have equal areas.

So, Area($\triangle ABE$) = Area($\triangle FEB$).

The area of parallelogram ABFE is the sum of the areas of these two triangles:

Area(parallelogram ABFE) = Area($\triangle ABE$) + Area($\triangle FEB$)

Since Area($\triangle ABE$) = Area($\triangle FEB$), we can write:

Area(parallelogram ABFE) = Area($\triangle ABE$) + Area($\triangle ABE$)

Area(parallelogram ABFE) = $2 \times$ Area($\triangle ABE$)

From this, we can express the area of the triangle:

Area($\triangle ABE$) = $\frac{1}{2} \times$ Area(parallelogram ABFE)

Since Area(parallelogram ABFE) = Area(parallelogram ABCD), we can substitute Area(parallelogram ABCD) into the equation:

Area($\triangle ABE$) = $\frac{1}{2} \times$ Area(parallelogram ABCD)

This proves that the area of the triangle is half the area of the parallelogram when they are on the same base and between the same parallels.

Theorem: Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

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Given:

Two triangles, $\triangle ABC$ and $\triangle PBC$, are on the same base BC.

They are between the same parallel lines, BC and AP (where A and P lie on the line parallel to BC).

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To Prove:

Area($\triangle ABC$) = Area($\triangle PBC$)

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Diagram:

Draw two parallel lines. Let the lower line be denoted by $l_1$ and the upper line by $l_2$. Let BC be a line segment on $l_1$. Draw points A and P on the line $l_2$. Draw line segments AB, AC, PB, and PC to form $\triangle ABC$ and $\triangle PBC$.

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Construction:

Through B, draw a line parallel to AC, meeting the line AP at point D. This forms parallelogram ABCD.

Through C, draw a line parallel to BP, meeting the line AP at point Q. This forms parallelogram PBCQ.

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Proof:

Consider $\triangle ABC$ and parallelogram ABCD. They are on the same base BC and between the same parallel lines BC and AD (since A and D are on the line AP, which is parallel to BC).

According to the property relating the area of a triangle to the area of a parallelogram on the same base and between the same parallels:

Now, consider $\triangle PBC$ and parallelogram PBCQ. They are on the same base BC and between the same parallel lines BC and PQ (since P and Q are on the line AP, which is parallel to BC).

Using the same property:

Now, consider the parallelograms ABCD and PBCQ. Both parallelograms are on the same base BC and between the same parallel lines BC and AP.

According to the theorem proved in Question 1, parallelograms on the same base and between the same parallels are equal in area.

From equations (i), (ii), and (iii), we can conclude that:

Area($\triangle ABC$) = $\frac{1}{2} \times$ Area(parallelogram ABCD)

Area($\triangle PBC$) = $\frac{1}{2} \times$ Area(parallelogram PBCQ)

Since Area(parallelogram ABCD) = Area(parallelogram PBCQ), it follows that $\frac{1}{2} \times$ Area(parallelogram ABCD) = $\frac{1}{2} \times$ Area(parallelogram PBCQ).

Therefore, Area($\triangle ABC$) = Area($\triangle PBC$).

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Extension to Equal Bases:

If two triangles are on equal bases (say $b$) and between the same parallels (meaning they have the same height $h$), their areas are $\frac{1}{2} \times b \times h$. Since both base and height are equal for both triangles, their areas will also be equal.

We will address the third version of Question 4, which asks us to prove a standard property about the areas of triangles formed by an interior point in a parallelogram.

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Theorem: If P is any point in the interior of a parallelogram ABCD, then Area($\triangle APB$) + Area($\triangle CPD$) = Area($\triangle APD$) + Area($\triangle BPC$). Also, Area($\triangle APB$) + Area($\triangle CPD$) = $\frac{1}{2}$ Area$(ABCD)$.

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Given:

ABCD is a parallelogram.

P is any point in the interior of the parallelogram.

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To Prove:

1. Area($\triangle APB$) + Area($\triangle CPD$) = Area($\triangle APD$) + Area($\triangle BPC$)

2. Area($\triangle APB$) + Area($\triangle CPD$) = $\frac{1}{2}$ Area$(ABCD)$

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Diagram:

Draw a parallelogram ABCD and mark a point P somewhere inside it. Draw line segments from P to each vertex A, B, C, and D.

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Construction:

Through point P, draw a line EF parallel to AB and DC, where E is on AD and F is on BC.

Through point P, draw a line GH parallel to AD and BC, where G is on AB and H is on DC.

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Proof:

Consider the line segment EF drawn through P parallel to AB and DC. Since EF is parallel to AB and DC, and AD is parallel to BC, the construction divides parallelogram ABCD into two smaller parallelograms: ABFE and EFCD.

Area(ABCD) = Area(parallelogram ABFE) + Area(parallelogram EFCD)

Consider $\triangle APB$. This triangle is on the base AB and between the parallel lines AB and EF (since P lies on EF). According to the theorem relating the area of a triangle to the area of a parallelogram on the same base and between the same parallels:

Consider $\triangle CPD$. This triangle is on the base CD and between the parallel lines CD and EF (since P lies on EF). CD = EF (opposite sides of parallelogram EFCD).

Adding equation (i) and equation (ii):

Area($\triangle APB$) + Area($\triangle CPD$) = $\frac{1}{2}$ Area(parallelogram ABFE) + $\frac{1}{2}$ Area(parallelogram EFCD)

Area($\triangle APB$) + Area($\triangle CPD$) = $\frac{1}{2}$ [Area(parallelogram ABFE) + Area(parallelogram EFCD)]

Since Area(parallelogram ABFE) + Area(parallelogram EFCD) = Area(ABCD), we have:

This proves the second part of the theorem.

Now consider the line segment GH drawn through P parallel to AD and BC. Since GH is parallel to AD and BC, and AB is parallel to DC, the construction divides parallelogram ABCD into two smaller parallelograms: AGHD and GBCF.

Area(ABCD) = Area(parallelogram AGHD) + Area(parallelogram GBCF)

Consider $\triangle APD$. This triangle is on the base AD and between the parallel lines AD and GH (since P lies on GH). According to the same theorem:

Consider $\triangle BPC$. This triangle is on the base BC and between the parallel lines BC and GH (since P lies on GH). BC = GH (opposite sides of parallelogram GBCF).

Adding equation (iv) and equation (v):

Area($\triangle APD$) + Area($\triangle BPC$) = $\frac{1}{2}$ Area(parallelogram AGHD) + $\frac{1}{2}$ Area(parallelogram GBCF)

Area($\triangle APD$) + Area($\triangle BPC$) = $\frac{1}{2}$ [Area(parallelogram AGHD) + Area(parallelogram GBCF)]

Since Area(parallelogram AGHD) + Area(parallelogram GBCF) = Area(ABCD), we have:

From equation (iii) and equation (vi), we see that both sums are equal to $\frac{1}{2}$ Area(ABCD).

Therefore, Area($\triangle APB$) + Area($\triangle CPD$) = Area($\triangle APD$) + Area($\triangle BPC$).

This proves the first part of the theorem.

Both parts of the theorem are thus proved.

This problem can be solved using the property that a median of a triangle divides it into two triangles of equal area, and the property that triangles on the same base and between the same parallels are equal in area (which implies that triangles with the same base and same height have equal areas, and triangles with common vertex and collinear bases have areas proportional to their bases).

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Given:

In $\triangle ABC$, D is the midpoint of BC.

E is the midpoint of AD.

BE is joined and extended to meet AC at F.

---

To Prove:

$AF = \frac{1}{3} AC$

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Diagram:

Draw $\triangle ABC$. Mark the midpoint D on BC. Draw the line segment AD. Mark the midpoint E on AD. Draw the line segment BE and extend it to meet AC at F.

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Proof:

Since D is the midpoint of BC, AD is a median of $\triangle ABC$.

A median divides a triangle into two triangles of equal area.

Since E is the midpoint of AD, BE is a line segment from B to E on AD, and CE is a line segment from C to E on AD.

Consider $\triangle ABD$. E is the midpoint of AD. Triangles $\triangle ABE$ and $\triangle BDE$ have equal bases (AE = ED) and share the common vertex B. Therefore, they have the same height from B to AD, and their areas are equal.

Similarly, consider $\triangle ACD$. E is the midpoint of AD. Triangles $\triangle ACE$ and $\triangle CDE$ have equal bases (AE = ED) and share the common vertex C. Therefore, they have the same height from C to AD, and their areas are equal.

From (i), Area($\triangle ABD$) = Area($\triangle ACD$).

Area($\triangle ABD$) can be written as Area($\triangle ABE$) + Area($\triangle BDE$).

Area($\triangle ACD$) can be written as Area($\triangle ACE$) + Area($\triangle CDE$).

So, Area($\triangle ABE$) + Area($\triangle BDE$) = Area($\triangle ACE$) + Area($\triangle CDE$).

Using (ii) and (iii), we can substitute: $2 \times$ Area($\triangle ABE$) = $2 \times$ Area($\triangle ACE$).

This implies:

Combining (ii), (iii), and (iv), we have: Area($\triangle ABE$) = Area($\triangle BDE$) = Area($\triangle ACE$) = Area($\triangle CDE$).

Let this common area be denoted by $k$. So, Area($\triangle ABE$) = $k$, Area($\triangle BDE$) = $k$, Area($\triangle ACE$) = $k$, Area($\triangle CDE$) = $k$.

Now consider the area of $\triangle CBE$.

Area($\triangle CBE$) = Area($\triangle BDE$) + Area($\triangle CDE$) = $k + k = 2k$.

Consider triangles $\triangle ABE$ and $\triangle CBE$. They share the common base BE. The ratio of their areas is equal to the ratio of the lengths of the perpendiculars drawn from A and C to the line BEF.

Now consider triangles $\triangle ABF$ and $\triangle CBF$. These triangles share the common base BF (since F lies on the extension of BE). The ratio of their areas is also equal to the ratio of the lengths of the perpendiculars drawn from A and C to the line BEF. This ratio is the same as the ratio of areas of $\triangle ABE$ and $\triangle CBE$.

From (vi), we have Area($\triangle CBF$) = $2 \times$ Area($\triangle ABF$).

Finally, consider triangles $\triangle ABF$ and $\triangle CBF$. They have a common vertex B, and their bases AF and FC are collinear along the line segment AC.

For triangles sharing a common vertex and having collinear bases, the ratio of their areas is equal to the ratio of their corresponding bases.

From (vi) and (vii), we equate the ratios:

$\frac{AF}{FC} = \frac{1}{2}$

This implies $FC = 2 \times AF$.

Now, look at the line segment AC. F is a point on AC such that AC = AF + FC.

Substitute the expression for FC:

$AC = AF + 2 \times AF$

$AC = 3 \times AF$

Dividing both sides by 3, we get:

$AF = \frac{1}{3} AC$

Hence, proved.

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Alternate Solution (Using Construction and BPT - not explicitly using area ratios as the primary tool, but a valid geometric proof):

Construction: Draw a line through D parallel to BF, which intersects AC at G.

Proof:

In $\triangle BCF$, D is the midpoint of BC, and by construction, DG $\parallel$ BF (and thus DG $\parallel$ CF since F is on the extension of BE). No, DG $\parallel$ BF means DG $\parallel$ BF. DG is parallel to BEF.

In $\triangle CBF$, D is the midpoint of BC, and DG $\parallel$ BF. By the Midpoint Theorem (or converse of BPT), a line through the midpoint of one side of a triangle parallel to another side bisects the third side. However, D is on BC and DG is parallel to BF which is a side of $\triangle CBF$. We should use BPT.

In $\triangle CBF$, D is on BC and G is on CF. DG $\parallel$ BF. By Basic Proportionality Theorem (BPT):

Since D is the midpoint of BC, CD = DB. So, $\frac{CD}{DB} = 1$.

Substituting this into (viii): $1 = \frac{CG}{GF} \implies CG = GF$.

Now consider $\triangle ADG$. E is the midpoint of AD, and EF is part of the line BEF. By construction, DG $\parallel$ BEF, so DG $\parallel$ EF. E is on AD and F is on AG (since G is on AC). No, F is on AC and G is on AC. F is on line BE extended. F is the intersection of BE and AC. G is the intersection of the line through D parallel to BF and AC.

Let's re-examine the construction and its implication.

In $\triangle ADG$, E is the midpoint of AD and EF is parallel to DG (since EF is part of BEF and DG $\parallel$ BEF by construction). Thus, a line segment through the midpoint of one side of $\triangle ADG$ (side AD), parallel to another side (DG), must bisect the third side (AG).

Therefore, F must be the midpoint of AG.

So, AF = FG.

We have AC = AF + FG + GC.

From the first part of the proof (using BPT in $\triangle CBF$), we got CG = GF.

Since AF = FG and CG = GF, we have AF = FG = GC.

AC = AF + AF + AF = 3AF.

Thus, AF = $\frac{1}{3} AC$.

This alternate solution also proves the result, primarily using BPT and properties of midpoints rather than focusing heavily on area ratios of triangles with common bases or heights from a common vertex.

We will address the second version of Question 6, which asks us to prove that a median of a triangle divides it into two triangles of equal areas.

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Theorem: A median of a triangle divides it into two triangles of equal areas.

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Given:

In $\triangle ABC$, AD is a median.

(This means D is the midpoint of the side BC).

---

To Prove:

Area($\triangle ABD$) = Area($\triangle ACD$)

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Diagram:

Draw a triangle ABC. Mark the midpoint D of the side BC. Draw the median AD.

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Construction:

Draw a perpendicular AM from vertex A to the side BC. AM represents the height of $\triangle ABC$, $\triangle ABD$, and $\triangle ACD$ with respect to their bases on the line BC.

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Proof:

Since AD is a median of $\triangle ABC$, D is the midpoint of BC.

Therefore, the length of the base BD is equal to the length of the base DC.

Now, consider the area of $\triangle ABD$. The base is BD and the corresponding height is AM (the perpendicular distance from A to the line containing BC).

Using the area formula for a triangle (Area $= \frac{1}{2} \times \text{base} \times \text{height}$):

Next, consider the area of $\triangle ACD$. The base is DC and the corresponding height is AM (the perpendicular distance from A to the line containing BC).

Using the area formula for a triangle:

From the fact that D is the midpoint of BC, we have BD = DC.

Substitute DC with BD in equation (ii):

Area($\triangle ACD$) = $\frac{1}{2} \times BD \times AM$

Comparing this with equation (i), we see that the right-hand sides are equal.

Therefore, Area($\triangle ABD$) = Area($\triangle ACD$).

This proves that the median AD divides the triangle ABC into two triangles of equal areas, $\triangle ABD$ and $\triangle ACD$.

We will address the second version of Question 7, which involves triangles on the same base or equal bases and between parallel lines.

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Given:

In $\triangle LMN$, X and Y are two points on the side LM such that LX = XY = YM.

A line is drawn through X parallel to LN, meeting MN at Z.

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To Prove:

Area$(\triangle LXZ) =$ Area$(\triangle XYZ) =$ Area$(\triangle YMZ)$

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Diagram:

Draw a triangle LMN. Mark points X and Y on side LM such that LX=XY=YM. Draw a line through X parallel to LN, intersecting MN at Z.

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Proof:

We are given that LX = XY = YM.

We are also given that XZ is parallel to LN.

Since XZ $\parallel$ LN, we can consider $\triangle LMN$ and the line segment XZ.

By the Basic Proportionality Theorem (BPT), if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides proportionally.

In $\triangle LMN$, XZ $\parallel$ LN. Therefore, by BPT:

We know that XM = XY + YM. Since LX = XY = YM, let LX = XY = YM = $a$ units.

Then XM = $a + a = 2a$. LX = $a$.

So, $\frac{LX}{XM} = \frac{a}{2a} = \frac{1}{2}$.

From (i), $\frac{LZ}{ZN} = \frac{1}{2}$. This implies $ZN = 2 \times LZ$.

Now, consider the areas of the triangles $\triangle LXZ$, $\triangle XYZ$, and $\triangle YMZ$.

Triangles $\triangle LXZ$ and $\triangle XYZ$ share the common altitude from Z to the line segment LM. Their bases are LX and XY, respectively.

The ratio of the areas of two triangles with the same altitude is equal to the ratio of their corresponding bases.

We are given LX = XY.

Substitute LX = XY into (ii):

$\frac{\text{Area}(\triangle LXZ)}{\text{Area}(\triangle XYZ)} = \frac{LX}{LX} = 1$

Similarly, triangles $\triangle XYZ$ and $\triangle YMZ$ share the common altitude from Z to the line segment LM. Their bases are XY and YM, respectively.

We are given XY = YM.

Substitute XY = YM into (iv):

$\frac{\text{Area}(\triangle XYZ)}{\text{Area}(\triangle YMZ)} = \frac{XY}{XY} = 1$

From (iii) and (v), we can conclude that:

Area$(\triangle LXZ) =$ Area$(\triangle XYZ) =$ Area$(\triangle YMZ)$

This proves the required result.

We are given a quadrilateral ABCD where the diagonals AC and BD intersect at O. We are also given that the areas of two triangles formed by the intersecting diagonals are equal.

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Given:

In quadrilateral ABCD, diagonals AC and BD intersect at O.

Area($\triangle AOD$) = Area($\triangle BOC$).

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To Prove:

ABCD is a trapezium (i.e., one pair of opposite sides is parallel).

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Diagram:

Draw a quadrilateral ABCD and its diagonals AC and BD intersecting at O. Label the triangles formed: $\triangle AOB$, $\triangle BOC$, $\triangle COD$, and $\triangle AOD$.

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Proof:

We are given that Area($\triangle AOD$) = Area($\triangle BOC$).

Add Area($\triangle AOB$) to both sides of this equation:

Area($\triangle AOD$) + Area($\triangle AOB$) = Area($\triangle BOC$) + Area($\triangle AOB$)

The sum of the areas of $\triangle AOD$ and $\triangle AOB$ gives the area of $\triangle ABD$.

The sum of the areas of $\triangle BOC$ and $\triangle AOB$ gives the area of $\triangle ABC$.

Since Area($\triangle AOD$) + Area($\triangle AOB$) = Area($\triangle BOC$) + Area($\triangle AOB$), we can conclude that:

Now, consider the two triangles $\triangle ABD$ and $\triangle ABC$. They share the same base AB.

Also, their areas are equal, as shown in (iii).

We know the property that if two triangles have the same base and equal areas, then they must lie between the same parallel lines.

Since $\triangle ABD$ and $\triangle ABC$ have the same base AB and Area($\triangle ABD$) = Area($\triangle ABC$), their vertices opposite to the base AB (which are D and C, respectively) must lie on a line parallel to AB.

Therefore, the line segment DC is parallel to the line segment AB.

A quadrilateral with at least one pair of opposite sides parallel is defined as a trapezium.

Since AB $\parallel$ DC, the quadrilateral ABCD is a trapezium.

Hence, proved.

We are given a parallelogram ABCD with diagonal BD. Points P and Q are on BD such that the lengths DP and BQ are equal.

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Given:

ABCD is a parallelogram.

P and Q are points on the diagonal BD.

DP = BQ.

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To Prove:

Area$(\triangle APD) =$ Area$(\triangle CQB)$.

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Diagram:

Draw a parallelogram ABCD. Draw the diagonal BD. Mark points P and Q on BD such that the segment from D to P has the same length as the segment from B to Q.

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Proof:

In a parallelogram ABCD, the diagonal BD divides it into two triangles of equal area, $\triangle ABD$ and $\triangle CDB$.

Area($\triangle ABD$) = Area($\triangle CDB$).

Let $h_A$ be the perpendicular distance from vertex A to the diagonal BD, and $h_C$ be the perpendicular distance from vertex C to the diagonal BD. These are the heights of $\triangle ABD$ and $\triangle CDB$ respectively, with respect to the base BD.

Area($\triangle ABD$) $= \frac{1}{2} \times BD \times h_A$

Area($\triangle CDB$) $= \frac{1}{2} \times BD \times h_C$

Since Area($\triangle ABD$) = Area($\triangle CDB$):

Since BD is a diagonal, $BD > 0$. We can cancel $\frac{1}{2} \times BD$ from both sides of equation (i).

This gives us $h_A = h_C$.

Let $h = h_A = h_C$. This height $h$ represents the perpendicular distance of both A and C from the diagonal line BD.

Now consider the triangle $\triangle APD$. Its base is PD and the corresponding height is $h_A = h$ (the perpendicular distance from A to the line containing PD, which is BD).

Next, consider the triangle $\triangle CQB$. Its base is BQ and the corresponding height is $h_C = h$ (the perpendicular distance from C to the line containing BQ, which is BD).

We are given that DP = BQ.

Substitute BQ with DP in equation (iii):

Area($\triangle CQB$) = $\frac{1}{2} \times DP \times h$

Comparing this modified equation (iii) with equation (ii):

Area($\triangle APD$) = $\frac{1}{2} \times PD \times h$

Area($\triangle CQB$) = $\frac{1}{2} \times DP \times h$

Since PD = DP, the expressions for the areas are identical.

Therefore, Area($\triangle APD$) = Area($\triangle CQB$).

Hence, proved.

We will address the second version of Question 10, which focuses on finding the area of a general quadrilateral by dividing it into two triangles.

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Explanation:

Any quadrilateral can be divided into two triangles by drawing one of its diagonals. The area of the quadrilateral is then the sum of the areas of the two triangles formed.

Consider a quadrilateral ABCD. Draw the diagonal AC. This divides the quadrilateral into two triangles: $\triangle ABC$ and $\triangle ADC$.

Area(quadrilateral ABCD) = Area($\triangle ABC$) + Area($\triangle ADC$)

To find the area of each triangle, we can use the formula Area $= \frac{1}{2} \times \text{base} \times \text{height}$. In this case, the diagonal serves as a common base for both triangles if we consider the heights perpendicular to the diagonal from the opposite vertices.

Draw a perpendicular from B to AC, let its length be $h_1$. This is the height of $\triangle ABC$ with base AC.

Draw a perpendicular from D to AC, let its length be $h_2$. This is the height of $\triangle ADC$ with base AC.

Area($\triangle ABC$) $= \frac{1}{2} \times AC \times h_1$

Area($\triangle ADC$) $= \frac{1}{2} \times AC \times h_2$

Area(quadrilateral ABCD) $= \frac{1}{2} \times AC \times h_1 + \frac{1}{2} \times AC \times h_2$

Area(quadrilateral ABCD) $= \frac{1}{2} \times AC \times (h_1 + h_2)$

So, the area of a quadrilateral is half the product of a diagonal and the sum of the lengths of the perpendiculars drawn from the opposite vertices to that diagonal.

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Illustration with an example:

Consider a quadrilateral ABCD with diagonal AC.

Given:

Length of the diagonal AC $= 10$ cm.

Length of the perpendicular from B to AC, $h_1 = 6$ cm.

Length of the perpendicular from D to AC, $h_2 = 8$ cm.

To Find:

The area of the quadrilateral ABCD.

Solution:

The diagonal AC divides the quadrilateral into two triangles, $\triangle ABC$ and $\triangle ADC$.

Area($\triangle ABC$) $= \frac{1}{2} \times \text{base AC} \times \text{height } h_1$

Area($\triangle ABC$) $= \frac{1}{2} \times 10 \text{ cm} \times 6 \text{ cm}$

Area($\triangle ABC$) $= \frac{1}{2} \times 60$ $cm^2$

Area($\triangle ABC$) $= 30$ $cm^2$

Area($\triangle ADC$) $= \frac{1}{2} \times \text{base AC} \times \text{height } h_2$

Area($\triangle ADC$) $= \frac{1}{2} \times 10 \text{ cm} \times 8 \text{ cm}$

Area($\triangle ADC$) $= \frac{1}{2} \times 80$ $cm^2$

Area($\triangle ADC$) $= 40$ $cm^2$

The area of the quadrilateral ABCD is the sum of the areas of $\triangle ABC$ and $\triangle ADC$.

Area(quadrilateral ABCD) = Area($\triangle ABC$) + Area($\triangle ADC$)

Area(quadrilateral ABCD) $= 30$ $cm^2$ + $40$ $cm^2$

Area(quadrilateral ABCD) $= 70$ $cm^2$

Alternatively, using the combined formula:

Area(quadrilateral ABCD) $= \frac{1}{2} \times AC \times (h_1 + h_2)$

Area(quadrilateral ABCD) $= \frac{1}{2} \times 10 \text{ cm} \times (6 \text{ cm} + 8 \text{ cm})$

Area(quadrilateral ABCD) $= \frac{1}{2} \times 10 \text{ cm} \times 14 \text{ cm}$

Area(quadrilateral ABCD) $= \frac{1}{2} \times 140$ $cm^2$

Area(quadrilateral ABCD) $= 70$ $cm^2$

The area of the quadrilateral is $70$ sq cm.

We are given a parallelogram ABCD and a rectangle EFCD in the same figure. We need to prove that their areas are equal.

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Given:

ABCD is a parallelogram.

EFCD is a rectangle.

AL $\perp$ DC (where L is a point on DC or its extension).

(From the diagram, it appears that the parallelogram and the rectangle share the base CD and are between the same parallel lines DC and AF/BE).

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To Prove:

Area$(ABCD) =$ Area$(EFCD)$.

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Diagram Analysis:

From the figure, we can see that both the parallelogram ABCD and the rectangle EFCD share the common base CD.

Since ABCD is a parallelogram, AB $\parallel$ DC. The vertices A and B lie on a line parallel to DC.

Since EFCD is a rectangle, EF $\parallel$ DC, and sides ED and FC are perpendicular to DC. The vertices E and F lie on a line parallel to DC.

From the diagram, it appears that the points A, B, E, and F lie on the same line which is parallel to DC. This means the parallelogram and the rectangle are between the same parallel lines DC and AF (or BE).

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Proof:

Parallelogram ABCD and rectangle EFCD are on the same base CD.

Since EFCD is a rectangle, EF is parallel to CD. The distance between the parallel lines EF and CD is the height of the rectangle.

Since ABCD is a parallelogram, AB is parallel to CD. The distance between the parallel lines AB and CD is the height of the parallelogram.

From the figure, the line containing A and B is the same as the line containing E and F, and this line is parallel to DC. Thus, the parallelogram ABCD and the rectangle EFCD are between the same parallel lines (DC and the line containing A, B, E, F).

We know that the area of a parallelogram is given by base $\times$ height.

The base of parallelogram ABCD is CD.

The height of parallelogram ABCD corresponding to base CD is the perpendicular distance between the parallel lines AB and DC. From the figure and the fact that AL $\perp$ DC, AL is the height of the parallelogram.

Area$(ABCD) = CD \times AL$

The area of a rectangle is given by length $\times$ width.

The length of rectangle EFCD is CD.

The width of rectangle EFCD is the perpendicular distance between the parallel sides CD and EF. This is equal to FC or ED.

Since EFCD is a rectangle, FC is perpendicular to DC. Thus, FC is the height of the rectangle with respect to base CD.

Area$(EFCD) = CD \times FC$

Since parallelogram ABCD and rectangle EFCD are between the same parallel lines DC and AF, the perpendicular distance between these parallel lines is the same. The height of the parallelogram AL is equal to the height of the rectangle FC.

Now substitute AL = FC into the area formula for the parallelogram:

Area$(ABCD) = CD \times FC$

Comparing this with the area of the rectangle, Area$(EFCD) = CD \times FC$, we see that the areas are equal.

Therefore, Area$(ABCD) =$ Area$(EFCD)$.

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This proof relies on the fundamental property that parallelograms and rectangles on the same base and between the same parallels have equal areas. A rectangle is just a special case of a parallelogram where the height is one of the sides.

We are given two triangles $\triangle ABC$ and $\triangle ABD$ on the same base AB. The line segment CD is bisected by AB at point O, which means O is the midpoint of CD.

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Given:

Triangles $\triangle ABC$ and $\triangle ABD$ are on the same base AB.

Line segment CD is bisected by AB at O, i.e., O is the midpoint of CD, so CO = OD.

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To Prove:

Area($\triangle ABC$) = Area($\triangle ABD$).

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Diagram:

Draw a line segment AB. Draw points C and D on opposite sides (or same side, but problem implies separate triangles) of AB. Draw lines AC, BC, AD, BD. Draw the line segment CD intersecting AB at O, such that O is the midpoint of CD.

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Proof:

Consider $\triangle BCD$. BO is a line segment from vertex B to the midpoint O of the side CD. Thus, BO is a median of $\triangle BCD$.

A median of a triangle divides it into two triangles of equal areas.

Similarly, consider $\triangle ACD$. AO is a line segment from vertex A to the midpoint O of the side CD. Thus, AO is a median of $\triangle ACD$.

Area($\triangle AOC$) = Area($\triangle AOD$)

Now, consider the area of $\triangle ABC$. It is the sum of the areas of $\triangle AOC$ and $\triangle BOC$.

Consider the area of $\triangle ABD$. It is the sum of the areas of $\triangle AOD$ and $\triangle BOD$.

From equation (i), Area($\triangle BOC$) = Area($\triangle BOD$).

From the proof in a previous question (Question 6, Redo), we showed that a median divides a triangle into two triangles of equal areas. Applying this to $\triangle ACD$ with median AO, we get Area($\triangle AOC$) = Area($\triangle AOD$).

Substitute Area($\triangle AOC$) with Area($\triangle AOD$) and Area($\triangle BOC$) with Area($\triangle BOD$) in equation (ii):

Area($\triangle ABC$) = Area($\triangle AOD$) + Area($\triangle BOD$)

Comparing this with equation (iii), we see that the right-hand sides are identical.

Therefore, Area($\triangle ABC$) = Area($\triangle ABD$).

Hence, proved.