Chapter 3 Understanding Quadrilaterals

We classify each figure based on the given properties:

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(a) Simple curve: A curve that does not cross itself.

Figures: 1, 2, 5, 6, and 7

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(b) Simple closed curve: A simple curve that begins and ends at the same point.

Figures: 1, 2, 5, 6, and 7

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(c) Polygon: A simple closed curve made up entirely of line segments.

Figures: 1 and 2

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(d) Convex polygon: A polygon where all its interior angles are less than $180^\circ$.

Figure: 2

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(e) Concave polygon: A polygon that has at least one interior angle greater than $180^\circ$.

Figure: 1

Definition:

A regular polygon is a polygon that is both equiangular and equilateral. This means that all its interior angles are equal in measure, and all its sides are equal in length.

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Names of regular polygons:

(i) A regular polygon of 3 sides is called an Equilateral Triangle.

(ii) A regular polygon of 4 sides is called a Square.

(iii) A regular polygon of 6 sides is called a Regular Hexagon.

Given:

A quadrilateral with interior angles $x^\circ$, $90^\circ$, $50^\circ$, and $110^\circ$. Note that the square symbol indicates a $90^\circ$ angle.

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To Find:

The measure of angle $x$.

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Solution:

The given figure is a quadrilateral.

We know that the sum of the interior angles of a quadrilateral is $360^\circ$.

Therefore, we can write the equation:

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Add the known angles on the left side:

$(90 + 50 + 110)^\circ = 140^\circ + 110^\circ = 250^\circ$

The equation becomes:

$x + 250^\circ = 360^\circ$

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To find $x$, subtract $250^\circ$ from both sides of the equation:

$x = 360^\circ - 250^\circ$

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Calculate the result:

$x = 110^\circ$

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The measure of angle $x$ is $\mathbf{110^\circ}$.

Given:

The polygon is regular.

Measure of each exterior angle $= 45^\circ$.

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To Find:

The number of sides of the regular polygon.

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Solution:

We know that the sum of the measures of the exterior angles of any convex polygon is $360^\circ$.

For a regular polygon, all exterior angles are equal in measure.

Let the number of sides of the regular polygon be $n$.

The measure of each exterior angle of a regular polygon with $n$ sides is given by the formula:

Measure of each exterior angle $= \frac{\text{Sum of exterior angles}}{\text{Number of sides}}$

Measure of each exterior angle $= \frac{360^\circ}{n}$

We are given that the measure of each exterior angle is $45^\circ$. So, we have the equation:

$45^\circ = \frac{360^\circ}{n}$

Perform the division:

$n = 8$

Therefore, the regular polygon has $8$ sides.

A regular polygon with $8$ sides is called a regular octagon.

We will use the properties of linear pairs and the sum of angles in a polygon to find the value of $x$ in each figure.

Recall that angles in a linear pair sum up to $180^\circ$. Also, the sum of interior angles of a polygon with $n$ sides is given by $(n-2) \times 180^\circ$.

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Figure (a):

The figure shows a triangle with exterior angles $125^\circ$, $125^\circ$, and $x^\circ$. Let the interior angles adjacent to the $125^\circ$ exterior angles be $m$ and $n$ respectively.

Using the linear pair property:

$m = 180^\circ - 125^\circ$

$m = 55^\circ$

Similarly, for the other angle:

$n = 180^\circ - 125^\circ$

$n = 55^\circ$

Now, we can use the exterior angle property of a triangle, which states that an exterior angle is equal to the sum of the two opposite interior angles.

In this figure, the exterior angle $x$ is opposite to the interior angles $m$ and $n$.

Substitute the values of $m$ and $n$:

$x = 55^\circ + 55^\circ$

$x = 110^\circ$

Therefore, the value of $x$ in Figure (a) is $110^\circ$.

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Figure (b):

The figure is a pentagon (5 sides). It has exterior angles $x^\circ$, $70^\circ$, $60^\circ$, and two $90^\circ$ (adjacent to the interior right angles). The angles given are $x^\circ$, $70^\circ$, and $60^\circ$ as exterior angles, and two right angles ($90^\circ$) as interior angles.

Let's find the interior angles adjacent to the exterior angles $70^\circ$ and $60^\circ$. Let them be $m$ and $n$ respectively.

Using the linear pair property for the $70^\circ$ exterior angle:

$m = 180^\circ - 70^\circ$

$m = 110^\circ$

Using the linear pair property for the $60^\circ$ exterior angle:

$n = 180^\circ - 60^\circ$

$n = 120^\circ$

The interior angles of the pentagon are $90^\circ$, $90^\circ$, $m (110^\circ)$, $n (120^\circ)$, and let the interior angle adjacent to the exterior angle $x^\circ$ be $y^\circ$.

The sum of interior angles of a pentagon (a polygon with $n=5$ sides) is $(5-2) \times 180^\circ = 3 \times 180^\circ = 540^\circ$.

Sum of interior angles = $90^\circ + 90^\circ + m + n + y = 540^\circ$

$90^\circ + 90^\circ + 110^\circ + 120^\circ + y = 540^\circ$

$(90+90+110+120)^\circ + y = 540^\circ$

$410^\circ + y = 540^\circ$

Subtract $410^\circ$ from both sides:

$y = 540^\circ - 410^\circ$

$y = 130^\circ$

Finally, the exterior angle $x$ and the interior angle $y$ form a linear pair.

Substitute the value of $y$:

$x + 130^\circ = 180^\circ$

Subtract $130^\circ$ from both sides:

$x = 180^\circ - 130^\circ$

$x = 50^\circ$

Therefore, the value of $x$ in Figure (b) is $50^\circ$.

We know that the sum of the measures of the exterior angles of any convex polygon is $360^\circ$.

For a regular polygon, all exterior angles are equal in measure.

The measure of each exterior angle of a regular polygon with $n$ sides is given by the formula:

Measure of each exterior angle $= \frac{360^\circ}{\text{Number of sides}}$

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(i) Regular polygon of 9 sides:

Given:

Number of sides, $n = 9$.

The polygon is regular.

To Find:

Measure of each exterior angle.

Solution:

Using the formula:

Measure of each exterior angle $= \frac{360^\circ}{n}$

Substitute $n = 9$:

Measure of each exterior angle $= \frac{360^\circ}{9}$

Performing the division:

$\frac{360}{9} = 40^\circ$

So, the measure of each exterior angle is $40^\circ$.

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(ii) Regular polygon of 15 sides:

Given:

Number of sides, $n = 15$.

The polygon is regular.

To Find:

Measure of each exterior angle.

Solution:

Using the formula:

Measure of each exterior angle $= \frac{360^\circ}{n}$

Substitute $n = 15$:

Measure of each exterior angle $= \frac{360^\circ}{15}$

Performing the division:

$\frac{360}{15} = 24^\circ$

So, the measure of each exterior angle is $24^\circ$.

Given:

The polygon is regular.

Measure of each exterior angle $= 24^\circ$.

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To Find:

The number of sides of the regular polygon.

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Solution:

We know that the sum of the measures of the exterior angles of any convex polygon is $360^\circ$.

For a regular polygon, all exterior angles are equal in measure.

Let the number of sides of the regular polygon be $n$.

The measure of each exterior angle of a regular polygon with $n$ sides is given by the formula:

Measure of each exterior angle $= \frac{\text{Sum of exterior angles}}{\text{Number of sides}}$

Measure of each exterior angle $= \frac{360^\circ}{n}$

We are given that the measure of each exterior angle is $24^\circ$. So, we have the equation:

$24^\circ = \frac{360^\circ}{n}$

To find $n$, we can rearrange the equation. Multiply both sides by $n$:

Now, divide both sides by $24^\circ$:

Perform the division:

$n = \frac{360}{24}$

$n = 15$

$n = 15$

Therefore, the regular polygon has $15$ sides.

Solution:

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Let the number of sides of the regular polygon be $n$.

The measure of each interior angle is given as $165^\circ$.

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For any polygon, the sum of an interior angle and its corresponding exterior angle is $180^\circ$.

So, the measure of each exterior angle of the regular polygon is:

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For a regular polygon with $n$ sides, the measure of each exterior angle is also given by the formula:

We can now set up an equation using the calculated exterior angle:

To find the number of sides $n$, we rearrange the equation:

Now, we calculate the value of $n$:

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Alternatively, we can use the formula for the interior angle of a regular polygon with $n$ sides, which is given by:

We are given that the interior angle is $165^\circ$. So, we have:

Now, we solve this equation for $n$:

Subtract $180n$ from both sides:

Divide both sides by $-15$:

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Both methods give the same result.

Thus, the regular polygon has 24 sides.

Solution:

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(a) Is it possible to have a regular polygon with measure of each exterior angle as 22°?

For a regular polygon with $n$ sides, the measure of each exterior angle is given by the formula:

If the measure of each exterior angle is $22^\circ$, then we have:

Solving for $n$, we get:

For a polygon to exist, the number of sides $n$ must be a positive integer. Since $\frac{180}{11}$ is not an integer (it's approximately 16.36), it is not possible to have a regular polygon with an exterior angle of $22^\circ$.

Answer for (a): No, it is not possible.

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(b) Can it be an interior angle of a regular polygon? Why?

If $22^\circ$ were the measure of each interior angle of a regular polygon, then the measure of each exterior angle would be:

Now, using the formula relating the exterior angle and the number of sides $n$:

We would have:

Solving for $n$, we get:

Since $\frac{180}{79}$ is not an integer, it is not possible to have a regular polygon with an interior angle of $22^\circ$.

Reason:

The smallest possible interior angle for a regular polygon occurs for the regular triangle (equilateral triangle) where $n=3$. The interior angle is $\frac{(3-2) \times 180^\circ}{3} = \frac{1 \times 180^\circ}{3} = 60^\circ$.

As the number of sides ($n$) of a regular polygon increases, the interior angle increases and approaches $180^\circ$.

Since $22^\circ$ is less than the smallest possible interior angle ($60^\circ$) of any regular polygon, it cannot be an interior angle of a regular polygon.

Answer for (b): No, it cannot be an interior angle of a regular polygon. The minimum interior angle for a regular polygon is $60^\circ$ (for an equilateral triangle), and $22^\circ$ is less than $60^\circ$.

Solution:

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(a) What is the minimum interior angle possible for a regular polygon? Why?

The interior angle of a regular polygon with $n$ sides is given by the formula:

For a polygon to exist, the number of sides $n$ must be an integer greater than or equal to 3. The smallest possible number of sides for a polygon is $n=3$ (a triangle).

Let's find the interior angle for $n=3$:

Why: As the number of sides $n$ increases, the value of $\frac{n-2}{n} = 1 - \frac{2}{n}$ increases (since $\frac{2}{n}$ decreases). Therefore, the interior angle, which is proportional to $1 - \frac{2}{n}$, also increases. The minimum interior angle occurs at the minimum possible value of $n$, which is 3.

The minimum interior angle possible for a regular polygon is $60^\circ$.

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(b) What is the maximum exterior angle possible for a regular polygon?

The exterior angle of a regular polygon is related to its interior angle by the formula:

To maximize the exterior angle, we need to minimize the interior angle. From part (a), we found that the minimum interior angle is $60^\circ$, which occurs for a regular polygon with $n=3$ sides (an equilateral triangle).

Using the formula for the exterior angle for $n=3$:

Alternatively, using the relationship with the minimum interior angle:

The maximum exterior angle possible for a regular polygon is $120^\circ$.

Given:

Parallelogram PQRS.

From the figure (Fig 3.16) and the context of the problem, the adjacent side lengths are SR = $12$ cm and PS = $7$ cm.

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To Find:

The perimeter of parallelogram PQRS.

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Solution:

In a parallelogram, the opposite sides have the same length.

Therefore, in parallelogram PQRS:

The perimeter of a parallelogram is the sum of the lengths of its four sides.

Perimeter of parallelogram PQRS = PQ + QR + RS + SP

Substitute the side lengths:

Perimeter = $12$ cm + $7$ cm + $12$ cm + $7$ cm

We can group the equal sides:

Perimeter = $(12 + 12)$ cm + $(7 + 7)$ cm

Perimeter = $24$ cm + $14$ cm

Add the values:

Perimeter = $38$ cm

Alternatively, the perimeter of a parallelogram is also given by $2 \times (\text{sum of adjacent sides})$.

Perimeter = $2 \times (SR + PS)$

Perimeter = $2 \times (12 + 7)$ cm

Perimeter = $2 \times (19)$ cm

Perimeter = $38$ cm

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The perimeter of the parallelogram PQRS is $38$ cm.

Solution:

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Given:

BEST is a parallelogram.

$\angle$B = $100^\circ$

$\angle$E = $z$

Exterior Angle adjacent to $\angle$E = $y$

$\angle$S = $x$

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To Find:

The values of $x$, $y$, and $z$.

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Solution:

In a parallelogram, opposite angles are equal.

S is opposite to B.

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Exterior Angle adjacent to $\angle$B = $y$

Since BT || SE, using the property of alternate interior angles:

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Using the concept of linear pair between $\angle y$ and $\angle z$:

Substitute the value of $y$ found in the previous step:

$z + 100^\circ = 180^\circ$

Subtract $100^\circ$ from both sides:

$z = 180^\circ - 100^\circ$

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Hence, $x = 100^\circ$, $y = 100^\circ$, and $z = 80^\circ$.

Given:

Parallelogram RING.

$m\angle R = 70^\circ$.

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To Find:

$m\angle I$, $m\angle N$, and $m\angle G$.

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Solution:

In a parallelogram, opposite angles are equal.

Therefore, $\angle$R and $\angle$N are opposite angles.

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Adjacent angles in a parallelogram are supplementary (their sum is $180^\circ$).

Consider adjacent angles $\angle$R and $\angle$I:

Substitute the given value of $\angle$R:

Subtract $70^\circ$ from both sides:

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Now, consider opposite angles $\angle$G and $\angle$I:

Substitute the value of $\angle$I we found:

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Thus, the other angles are:

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The other angles of parallelogram RING are $m\angle I = 110^\circ$, $m\angle N = 70^\circ$, and $m\angle G = 110^\circ$.

Given:

HELP is a parallelogram.

O is the point of intersection of the diagonals HE and PL.

OE = 4 cm.

Length of diagonal HL is 5 more than the length of diagonal PE.

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To Find:

The length of OH.

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Solution:

In a parallelogram, the diagonals bisect each other. This means the point of intersection O divides each diagonal into two equal parts.

Given OE = $4$. According to the property of diagonals bisecting each other, O is the midpoint of the diagonal PE. Thus, OP = OE.

The length of the diagonal PE is the sum of OP and OE:

The length of the diagonal PE is $8$ cm.

We are given that HL is 5 more than PE.

Substitute the value of PE:

HL = $8 + 5 = 13$

The length of the diagonal HL is $13$ cm.

The question asks for the length of OH. OH is half of the diagonal HL.

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The value of OH is $6.5$ cm.

(i) AD = BC

Property: Opposite sides of a parallelogram are of equal length.

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(ii) $\angle$DCB = $\angle$DAB

Property: Opposite angles of a parallelogram are of equal measure.

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(iii) OC = OA

Property: Diagonals of a parallelogram bisect each other.

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(iv) m$\angle$DAB + m$\angle$CDA = $180^\circ$

Property: Adjacent angles in a parallelogram are supplementary.

Given:

Various parallelograms with unknown angles $x$, $y$, and $z$.

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To Find:

The values of $x$, $y$, and $z$ for each parallelogram.

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Solution:

We will use the properties of parallelograms: opposite angles are equal, adjacent angles are supplementary, and diagonals bisect each other. We will also use properties of angles formed by parallel lines and angle sum property of a triangle.

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Figure (i):

From the figure, the angle marked $y$ is opposite to the angle with measure $100^\circ$.

The angle marked $x$ and the angle with measure $100^\circ$ are adjacent angles.

Subtract $100^\circ$ from both sides:

$x = 180^\circ - 100^\circ$

$x = 80^\circ$

The angle marked $z$ is opposite to the angle marked $x$.

Since $x = 80^\circ$, we have:

Thus, for Figure (i), $x = 80^\circ$, $y = 100^\circ$, and $z = 80^\circ$.

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Figure (ii):

The angle with measure $50^\circ$ and the angle marked $x$ are adjacent angles.

Subtract $50^\circ$ from both sides:

$x = 180^\circ - 50^\circ$

$x = 130^\circ$

The angle marked $y$ is opposite to the angle marked $x$.

Since $x = 130^\circ$, we have:

The angle marked $z$ is a corresponding angle to the angle marked $x$ (considering the parallel sides and the transversal).

Since $x = 130^\circ$, we have:

Thus, for Figure (ii), $x = 130^\circ$, $y = 130^\circ$, and $z = 130^\circ$.

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Figure (iii):

The angle marked $x$ and the angle marked $90^\circ$ are vertically opposite angles.

Consider the triangle formed by the intersection of the diagonals. The angles of this triangle are $x$, $y$, and $30^\circ$.

The sum of angles in a triangle is $180^\circ$.

Substitute the value of $x = 90^\circ$:

$90^\circ + y + 30^\circ = 180^\circ$

$120^\circ + y = 180^\circ$

Subtract $120^\circ$ from both sides:

$y = 180^\circ - 120^\circ$

$y = 60^\circ$

The angle marked $y$ and the angle marked $z$ are alternate interior angles formed by the transversal and the parallel sides of the parallelogram.

Since $y = 60^\circ$, we have:

Thus, for Figure (iii), $x = 90^\circ$, $y = 60^\circ$, and $z = 60^\circ$.

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Figure (iv):

The angle marked $z$ is a corresponding angle to the angle with measure $80^\circ$ (considering the parallel sides and the transversal).

The angle marked $y$ and the angle marked $z$ are alternate interior angles formed by the transversal and the parallel sides.

Since $z = 80^\circ$, we have:

The angle marked $x$ and the angle marked $y$ are adjacent angles.

Substitute the value of $y = 80^\circ$:

$x + 80^\circ = 180^\circ$

Subtract $80^\circ$ from both sides:

$x = 180^\circ - 80^\circ$

$x = 100^\circ$

Thus, for Figure (iv), $x = 100^\circ$, $y = 80^\circ$, and $z = 80^\circ$.

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Figure (v):

In a parallelogram, opposite angles are equal.

The angle marked $y$ is opposite to the angle with measure $112^\circ$.

Consider the triangle formed by one diagonal and two sides. The angles of this triangle are $x$, $40^\circ$, and the angle with measure $112^\circ$.

The sum of angles in a triangle is $180^\circ$.

Combine the known angles:

$x + 152^\circ = 180^\circ$

Subtract $152^\circ$ from both sides:

$x = 180^\circ - 152^\circ$

$x = 28^\circ$

The angle marked $x$ and the angle marked $z$ form a pair of alternate interior angles (considering one of the parallel sides and the diagonal as a transversal).

Since $x = 28^\circ$, we have:

Thus, for Figure (v), $x = 28^\circ$, $y = 112^\circ$, and $z = 28^\circ$.

Let's analyse each condition for a quadrilateral ABCD to be a parallelogram.

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(i) $\angle$D + $\angle$B = $180^\circ$

In a parallelogram, opposite angles are equal. So, $\angle$B = $\angle$D.

If $\angle$D + $\angle$B = $180^\circ$ and $\angle$B = $\angle$D, then $2\angle$B = $180^\circ$, which implies $\angle$B = $90^\circ$. Similarly, $\angle$D = $90^\circ$.

If it were a parallelogram, the adjacent angles would be supplementary. So, $\angle$A + $\angle$B = $180^\circ$ and $\angle$B + $\angle$C = $180^\circ$. If $\angle$B = $90^\circ$, then $\angle$A = $180^\circ - 90^\circ = 90^\circ$ and $\angle$C = $180^\circ - 90^\circ = 90^\circ$.

In this case, all angles are $90^\circ$. A quadrilateral with all angles equal to $90^\circ$ is a rectangle, and a rectangle is a parallelogram.

Therefore, a quadrilateral ABCD can be a parallelogram if $\angle$D + $\angle$B = $180^\circ$ (for example, if it is a rectangle).

However, it is important to note that having one pair of opposite angles summing to $180^\circ$ does not guarantee that the quadrilateral is a parallelogram. For instance, in a cyclic quadrilateral, opposite angles sum to $180^\circ$, but a cyclic quadrilateral is not always a parallelogram.

So, the answer is Yes, it can be a parallelogram, but it is not always the case for any quadrilateral satisfying this condition.

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(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm

In a parallelogram, opposite sides must be equal in length.

We are given that one pair of opposite sides AB and DC are equal in length ($8$ cm).

We are given that the other pair of opposite sides AD and BC have lengths $4$ cm and $4.4$ cm respectively.

For ABCD to be a parallelogram, AD must be equal to BC.

Here, $AD = 4$ cm and $BC = 4.4$ cm.

Since $AD \neq BC$, the second pair of opposite sides are not equal.

Therefore, a quadrilateral with these side lengths cannot be a parallelogram.

The answer is No.

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(iii) $\angle$A = $70^\circ$ and $\angle$C = $65^\circ$

In a parallelogram, opposite angles must be equal.

Assuming $\angle$A and $\angle$C are opposite angles in the quadrilateral ABCD, for it to be a parallelogram, $\angle$A must be equal to $\angle$C.

We are given $\angle$A = $70^\circ$ and $\angle$C = $65^\circ$.

Since $\angle$A $\neq$ $\angle$C ($70^\circ \neq 65^\circ$), the opposite angles are not equal.

Therefore, a quadrilateral with these angle measures cannot be a parallelogram.

The answer is No.

Solution:

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We need to draw a quadrilateral that is not a parallelogram, but has exactly two opposite angles of equal measure.

A Kite is a quadrilateral that satisfies this property.

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Properties of a Kite:

• Two pairs of adjacent sides are equal in length.
• One pair of opposite angles is equal.
• The diagonals are perpendicular to each other.
• One diagonal bisects the other diagonal.
• One diagonal bisects the pair of opposite angles that are equal.

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Consider a kite ABCD where AB = AD and CB = CD. The angles between the unequal sides are equal.

In Kite ABCD, $\angle$ABC and $\angle$ADC are the angles formed between unequal pairs of adjacent sides (AB and BC, AD and DC, assuming AB $\neq$ BC). These opposite angles are equal.

The other pair of opposite angles, $\angle$BAD and $\angle$BCD, are generally not equal in a kite unless it is a rhombus (which is a parallelogram). Since we need a quadrilateral that is NOT a parallelogram, we consider a kite where adjacent sides are not all equal (i.e., AB $\neq$ BC).

In such a kite, exactly one pair of opposite angles is equal ($\angle$ABC = $\angle$ADC), and the other pair is unequal ($\angle$BAD $\neq$ $\angle$BCD).

This fits the conditions: it is a quadrilateral, it is not a parallelogram (because not both pairs of opposite angles are equal), and it has exactly two opposite angles of equal measure.

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Rough Figure:

(Imagine a figure of a kite with vertices labelled A, B, C, D in clockwise order. Sides AB and AD are equal, sides BC and CD are equal. The angles at B and D are equal, and the diagonal AC bisects the angle at A and the angle at C, and is perpendicular to the diagonal BD).

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Thus, a kite is an example of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.

Let the measures of the two adjacent angles of the parallelogram be $3x$ and $2x$, according to the given ratio.

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We know that adjacent angles in a parallelogram are supplementary. This means their sum is $180^\circ$.

So, we can write the equation:

$3x + 2x = 180^\circ$

$5x = 180^\circ$

Now, we solve for $x$:

$x = \frac{180^\circ}{5}$

$x = 36^\circ$

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Now, we can find the measure of each of the adjacent angles:

First adjacent angle = $3x = 3 \times 36^\circ = 108^\circ$

Second adjacent angle = $2x = 2 \times 36^\circ = 72^\circ$

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In a parallelogram, opposite angles are equal.

Therefore, the other two angles of the parallelogram will have measures equal to these adjacent angles.

The four angles of the parallelogram are $108^\circ$, $72^\circ$, $108^\circ$, and $72^\circ$.

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Let's verify the sum of adjacent angles: $108^\circ + 72^\circ = 180^\circ$. This is correct.

Let's verify the sum of all angles: $108^\circ + 72^\circ + 108^\circ + 72^\circ = 360^\circ$. This is correct for a quadrilateral.

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The measures of the angles of the parallelogram are $108^\circ$, $72^\circ$, $108^\circ$, and $72^\circ$.

Let the measure of each of the two equal adjacent angles of the parallelogram be $a$.

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We know that adjacent angles in a parallelogram are supplementary.

Therefore, the sum of the measures of these two adjacent angles is $180^\circ$.

So, we can write the equation:

$a + a = 180^\circ$

$2a = 180^\circ$

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Now, we solve for $a$:

$a = \frac{180^\circ}{2}$

$a = 90^\circ$

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So, the measure of each of the two adjacent angles is $90^\circ$.

In a parallelogram, opposite angles are equal.

Since one angle is $90^\circ$, its opposite angle is also $90^\circ$.

Since the adjacent angle is also $90^\circ$, its opposite angle is also $90^\circ$.

Thus, all four angles of the parallelogram measure $90^\circ$.

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The measure of each of the angles of the parallelogram is $90^\circ$.

Note that a parallelogram with all angles equal to $90^\circ$ is a rectangle.

Given:

HOPE is a parallelogram.

Angles are marked as $\angle PEH = x$, $\angle HPO = y$, $\angle OHP = z$, and $\angle PHE = 40^\circ$.

Exterior angle at vertex O is $70^\circ$.

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To Find:

The values of angle measures $x$, $y$, and $z$.

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Solution:

We will use the properties of parallelograms and angles formed by parallel lines and transversals.

The exterior angle at vertex O is $70^\circ$. The interior angle $\angle$HOP at vertex O and this exterior angle form a linear pair. Linear pairs are supplementary.

Subtract $70^\circ$ from both sides:

$\angle\text{HOP} = 180^\circ - 70^\circ$

$\angle\text{HOP} = 110^\circ$

In a parallelogram, opposite angles are equal.

The angle marked $x$ is the interior angle at vertex E ($\angle$PEH). This angle is opposite to the interior angle at vertex O ($\angle$HOP).

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In a parallelogram, opposite sides are parallel. So, HE is parallel to OP (HE || OP), and HO is parallel to PE (HO || PE).

Consider the parallel lines HE and OP, and the transversal HP.

The angle $\angle$HPO (marked as $y$) and the angle $\angle$PHE (marked as $40^\circ$) are alternate interior angles.

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Now, consider the triangle POH formed by the intersection of the diagonals.

The angles of triangle POH are $\angle$HPO (marked as $y$), $\angle$OHP (marked as $z$), and $\angle$POH (which is the same as $\angle$HOP = $110^\circ$).

The sum of angles in a triangle is $180^\circ$ (Angle Sum Property of a Triangle).

Substitute the known values ($y = 40^\circ$, $\angle$POH = $110^\circ$) and $z$:

Combine the known angles:

$150^\circ + z = 180^\circ$

Subtract $150^\circ$ from both sides:

$z = 180^\circ - 150^\circ$

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Thus, the values of the angle measures are:

$x = 110^\circ$

$y = 40^\circ$

$z = 30^\circ$

We will find the values of $x$ and $y$ for each parallelogram using the properties of parallelograms.

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Figure 1: Parallelogram GUNS

Given:

GUNS is a parallelogram with side lengths GU = $(3y - 1)$ cm, SN = $26$ cm, GS = $(3x)$ cm, and UN = $18$ cm.

To Find:

The values of $x$ and $y$.

Solution:

In a parallelogram, opposite sides are equal in length.

Applying this property to parallelogram GUNS:

Substitute the given expressions for the side lengths:

$3y - 1 = 26$

Add 1 to both sides of the equation:

$3y = 26 + 1$

$3y = 27$

Divide both sides by 3:

$y = \frac{27}{3}$

$y = 9$

Also,

Substitute the given expressions:

$3x = 18$

Divide both sides by 3:

$x = \frac{18}{3}$

$x = 6$

For parallelogram GUNS, $x = 6$ and $y = 9$.

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Figure 2: Parallelogram RUNS

Given:

RUNS is a parallelogram with diagonals intersecting at O. Segments of diagonals are RO = $20$ cm, ON = $(y + 7)$ cm, SO = $16$ cm, and OU = $(x + y)$ cm.

To Find:

The values of $x$ and $y$.

Solution:

In a parallelogram, the diagonals bisect each other. This means the point of intersection O divides each diagonal into two equal parts.

Applying this property to the diagonal RN in parallelogram RUNS:

Substitute the given expressions for the segments:

$20 = y + 7$

Subtract 7 from both sides:

$y = 20 - 7$

$y = 13$

Applying the property to the diagonal SU:

Substitute the given expressions:

$16 = x + y$

Substitute the value of $y = 13$ into this equation:

$16 = x + 13$

Subtract 13 from both sides:

$x = 16 - 13$

$x = 3$

For parallelogram RUNS, $x = 3$ and $y = 13$.

Given:

RISK is a parallelogram.

CLUE is a parallelogram.

$\angle$K = $120^\circ$ in parallelogram RISK.

$\angle$L = $70^\circ$ in parallelogram CLUE.

x is the angle formed at the intersection of lines SI and CE.

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To Find:

The value of x.

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Solution:

Consider the parallelogram RISK.

In a parallelogram, adjacent angles are supplementary.

So, $\angle$S + $\angle$K = $180^\circ$.

Given $\angle$K = $120^\circ$.

$\angle$S = $60^\circ$.

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Consider the parallelogram CLUE.

In a parallelogram, opposite angles are equal.

So, $\angle$E = $\angle$L.

Given $\angle$L = $70^\circ$.

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Let the intersection point of lines SI and CE be O.

The triangle formed by the intersection of these lines is $\triangle$SOE.

The angles of $\triangle$SOE are $\angle$OSE, $\angle$SEO, and $\angle$SOE.

The angle $\angle$OSE is part of $\angle$S from parallelogram RISK.

$\angle$OSE = $\angle$S = $60^\circ$.

The angle $\angle$SEO is part of $\angle$E from parallelogram CLUE.

$\angle$SEO = $\angle$E = $70^\circ$.

The angle $\angle$SOE and the angle marked as x are vertically opposite angles.

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In $\triangle$SOE, the sum of interior angles is $180^\circ$.

Substitute the values we found:

$60^\circ + 70^\circ + x = 180^\circ$

$130^\circ + x = 180^\circ$

Subtract $130^\circ$ from both sides:

$x = 180^\circ - 130^\circ$

$x = 50^\circ$

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The value of x is $50^\circ$.

Given:

A quadrilateral KLMN with $\angle$L = $80^\circ$ and $\angle$M = $100^\circ$.

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To Explain and Find:

How the figure KLMN is a trapezium, and which of its two sides are parallel.

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Solution:

A trapezium is a quadrilateral that has at least one pair of parallel sides.

We need to check if any pair of opposite sides in quadrilateral KLMN are parallel.

Consider the sides KL and NM. The line segment LM is a transversal intersecting these two sides.

The angles $\angle$L and $\angle$M are interior angles on the same side of the transversal LM.

Let's find the sum of these two angles:

Sum of angles = $\angle$L + $\angle$M

Substitute the given values:

Sum of angles = $80^\circ$ + $100^\circ$

Sum of angles = $180^\circ$

We know that if a transversal intersects two lines such that the sum of the interior angles on the same side of the transversal is $180^\circ$, then the two lines are parallel.

In this case, the transversal LM intersects lines KL and NM, and the sum of consecutive interior angles $\angle$L and $\angle$M is $180^\circ$.

Therefore,

Since the quadrilateral KLMN has a pair of parallel sides (KL and NM), it is a trapezium.

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Conclusion:

The figure KLMN is a trapezium because the sum of consecutive interior angles $\angle$L and $\angle$M is $180^\circ$, which implies that the sides KL and NM are parallel.

The two parallel sides are KL and NM.

Given:

A quadrilateral ABCD.

Side $\overline{AB}$ is parallel to side $\overline{DC}$ ($\overline{AB}$ || $\overline{DC}$).

$\text{m}\angle$B = $120^\circ$.

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To Find:

The measure of angle C ($\text{m}\angle$C).

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Solution:

The given figure is a quadrilateral where $\overline{AB}$ || $\overline{DC}$. This means the quadrilateral is a trapezium.

When two parallel lines are intersected by a transversal, the interior angles on the same side of the transversal are supplementary (their sum is $180^\circ$).

In the figure, AB and DC are parallel lines, and BC is a transversal line intersecting them.

The angles $\angle$B and $\angle$C are interior angles on the same side of the transversal BC.

Therefore, the sum of $\text{m}\angle$B and $\text{m}\angle$C must be $180^\circ$.

Substitute the given value of $\text{m}\angle$B = $120^\circ$:

Subtract $120^\circ$ from both sides to find $\text{m}\angle$C:

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The measure of angle C is $60^\circ$.

Given:

Quadrilateral PQRS.

Side $\overline{SP}$ is parallel to side $\overline{RQ}$ ($\overline{SP}$ || $\overline{RQ}$).

$\text{m}\angle$Q = $130^\circ$.

$\text{m}\angle$R = $90^\circ$. (Indicated by the square symbol)

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To Find:

The measure of $\angle$P and $\angle$S.

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Solution:

Since $\overline{SP}$ || $\overline{RQ}$, PQ and SR are transversals. The quadrilateral PQRS is a trapezium.

When two parallel lines are intersected by a transversal, the interior angles on the same side of the transversal are supplementary (their sum is $180^\circ$).

Consider the parallel lines SP and RQ, and the transversal PQ.

The angles $\angle$P and $\angle$Q are interior angles on the same side of the transversal PQ.

Therefore, the sum of $\text{m}\angle$P and $\text{m}\angle$Q must be $180^\circ$.

Substitute the given value of $\text{m}\angle$Q = $130^\circ$:

$\text{m}\angle\text{P} + 130^\circ = 180^\circ$

Subtract $130^\circ$ from both sides:

$\text{m}\angle\text{P} = 180^\circ - 130^\circ$

Consider the parallel lines SP and RQ, and the transversal SR.

The angles $\angle$S and $\angle$R are interior angles on the same side of the transversal SR.

Therefore, the sum of $\text{m}\angle$S and $\text{m}\angle$R must be $180^\circ$.

Substitute the given value of $\text{m}\angle$R = $90^\circ$:

$\text{m}\angle\text{S} + 90^\circ = 180^\circ$

Subtract $90^\circ$ from both sides:

$\text{m}\angle\text{S} = 180^\circ - 90^\circ$

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Second Part of the Question:

If you find $\text{m}\angle$R, is there more than one method to find $\text{m}\angle$P?

Yes, there is more than one method to find $\text{m}\angle$P once we know the measures of $\angle$Q, $\angle$R, and $\angle$S.

Method 1 (already used): Using the property of interior angles on the same side of the transversal PQ being supplementary, since SP || RQ:

$\text{m}\angle\text{P} + \text{m}\angle\text{Q} = 180^\circ$

$\text{m}\angle\text{P} + 130^\circ = 180^\circ$

$\text{m}\angle\text{P} = 50^\circ$

Method 2: Using the angle sum property of a quadrilateral.

The sum of interior angles of a quadrilateral is $360^\circ$.

Substitute the known values $\text{m}\angle$Q = $130^\circ$, $\text{m}\angle$R = $90^\circ$, and $\text{m}\angle$S = $90^\circ$:

$\text{m}\angle\text{P} + 130^\circ + 90^\circ + 90^\circ = 360^\circ$

$\text{m}\angle\text{P} + 310^\circ = 360^\circ$

Subtract $310^\circ$ from both sides:

$\text{m}\angle\text{P} = 360^\circ - 310^\circ$

$\text{m}\angle\text{P} = 50^\circ$

Both methods give the same result for $\text{m}\angle$P.

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The measure of $\angle$P is $50^\circ$ and the measure of $\angle$S is $90^\circ$. Yes, there is more than one method to find $\text{m}\angle$P if $\text{m}\angle$R is known.

Given:

RICE is a rhombus.

The diagonals RC and IE intersect at O.

Lengths of segments are labelled as: RO = y, OC = 12, IO = 5, OE = x.

Side length RE = 13.

Side length RI = z.

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To Find:

The values of $x$, $y$, and $z$.

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Solution:

We will use the properties of a rhombus to find the unknown values.

Properties of a Rhombus:

• All sides are equal in length.
• Opposite angles are equal.
• Adjacent angles are supplementary.
• The diagonals bisect each other at right angles.
• The diagonals bisect the opposite angles.

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In a rhombus, the diagonals bisect each other.

This means that the point of intersection O divides each diagonal into two equal parts.

Consider the diagonal IE. O is the midpoint of IE.

From the figure, IO = 5 and OE = x.

Consider the diagonal RC. O is the midpoint of RC.

From the figure, RO = y and OC = 12.

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In a rhombus, all sides are equal in length.

From the figure, we are given that RE = 13.

Therefore, all sides of the rhombus RICE have a length of 13.

From the figure, RI = z and RE = 13.

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Justification of findings:

• $x=5$ because the diagonals of a rhombus bisect each other, so OE = IO = 5.
• $y=12$ because the diagonals of a rhombus bisect each other, so RO = OC = 12.
• $z=13$ because all sides of a rhombus are equal in length, and one side RE is given as 13.

Given:

RENT is a rectangle.

Diagonals RN and TE intersect at O.

OR = 2x + 4.

OT = 3x + 1.

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To Find:

The value of x.

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Solution:

We know that in a rectangle, the diagonals are equal in length and bisect each other.

Since the diagonals bisect each other at O, O is the midpoint of both diagonals RN and TE. This means RO = ON and TO = OE.

Also, since the diagonals are equal in length (RN = TE), their halves are also equal.

Therefore, OR = ON = OT = OE.

From the given information, we have expressions for OR and OT. Since OR = OT, we can set up an equation:

Substitute the given expressions:

$2x + 4 = 3x + 1$

Now, we solve this linear equation for $x$.

Subtract $2x$ from both sides of the equation:

$4 = 3x - 2x + 1$

$4 = x + 1$

Subtract 1 from both sides of the equation:

$4 - 1 = x$

$3 = x$

So, the value of $x$ is 3.

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The value of $x$ is $3$.

Solution:

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Let's examine each statement:

(a) All rectangles are squares

False. A square is a special type of rectangle where all sides are equal. However, a rectangle only requires opposite sides to be equal and all angles to be $90^\circ$. A rectangle does not necessarily have all sides equal.

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(b) All rhombuses are parallelograms

True. A rhombus is defined as a quadrilateral with all four sides of equal length. This property automatically implies that opposite sides are parallel (a property of a parallelogram). Therefore, every rhombus is also a parallelogram.

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(c) All squares are rhombuses and also rectangles

True. A square has all sides equal, which is the defining property of a rhombus. A square also has all interior angles equal to $90^\circ$ (right angles), which is the defining property of a rectangle. Thus, a square is both a rhombus and a rectangle.

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(d) All squares are not parallelograms.

False. A square has both pairs of opposite sides parallel and equal, which is the definition of a parallelogram. So, every square is indeed a parallelogram.

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(e) All kites are rhombuses.

False. A rhombus is a special type of kite where all four sides are equal. A general kite only requires two distinct pairs of equal adjacent sides. A kite does not necessarily have all sides equal.

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(f) All rhombuses are kites.

True. A rhombus has all four sides equal. This means it has two pairs of equal adjacent sides (since all sides are equal, any two adjacent sides form a pair of equal adjacent sides), which satisfies the definition of a kite.

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(g) All parallelograms are trapeziums.

True. A parallelogram is a quadrilateral with two pairs of parallel sides. A trapezium is defined as a quadrilateral with at least one pair of parallel sides. Since a parallelogram has two pairs of parallel sides, it certainly has at least one pair of parallel sides. Therefore, every parallelogram is a trapezium.

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(h) All squares are trapeziums.

True. From part (d), all squares are parallelograms. From part (g), all parallelograms are trapeziums. Therefore, all squares are also trapeziums.

Solution:

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Let's identify the quadrilaterals based on the given properties:

(a) quadrilaterals that have four sides of equal length

A quadrilateral with all four sides of equal length is called a Rhombus.

A square is a special type of rhombus where all angles are right angles. So, a Square also has four sides of equal length.

The quadrilaterals with four sides of equal length are: Rhombus and Square.

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(b) quadrilaterals that have four right angles

A quadrilateral with all four interior angles equal to $90^\circ$ is called a Rectangle.

A square is a special type of rectangle where all sides are equal. So, a Square also has four right angles.

The quadrilaterals with four right angles are: Rectangle and Square.

Solution:

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Let's explain how a square fits the definition of each type of quadrilateral:

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(i) a quadrilateral

A quadrilateral is any closed figure formed by four line segments (sides). A square has four sides and four vertices, making it a four-sided polygon.

Thus, a square is a quadrilateral because it is a closed figure with four sides.

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(ii) a parallelogram

A parallelogram is a quadrilateral where both pairs of opposite sides are parallel. In a square, all adjacent sides are perpendicular (form $90^\circ$ angles), which means opposite sides are parallel to each other. For example, if side AB is perpendicular to BC and DC is perpendicular to BC, then AB is parallel to DC. Similarly, AD is parallel to BC.

Thus, a square is a parallelogram because its opposite sides are parallel.

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(iii) a rhombus

A rhombus is a quadrilateral where all four sides are equal in length. By definition, a square has all four sides equal in length.

Thus, a square is a rhombus because all its sides are equal.

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(iv) a rectangle

A rectangle is a quadrilateral where all four interior angles are right angles ($90^\circ$). By definition, a square has all four interior angles equal to $90^\circ$. A rectangle is also a parallelogram, and a square is a parallelogram with right angles.

Thus, a square is a rectangle because all its interior angles are $90^\circ$.

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In conclusion, a square possesses the defining properties of a quadrilateral, a parallelogram, a rhombus, and a rectangle, making it a special case of each of these figures.

Solution:

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Let's name the quadrilaterals whose diagonals satisfy the given conditions:

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(i) diagonals bisect each other

When the diagonals of a quadrilateral cut each other into two equal halves at their point of intersection, we say they bisect each other. This property holds true for all parallelograms.

The quadrilaterals whose diagonals bisect each other are:

Parallelogram

Rhombus (since it's a parallelogram)

Rectangle (since it's a parallelogram)

Square (since it's a parallelogram)

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(ii) diagonals are perpendicular bisectors of each other

This means the diagonals not only bisect each other but also intersect at a right angle ($90^\circ$). This property is specific to rhombuses and squares.

The quadrilaterals whose diagonals are perpendicular bisectors of each other are:

Rhombus

Square (since it's a rhombus with right angles)

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(iii) diagonals are equal

When the lengths of the two diagonals are the same, the quadrilaterals are rectangles and squares. In some special trapeziums (isosceles trapeziums), the diagonals are also equal.

The quadrilaterals whose diagonals are equal are:

Rectangle

Square (since it's a rectangle)

Isosceles Trapezium

Solution:

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A convex quadrilateral is defined as a quadrilateral where all its interior angles are less than $180^\circ$.

In a rectangle, all four interior angles are right angles. The measure of a right angle is $90^\circ$.

Since $90^\circ < 180^\circ$, all interior angles of a rectangle are less than $180^\circ$.

Another way to understand a convex quadrilateral is that if you connect any two points inside the quadrilateral with a straight line segment, the entire segment lies inside the quadrilateral.

In a rectangle, no interior angle "points inwards". All vertices "point outwards". Any line segment joining two points within a rectangle will remain within its boundary.

Therefore, because all its interior angles are $90^\circ$ (which is less than $180^\circ$), a rectangle is a convex quadrilateral.

Solution:

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Given:

ABC is a right-angled triangle with the right angle at vertex B ($\angle$B = $90^\circ$).

O is the midpoint of the hypotenuse AC.

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To Explain:

Why O is equidistant from vertices A, B, and C.

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Solution:

To explain this, we can use the additional dotted lines shown in the figure. These lines are drawn to form a quadrilateral ABCD by extending BO to D such that BO = OD and joining AD and CD.

Since O is the midpoint of AC (given) and O is the midpoint of BD (by construction), the diagonals AC and BD of the quadrilateral ABCD bisect each other at O.

A quadrilateral whose diagonals bisect each other is a parallelogram.

Therefore, ABCD is a parallelogram.

We are given that $\angle$B = $90^\circ$.

A parallelogram with one interior angle equal to $90^\circ$ is a rectangle.

Therefore, ABCD is a rectangle.

In a rectangle, the diagonals are equal in length and they bisect each other.

The diagonals of rectangle ABCD are AC and BD.

Since the diagonals bisect each other at O, O is the midpoint of both diagonals.

Since AC = BD, it follows that their halves are also equal:

Substituting the segments, we get:

This shows that the distance from O to A, O to B, and O to C are all equal (AO = BO = OC).

Therefore, O is equidistant from A, B, and C.

This is a general property that the midpoint of the hypotenuse of a right-angled triangle is always equidistant from all three vertices of the triangle.