Chapter 6 The Triangle and its Properties

In the given figure of $\triangle$ PQR:

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$\overline{PM}$ is the segment from vertex P to the side QR, such that $\overline{PM}$ is perpendicular to $\overline{QR}$. A segment from a vertex perpendicular to the opposite side is called an altitude.

Therefore, $\overline{PM}$ is an altitude.

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D is given as the mid-point of $\overline{QR}$. $\overline{PD}$ is the segment connecting the vertex P to the mid-point D of the opposite side $\overline{QR}$. A segment connecting a vertex to the midpoint of the opposite side is called a median.

Therefore, $\overline{PD}$ is a median.

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The question asks "Is QM = MR?".

D is the mid-point of $\overline{QR}$, which means QD = DR. This implies that the point D divides the segment QR into two equal halves.

M is the foot of the altitude from P to QR. M is the point where the perpendicular from P meets QR.

In a general triangle, the foot of the altitude (M) and the midpoint of the base (D) are different points.

QM would be equal to MR only if M and D coincide. This happens in special types of triangles:

• In an isosceles triangle where PQ = PR.
• In an equilateral triangle.

Since $\triangle$ PQR is a general triangle and not specified as isosceles or equilateral, M and D are generally different points on QR.

Therefore, QM is generally not equal to MR.

The answer to "Is QM = MR?" is No (in a general triangle).

Solution:

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Here are the descriptions and required image placeholders for rough sketches of the given scenarios:

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(a) In $\Delta$ABC, BE is a median.

Sketch description:

Draw any triangle and label its vertices as A, B, and C.

A median connects a vertex to the midpoint of the opposite side.

BE is the median from vertex B, so it connects B to the midpoint of the side opposite to B, which is side AC.

Find the middle point of the side AC. Mark this point as E.

Draw a line segment from vertex B to point E.

This segment BE is the median of $\Delta$ABC.

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(b) In $\Delta$PQR, PQ and PR are altitudes of the triangle.

Sketch description:

An altitude is a line segment from a vertex perpendicular to the opposite side (or its extension).

If PQ is an altitude, it means PQ is perpendicular to the side opposite to Q, which is PR (or its extension). If PR is an altitude, it means PR is perpendicular to the side opposite to R, which is PQ (or its extension).

This situation (where two sides are also altitudes) happens only in a right-angled triangle, and the right angle is at the common vertex P.

Draw two line segments perpendicular to each other at a point P. Label one line segment as PQ and the other as PR.

Connect points Q and R with a line segment.

The figure formed is $\Delta$PQR, which is a right-angled triangle at P. In this triangle, PQ is the altitude from Q to PR, and PR is the altitude from R to PQ.

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(c) In $\Delta$XYZ, YL is an altitude in the exterior of the triangle.

Sketch description:

An altitude lies in the exterior of a triangle only when the triangle is obtuse.

YL is the altitude from vertex Y to the side XZ (or its extension).

Draw an obtuse triangle and label its vertices as X, Y, and Z. Make sure one angle is obtuse (e.g., angle at Z).

Extend the side XZ beyond the obtuse angle vertex Z using a dashed line.

From vertex Y, draw a line segment that is perpendicular to the extended line XZ.

The point where this perpendicular line segment meets the extended line XZ is L.

The segment YL is the altitude from Y, and it is outside the triangle $\Delta$XYZ.

Given:

An isosceles triangle.

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To Verify:

If the median and altitude of an isosceles triangle can be the same by drawing a diagram.

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Solution:

Let us consider an isosceles triangle, say $\Delta$ABC, where sides AB and AC are equal in length (AB = AC).

We need to verify if the median and altitude from the vertex A to the opposite side BC can be the same.

Step 1: Draw the triangle.

Draw an isosceles triangle ABC with AB = AC.

Step 2: Draw the median.

A median is a line segment from a vertex to the midpoint of the opposite side.

To draw the median from vertex A to side BC, find the midpoint of BC. Let this midpoint be D.

Draw the line segment AD. AD is the median of $\Delta$ABC from vertex A.

Step 3: Draw the altitude.

An altitude is a perpendicular line segment from a vertex to the opposite side.

To draw the altitude from vertex A to side BC, draw a line segment from A perpendicular to BC. Let the point where this perpendicular intersects BC be E.

Draw the line segment AE such that $\angle$AEB = $\angle$AEC = $90^\circ$. AE is the altitude of $\Delta$ABC from vertex A.

Step 4: Observe the diagram.

Upon drawing the isosceles triangle and constructing both the median AD and the altitude AE from the vertex A to the base BC, you will observe that point D and point E coincide.

This means the line segment AD (the median) and the line segment AE (the altitude) are the same line segment.

Verification:

In an isosceles triangle, the median drawn from the vertex where the two equal sides meet (the vertex angle) has special properties when drawn to the base (the side opposite the vertex angle).

This line segment:

• Connects the vertex to the midpoint of the opposite side (it's a median).
• Is perpendicular to the opposite side (it's an altitude).
• Divides the vertex angle into two equal parts (it's an angle bisector).
• Divides the opposite side into two equal parts and is perpendicular to it (it's a perpendicular bisector).

Because it serves as both the median and the altitude from the vertex angle to the base, they are the same line segment in this specific case.

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Conclusion:

Yes, the median and altitude of an isosceles triangle can be the same. This happens when both are drawn from the vertex between the two equal sides to the unequal side (the base).

Solution:

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We use the property of the Exterior Angle of a Triangle.

The Exterior Angle Property states that the measure of an exterior angle of a triangle is equal to the sum of the measures of its two opposite interior angles.

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Based on the information provided in the correction:

We are given an equation based on this property:

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To find the value of $\angle$x, we need to isolate $\angle$x in the equation.

Subtract $50^\circ$ from both sides of the equation:

$\angle x = 110^\circ - 50^\circ$

$\angle x = 60^\circ$

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Answer:

The value of x is $60^\circ$.

We will use the property that the exterior angle of a triangle is equal to the sum of the two opposite interior angles.

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Figure 1:

Given: Interior opposite angles are $50^\circ$ and $70^\circ$. Exterior angle is $\angle$x.

To Find: The value of $\angle$x.

Solution:

$\angle$x = $120^\circ$

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Figure 2:

Given: Interior opposite angles are $65^\circ$ and $45^\circ$. Exterior angle is $\angle$x.

To Find: The value of $\angle$x.

Solution:

$\angle$x = $110^\circ$

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Figure 3:

Given: Interior opposite angles are $30^\circ$ and $40^\circ$. Exterior angle is $\angle$x.

To Find: The value of $\angle$x.

Solution:

$\angle$x = $70^\circ$

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Figure 4:

Given: Interior opposite angles are $40^\circ$ and $60^\circ$. Exterior angle is $\angle$x.

To Find: The value of $\angle$x.

Solution:

$\angle$x = $100^\circ$

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Figure 5:

Given: Interior opposite angles are $50^\circ$ and $50^\circ$. Exterior angle is $\angle$x.

To Find: The value of $\angle$x.

Solution:

$\angle$x = $100^\circ$

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Figure 6:

Given: Interior opposite angles are $30^\circ$ and $60^\circ$. Exterior angle is $\angle$x.

To Find: The value of $\angle$x.

Solution:

$\angle$x = $90^\circ$

We will use the property that the exterior angle of a triangle is equal to the sum of the two opposite interior angles.

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Figure 1:

Given: Exterior angle = $115^\circ$. One interior opposite angle = $50^\circ$. Unknown interior opposite angle = $\angle$x.

To Find: The value of $\angle$x.

Solution:

Subtract $50^\circ$ from both sides:

$\angle$x = $115^\circ - 50^\circ$

$\angle$x = $65^\circ$

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Figure 2:

Given: Exterior angle = $100^\circ$. One interior opposite angle = $70^\circ$. Unknown interior opposite angle = $\angle$x.

To Find: The value of $\angle$x.

Solution:

Subtract $70^\circ$ from both sides:

$\angle$x = $100^\circ - 70^\circ$

$\angle$x = $30^\circ$

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Figure 3:

Given: Exterior angle = $125^\circ$. One interior opposite angle = $90^\circ$ (right angle). Unknown interior opposite angle = $\angle$x.

To Find: The value of $\angle$x.

Solution:

Subtract $90^\circ$ from both sides:

$\angle$x = $125^\circ - 90^\circ$

$\angle$x = $35^\circ$

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Figure 4:

Given: Exterior angle = $120^\circ$. One interior opposite angle = $60^\circ$. Unknown interior opposite angle = $\angle$x.

To Find: The value of $\angle$x.

Solution:

Subtract $60^\circ$ from both sides:

$\angle$x = $120^\circ - 60^\circ$

$\angle$x = $60^\circ$

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Figure 5:

Given: Exterior angle = $80^\circ$. One interior opposite angle = $30^\circ$. Unknown interior opposite angle = $\angle$x.

To Find: The value of $\angle$x.

Solution:

Subtract $30^\circ$ from both sides:

$\angle$x = $80^\circ - 30^\circ$

$\angle$x = $50^\circ$

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Figure 6:

Given: Exterior angle = $75^\circ$. One interior opposite angle = $35^\circ$. Unknown interior opposite angle = $\angle$x.

To Find: The value of $\angle$x.

Solution:

Subtract $35^\circ$ from both sides:

$\angle$x = $75^\circ - 35^\circ$

$\angle$x = $40^\circ$

Solution:

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In $\Delta$ PQR shown in the figure, we are given the measures of two angles: $47^\circ$ and $52^\circ$. We need to find the measure of $\angle$P (m$\angle$P).

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We can find the measure of the unknown angle $\angle$P using the Angle Sum Property of a Triangle.

This property states that the sum of the interior angles in any triangle is always $180^\circ$.

By the angle sum property of a triangle, we have:

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To find m$\angle$P, we first add the known angles:

47$^\circ$ + 52$^\circ$ = 99$^\circ$

Now, substitute this sum back into the equation:

m$\angle$P + 99$^\circ$ = 180$^\circ$

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To isolate m$\angle$P, subtract 99$^\circ$ from both sides of the equation:

m$\angle$P = 180$^\circ$ – 99$^\circ$

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Perform the subtraction:

m$\angle$P = 81$^\circ$

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Answer:

The measure of $\angle$P is $81^\circ$.

Solution:

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We know that the sum of all the interior angles of a triangle is $180^\circ$. We will use this property to find the value of the unknown angle x in each figure.

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(i) Figure 1:

In this triangle, the interior angles are x, $50^\circ$, and $60^\circ$.

By the Angle Sum Property of a triangle:

Combine the known angles:

$\angle x + 110^\circ = 180^\circ$

By transposing $110^\circ$ from the left side (LHS) to the right side (RHS), it becomes $-110^\circ$.

$\angle x = 180^\circ - 110^\circ$

$\angle x = 70^\circ$

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(ii) Figure 2:

This is a right-angled triangle. The angles in the triangle are x, $90^\circ$ (indicated by the right angle symbol), and $30^\circ$.

By the Angle Sum Property of a triangle:

Combine the known angles:

$\angle x + 120^\circ = 180^\circ$

By transposing $120^\circ$ from LHS to RHS, it becomes $-120^\circ$.

$\angle x = 180^\circ - 120^\circ$

$\angle x = 60^\circ$

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(iii) Figure 3:

In this triangle, the interior angles are x, $110^\circ$, and $30^\circ$.

By the Angle Sum Property of a triangle:

Combine the known angles:

$\angle x + 140^\circ = 180^\circ$

By transposing $140^\circ$ from LHS to RHS, it becomes $-140^\circ$.

$\angle x = 180^\circ - 140^\circ$

$\angle x = 40^\circ$

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(iv) Figure 4:

In this triangle, the interior angles are $50^\circ$, $\angle$x, and $\angle$x. This is an isosceles triangle as two angles are equal.

By the Angle Sum Property of a triangle:

$50^\circ + 2\angle x = 180^\circ$

By transposing $50^\circ$ from LHS to RHS, it becomes $-50^\circ$.

$2\angle x = 180^\circ - 50^\circ$

$2\angle x = 130^\circ$

Divide both sides by 2:

$\angle x = \frac{130^\circ}{2}$

$\angle x = 65^\circ$

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(v) Figure 5:

In this triangle, all three interior angles are $\angle$x, $\angle$x, and $\angle$x. This is an equiangular triangle.

By the Angle Sum Property of a triangle:

$3\angle x = 180^\circ$

Divide both sides by 3:

$\angle x = \frac{180^\circ}{3}$

$\angle x = 60^\circ$

Therefore, the given triangle is an equiangular triangle, and each angle measures $60^\circ$.

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(vi) Figure 6:

In this triangle, the interior angles are $90^\circ$ (indicated by the right angle symbol), $2\angle$x, and $\angle$x.

By the Angle Sum Property of a triangle:

Combine the terms with $\angle$x:

$90^\circ + 3\angle x = 180^\circ$

By transposing $90^\circ$ from LHS to RHS, it becomes $-90^\circ$.

$3\angle x = 180^\circ - 90^\circ$

$3\angle x = 90^\circ$

Divide both sides by 3:

$\angle x = \frac{90^\circ}{3}$

$\angle x = 30^\circ$

Then, the other angle is $2\angle x = 2 \times 30^\circ = 60^\circ$.

The angles of the triangle are $90^\circ$, $60^\circ$, and $30^\circ$.

Solution:

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We will use the properties of angles in a triangle and angles formed on a straight line to find the values of the unknown angles x and y in each figure.

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(i) Figure 1:

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

In this figure, the exterior angle is $120^\circ$, and the two interior opposite angles are $50^\circ$ and $\angle$x.

By transposing $50^\circ$ from the left side (LHS) to the right side (RHS), it becomes $-50^\circ$.

$\angle$x = 120$^\circ$ – 50$^\circ$

$\angle$x = 70$^\circ$

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We also know that the sum of all the interior angles of a triangle is $180^\circ$.

The interior angles are $50^\circ$, $\angle$x, and $\angle$y.

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

Combine the known angles:

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

By transposing $120^\circ$ from LHS to RHS, it becomes $-120^\circ$.

$\angle$y = 180$^\circ$ – 120$^\circ$

$\angle$y = 60$^\circ$

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So, $\angle$x = $70^\circ$ and $\angle$y = $60^\circ$.

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(ii) Figure 2:

From the rule of vertically opposite angles:

The angle $\angle$y and the $80^\circ$ angle are vertically opposite angles.

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We know that the sum of all the interior angles of a triangle is $180^\circ$.

The interior angles are $50^\circ$, $\angle$y, and $\angle$x.

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

Combine the known angles:

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

By transposing $130^\circ$ from LHS to RHS, it becomes $-130^\circ$.

$\angle$x = 180$^\circ$ – 130$^\circ$

$\angle$x = 50$^\circ$

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So, $\angle$x = $50^\circ$ and $\angle$y = $80^\circ$.

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(iii) Figure 3:

We know that the sum of all the interior angles of a triangle is $180^\circ$.

The interior angles are $50^\circ$, $60^\circ$, and $\angle$y.

Combine the known angles:

110$^\circ$ + $\angle$y = 180$^\circ$

By transposing $110^\circ$ from LHS to RHS, it becomes $-110^\circ$.

$\angle$y = 180$^\circ$ – 110$^\circ$

$\angle$y = 70$^\circ$

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Now, from the rule of linear pair:

The angle $\angle$x and angle $\angle$y are adjacent angles on a straight line, so they form a linear pair and their sum is $180^\circ$.

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

By transposing $70^\circ$ from LHS to RHS, it becomes $-70^\circ$.

$\angle$x = 180$^\circ$ – 70$^\circ$

$\angle$x = 110$^\circ$

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So, $\angle$x = $110^\circ$ and $\angle$y = $70^\circ$.

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(iv) Figure 4:

From the rule of vertically opposite angles:

The angle $\angle$x and the $60^\circ$ angle are vertically opposite angles.

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We know that the sum of all the interior angles of a triangle is $180^\circ$.

The interior angles are $30^\circ$, $\angle$x, and $\angle$y.

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

Combine the known angles:

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

By transposing $90^\circ$ from LHS to RHS, it becomes $-90^\circ$.

$\angle$y = 180$^\circ$ – 90$^\circ$

$\angle$y = 90$^\circ$

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So, $\angle$x = $60^\circ$ and $\angle$y = $90^\circ$.

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(v) Figure 5:

From the rule of vertically opposite angles:

The angle $\angle$y and the $90^\circ$ angle (indicated by the symbol) are vertically opposite angles.

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We know that the sum of all the interior angles of a triangle is $180^\circ$.

The interior angles are $\angle$x, $\angle$x, and $\angle$y.

Combine the terms with $\angle$x and substitute the value of $\angle$y:

By transposing $90^\circ$ from LHS to RHS, it becomes $-90^\circ$.

2$\angle$x = 180$^\circ$ – 90$^\circ$

2$\angle$x = 90$^\circ$

Divide both sides by 2:

$\angle$x = $\frac{90^\circ}{2}$

$\angle$x = 45$^\circ$

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So, $\angle$x = $45^\circ$ and $\angle$y = $90^\circ$.

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(vi) Figure 6:

In this figure, one interior angle of the triangle is labelled $\angle$y.

Since, all the interior angles of the triangle are vertically opposite angles of x. So we can conclude that all the angles in the provided triangle are equal to $\angle$ y.

Now, We will use the Angle Sum Property of a triangle.

Combine the terms:

3$\angle$y = 180$^\circ$

Divide both sides by 3:

$\angle$y = $\frac{180^\circ}{3}$

$\angle$y = 60$^\circ$

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From the statement $\angle$x = $\angle$y (Vertically Opposite Angles), we substitute the value of $\angle$y:

$\angle$x = 60$^\circ$

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Therefore, the values of the unknowns are:

$\angle$x = 60$^\circ$

$\angle$y = 60$^\circ$

Given:

The lengths of three segments are 10.2 cm, 5.8 cm, and 4.5 cm.

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To Determine:

Whether a triangle can be formed with these side lengths.

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Solution:

To determine if a triangle can be formed with given side lengths, we use the Triangle Inequality Property.

The Triangle Inequality Property states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

Let the given side lengths be $a = 10.2$ cm, $b = 5.8$ cm, and $c = 4.5$ cm.

We need to check the following three inequalities:

1. Is $a + b > c$ ?

$10.2 + 5.8 = 16.0$

Is $16.0 > 4.5$ ? Yes.

This condition is satisfied.

2. Is $a + c > b$ ?

$10.2 + 4.5 = 14.7$

Is $14.7 > 5.8$ ? Yes.

This condition is satisfied.

3. Is $b + c > a$ ?

$5.8 + 4.5 = 10.3$

Is $10.3 > 10.2$ ? Yes.

This condition is satisfied.

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Conclusion:

Since the sum of the lengths of any two sides is greater than the length of the third side in all three cases, the Triangle Inequality Property is satisfied.

Therefore, a triangle whose sides have lengths 10.2 cm, 5.8 cm and 4.5 cm can be formed.

Given:

The lengths of two sides of a triangle are 6 cm and 8 cm.

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To Find:

The range of possible lengths for the third side.

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Solution:

Let the lengths of the two given sides be $a = 6$ cm and $b = 8$ cm.

Let the length of the unknown third side be $c$ cm.

To determine the possible length of the third side, we use the Triangle Inequality Property.

The Triangle Inequality Property states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

Applying this property to the three sides of the triangle, we get three inequalities:

1. The sum of sides a and b must be greater than side c:

$6 + 8 > c$

$14 > c$, or $c < 14$

2. The sum of sides a and c must be greater than side b:

$6 + c > 8$

Subtract 6 from both sides:

$c > 8 - 6$

$c > 2$

3. The sum of sides b and c must be greater than side a:

$8 + c > 6$

Subtract 8 from both sides:

$c > 6 - 8$

$c > -2$

Since the length of a side must be positive, the inequality $c > -2$ is always satisfied if we find a valid range for $c$.

Combining the two valid inequalities, $c < 14$ and $c > 2$, we find the range for the length of the third side, c.

The length of the third side must be greater than 2 cm and less than 14 cm.

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Answer:

The length of the third side must fall between 2 cm and 14 cm.

To determine if it is possible to form a triangle with the given side lengths, we use the Triangle Inequality Property. This property states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

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(i) Sides: 2 cm, 3 cm, 5 cm

Given: Side lengths are $a = 2$ cm, $b = 3$ cm, and $c = 5$ cm.

To Determine: If a triangle can be formed.

Solution:

Check the Triangle Inequality Property:

1. Is $a + b > c$ ?

$2 + 3 = 5$

Is $5 > 5$ ? No.

Since $2 + 3$ is not greater than 5, the Triangle Inequality Property is not satisfied for this combination of sides.

Conclusion:

It is not possible to have a triangle with sides 2 cm, 3 cm, and 5 cm.

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(ii) Sides: 3 cm, 6 cm, 7 cm

Given: Side lengths are $a = 3$ cm, $b = 6$ cm, and $c = 7$ cm.

To Determine: If a triangle can be formed.

Solution:

Check the Triangle Inequality Property:

1. Is $a + b > c$ ?

$3 + 6 = 9$

Is $9 > 7$ ? Yes.

2. Is $a + c > b$ ?

$3 + 7 = 10$

Is $10 > 6$ ? Yes.

3. Is $b + c > a$ ?

$6 + 7 = 13$

Is $13 > 3$ ? Yes.

Since the sum of the lengths of any two sides is greater than the length of the third side in all three cases, the Triangle Inequality Property is satisfied.

Conclusion:

It is possible to have a triangle with sides 3 cm, 6 cm, and 7 cm.

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(iii) Sides: 6 cm, 3 cm, 2 cm

Given: Side lengths are $a = 6$ cm, $b = 3$ cm, and $c = 2$ cm.

To Determine: If a triangle can be formed.

Solution:

Check the Triangle Inequality Property:

1. Is $a + b > c$ ?

$6 + 3 = 9$

Is $9 > 2$ ? Yes.

2. Is $a + c > b$ ?

$6 + 2 = 8$

Is $8 > 3$ ? Yes.

3. Is $b + c > a$ ?

$3 + 2 = 5$

Is $5 > 6$ ? No.

Since $3 + 2$ is not greater than 6, the Triangle Inequality Property is not satisfied for this combination of sides.

Conclusion:

It is not possible to have a triangle with sides 6 cm, 3 cm, and 2 cm.

Given:

Triangle PQR and a point O in its interior.

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To Determine:

Whether the given inequalities are true.

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Solution:

We will use the Triangle Inequality Property, which states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.

By joining the point O to the vertices P, Q, and R, we form three smaller triangles within $\triangle$PQR: $\triangle$OPQ, $\triangle$OQR, and $\triangle$ORP.

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(i) Is OP + OQ > PQ?

Consider $\triangle$OPQ.

The sides of this triangle are OP, OQ, and PQ.

According to the Triangle Inequality Property, the sum of the lengths of any two sides must be greater than the length of the third side.

Applying this to $\triangle$OPQ:

OP + OQ > PQ

OP + PQ > OQ

OQ + PQ > OP

The inequality OP + OQ > PQ is one of the conditions from the Triangle Inequality Property for $\triangle$OPQ.

Therefore, the statement is true.

Answer: Yes

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(ii) Is OQ + OR > QR?

Consider $\triangle$OQR.

The sides of this triangle are OQ, OR, and QR.

According to the Triangle Inequality Property, the sum of the lengths of any two sides must be greater than the length of the third side.

Applying this to $\triangle$OQR:

OQ + OR > QR

OQ + QR > OR

OR + QR > OQ

The inequality OQ + OR > QR is one of the conditions from the Triangle Inequality Property for $\triangle$OQR.

Therefore, the statement is true.

Answer: Yes

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(iii) Is OR + OP > RP?

Consider $\triangle$ORP.

The sides of this triangle are OR, OP, and RP.

According to the Triangle Inequality Property, the sum of the lengths of any two sides must be greater than the length of the third side.

Applying this to $\triangle$ORP:

OR + OP > RP

OR + RP > OP

OP + RP > OR

The inequality OR + OP > RP is one of the conditions from the Triangle Inequality Property for $\triangle$ORP.

Therefore, the statement is true.

Answer: Yes

Given:

AM is a median of $\triangle$ ABC.

This means M is the midpoint of BC, so BM = MC.

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To Verify:

Is AB + BC + CA > 2 AM ?

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Solution:

We will consider the two triangles formed by the median AM: $\triangle$ABM and $\triangle$AMC.

Using the Triangle Inequality Property in $\triangle$ABM:

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

Applying this to $\triangle$ABM, considering the sides AB, BM, and AM:

In $\triangle$ABM:

AB + BM > AM

Using the Triangle Inequality Property in $\triangle$AMC:

Applying this to $\triangle$AMC, considering the sides AC, MC, and AM:

In $\triangle$AMC:

AC + MC > AM

Now, add inequality (1) and inequality (2):

(AB + BM) + (AC + MC) > AM + AM

AB + BM + AC + MC > 2 AM

Rearrange the terms:

AB + AC + (BM + MC) > 2 AM

Since AM is the median, M is the midpoint of BC. Therefore, BM = MC.

The sum of BM and MC is equal to the length of the side BC:

Substitute (BM + MC) with BC in the inequality:

AB + AC + BC > 2 AM

This inequality can be written as:

AB + BC + CA > 2 AM

This matches the statement we needed to verify.

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Conclusion:

Yes, the sum of the lengths of the three sides of the triangle (AB + BC + CA) is greater than twice the length of the median (2 AM).

So, AB + BC + CA > 2 AM is True.

Given:

ABCD is a quadrilateral.

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To Verify:

Is AB + BC + CD + DA > AC + BD?

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Solution:

We will use the Triangle Inequality Property, which states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.

Consider the triangles formed by the sides and the diagonals of the quadrilateral.

In $\triangle$ ABC, the sides are AB, BC, and AC.

Applying the Triangle Inequality Property to $\triangle$ ABC:

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In $\triangle$ ADC, the sides are AD, DC, and AC.

Applying the Triangle Inequality Property to $\triangle$ ADC:

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In $\triangle$ ABD, the sides are AB, AD, and BD.

Applying the Triangle Inequality Property to $\triangle$ ABD:

---

In $\triangle$ BCD, the sides are BC, CD, and BD.

Applying the Triangle Inequality Property to $\triangle$ BCD:

---

Now, add all four inequalities (i), (ii), (iii), and (iv):

(AB + BC) + (AD + DC) + (AB + AD) + (BC + CD) > AC + AC + BD + BD

Combine like terms on the left side:

AB + BC + AD + DC + AB + AD + BC + CD > 2AC + 2BD

Rearrange and group the side lengths:

(AB + AB) + (BC + BC) + (CD + CD) + (AD + AD) > 2AC + 2BD

$2 \times$ AB + $2 \times$ BC + $2 \times$ CD + $2 \times$ DA > 2AC + 2BD

Factor out 2 from both sides of the inequality:

2 (AB + BC + CD + DA) > 2 (AC + BD)

Divide both sides of the inequality by 2:

AB + BC + CD + DA > AC + BD

This result matches the statement we needed to verify.

---

Conclusion:

Yes, the sum of the lengths of the sides of the quadrilateral ABCD (AB + BC + CD + DA) is greater than the sum of the lengths of its diagonals (AC + BD).

So, AB + BC + CD + DA > AC + BD is True.

Solution:

---

Given:

ABCD is a quadrilateral.

---

To Verify:

Is AB + BC + CD + DA < 2 (AC + BD) ?

---

Construction Required:

Draw the quadrilateral ABCD and its diagonals AC and BD intersecting at point O.

---

Proof:

We will use the Triangle Inequality Property, which states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.

Consider the four triangles formed by the point of intersection O of the diagonals and the vertices of the quadrilateral: $\triangle$OAB, $\triangle$OBC, $\triangle$OCD, and $\triangle$ODA.

Applying the Triangle Inequality Property to each of these triangles:

In $\triangle$OAB:

---

In $\triangle$OBC:

---

In $\triangle$OCD:

---

In $\triangle$ODA:

---

Now, add the four inequalities (1), (2), (3), and (4):

(OA + OB) + (OB + OC) + (OC + OD) + (OD + OA) > AB + BC + CD + DA

Group the terms on the left side:

OA + OA + OB + OB + OC + OC + OD + OD > AB + BC + CD + DA

$2 \times$ OA + $2 \times$ OB + $2 \times$ OC + $2 \times$ OD > AB + BC + CD + DA

Factor out 2:

2 (OA + OB + OC + OD) > AB + BC + CD + DA

Rearrange the terms within the parenthesis:

2 (OA + OC + OB + OD) > AB + BC + CD + DA

From the figure, OA + OC represents the length of the diagonal AC, and OB + OD represents the length of the diagonal BD.

Substitute these into the inequality:

2 (AC + BD) > AB + BC + CD + DA

This inequality can also be written as:

AB + BC + CD + DA < 2 (AC + BD)

This matches the statement we needed to verify.

---

Conclusion:

Yes, the sum of the lengths of the sides of the quadrilateral ABCD is less than twice the sum of the lengths of its diagonals.

So, AB + BC + CD + DA < 2 (AC + BD) is True.

Given:

The lengths of two sides of a triangle are 12 cm and 15 cm.

---

To Find:

Between which two measures the length of the third side should fall.

---

Solution:

Let the lengths of the two given sides be $a = 12$ cm and $b = 15$ cm.

Let the length of the unknown third side be $c$ cm.

To determine the possible length of the third side, we use the Triangle Inequality Property.

The Triangle Inequality Property states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

Applying this property, we must satisfy the following conditions:

1. The sum of sides a and b must be greater than side c:

$a + b > c$

$12 + 15 > c$

$27 > c$, or $c < 27$

2. The sum of sides a and c must be greater than side b:

$a + c > b$

$12 + c > 15$

Subtract 12 from both sides:

$c > 15 - 12$

$c > 3$

3. The sum of sides b and c must be greater than side a:

$b + c > a$

$15 + c > 12$

Subtract 15 from both sides:

$c > 12 - 15$

$c > -3$

Since the length of a side must be a positive value, the condition $c > -3$ is automatically satisfied if $c$ is greater than 3.

Combining the conditions $c < 27$ and $c > 3$, we find the range for the length of the third side, c.

The length of the third side must be greater than 3 cm and less than 27 cm.

---

Answer:

The length of the third side should fall between 3 cm and 27 cm.

Given:

The lengths of the sides of a triangle are 3 cm, 4 cm, and 5 cm.

---

To Determine:

Whether the triangle is a right-angled triangle.

---

Solution:

We can determine if a triangle is a right-angled triangle by using the Converse of the Pythagorean Property.

The Pythagorean Property states that in a right-angled triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides (legs).

Conversely, if the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle, and the triangle is a right-angled triangle.

Let the side lengths be $a = 3$ cm, $b = 4$ cm, and $c = 5$ cm.

The longest side is 5 cm, so this is the potential hypotenuse.

Let's calculate the square of each side length:

$a^2 = 3^2 = 3 \times 3 = 9$

$b^2 = 4^2 = 4 \times 4 = 16$

$c^2 = 5^2 = 5 \times 5 = 25$

Now, let's check if the square of the longest side ($c^2$) is equal to the sum of the squares of the other two sides ($a^2 + b^2$).

Is $c^2 = a^2 + b^2$ ?

Substitute the calculated values:

$25 = 9 + 16$

$25 = 25$

The equation is true.

Since the square of the longest side is equal to the sum of the squares of the other two sides, the Converse of the Pythagorean Property is satisfied.

This means the angle opposite the side with length 5 cm is a right angle ($90^\circ$).

---

Conclusion:

Yes, the triangle whose lengths of sides are 3 cm, 4 cm, and 5 cm is a right-angled triangle.

Solution:

---

Given:

In $\triangle$ ABC, the angle at C is a right angle ($\angle$C = $90^\circ$).

The lengths of the two sides forming the right angle are:

AC = 5 cm

BC = 12 cm

---

To Find:

The length of the side AB (the hypotenuse).

---

Solution:

Since $\triangle$ ABC is a right-angled triangle at C, we can use the Pythagorean Property.

The Pythagorean Property states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides (legs).

In $\triangle$ ABC, AB is the hypotenuse (opposite the right angle at C), and AC and BC are the legs.

According to the Pythagorean Property:

Substitute the given values of AC and BC into the equation:

$\text{AB}^2 = (5)^2 + (12)^2$

Calculate the squares of the lengths:

$5^2 = 5 \times 5 = 25$

$12^2 = 12 \times 12 = 144$

Substitute these values back into equation (i):

$\text{AB}^2 = 25 + 144$

Add the values on the right side:

$\text{AB}^2 = 169$

To find the length of AB, take the square root of both sides of the equation:

$\text{AB} = \sqrt{169}$

The square root of 169 is 13 (since $13 \times 13 = 169$).

$\text{AB} = 13$

Since AB is a length, it must be a positive value. The unit of length is cm.

AB = 13 cm

---

Answer:

The length of AB is 13 cm.

Given:

In $\triangle$ PQR, it is right-angled at P.

The length of side PQ = 10 cm.

The length of side PR = 24 cm.

---

To Find:

The length of side QR.

---

Solution:

Since $\triangle$ PQR is a right-angled triangle at P, we can use the Pythagorean Property (also known as the Pythagorean Theorem).

The Pythagorean Property states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides (legs).

In $\triangle$ PQR, the side opposite the right angle at P is QR, which is the hypotenuse.

The sides PQ and PR are the legs forming the right angle.

According to the Pythagorean Property:

Applying this to $\triangle$ PQR:

Substitute the given lengths of PQ and PR into the equation:

$\text{QR}^2 = (10 \text{ cm})^2 + (24 \text{ cm})^2$

Calculate the squares of the given lengths:

$10^2 = 10 \times 10 = 100$

$24^2 = 24 \times 24 = 576$

Substitute these values back into the equation:

$\text{QR}^2 = 100 + 576$

Add the values on the right side:

$\text{QR}^2 = 676$

To find the length of QR, take the square root of both sides of the equation:

$\text{QR} = \sqrt{676}$

The square root of 676 is 26.

$\text{QR} = 26$

Since QR represents a length, we consider the positive square root. The unit of length is cm.

QR = 26 cm

---

Answer:

The length of QR is 26 cm.

Given:

In $\triangle$ ABC, it is right-angled at C ($\angle$C = $90^\circ$).

The length of side AB (Hypotenuse) = 25 cm.

The length of side AC (Leg) = 7 cm.

---

To Find:

The length of side BC (the other Leg).

---

Solution:

Since $\triangle$ ABC is a right-angled triangle at C, we can use the Pythagorean Property (Pythagorean Theorem).

The Pythagorean Property states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In $\triangle$ ABC, AB is the hypotenuse, and AC and BC are the legs.

According to the Pythagorean Property:

Substitute the given values into the equation:

$(25 \text{ cm})^2 = (7 \text{ cm})^2 + \text{BC}^2$

Calculate the squares:

$25^2 = 25 \times 25 = 625$

$7^2 = 7 \times 7 = 49$

Substitute these values into the equation:

$625 = 49 + \text{BC}^2$

To find $\text{BC}^2$, subtract 49 from both sides of the equation:

$\text{BC}^2 = 625 - 49$

$\text{BC}^2 = 576$

To find the length of BC, take the square root of both sides of the equation:

$\text{BC} = \sqrt{576}$

We know that $24 \times 24 = 576$.

So,

$\text{BC} = 24$

Since BC is a length, it is a positive value. The unit is centimeters.

BC = 24 cm

---

Answer:

The length of BC is 24 cm.

Given:

Length of the ladder (hypotenuse) = 15 m.

Height of the window from the ground (one leg of the right triangle) = 12 m.

Distance of the foot of the ladder from the wall (other leg of the right triangle) = a.

The wall is perpendicular to the ground, so a right-angled triangle is formed by the ground, the wall, and the ladder.

---

To Find:

The distance 'a' (the distance of the foot of the ladder from the wall).

---

Solution:

Let the wall, the ground, and the ladder form a right-angled triangle. The right angle is between the wall and the ground.

The ladder is the hypotenuse, as it is opposite the right angle.

The height of the window from the ground and the distance of the foot of the ladder from the wall are the legs of the right-angled triangle.

We can use the Pythagorean Property (Pythagorean Theorem) for this right-angled triangle.

The Pythagorean Property states:

In this case:

Calculate the squares of the known lengths:

$15^2 = 15 \times 15 = 225$

$12^2 = 12 \times 12 = 144$

Substitute these values back into the equation:

$225 = 144 + a^2$

To find $a^2$, subtract 144 from both sides of the equation:

$a^2 = 225 - 144$

Perform the subtraction:

$a^2 = 81$

To find the value of 'a', take the square root of both sides of the equation:

$a = \sqrt{81}$

The square root of 81 is 9.

$a = 9$

Since 'a' represents a distance, it must be a positive value. The unit is meters.

a = 9 m

---

Answer:

The distance of the foot of the ladder from the wall is 9 m.

To determine if a triangle with the given side lengths is a right triangle, we will use the Converse of the Pythagorean Property. This property states that if the square of the length of the longest side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right-angled triangle, and the angle opposite the longest side is the right angle.

---

(i) Sides: 2.5 cm, 6.5 cm, 6 cm

Given: The lengths of the sides are 2.5 cm, 6.5 cm, and 6 cm.

To Determine: If these sides form a right triangle and identify the right angle.

Solution:

The longest side is 6.5 cm. Let's check if the square of the longest side is equal to the sum of the squares of the other two sides.

Square of the longest side: $(6.5)^2 = 6.5 \times 6.5 = 42.25$

Sum of the squares of the other two sides:

$(2.5)^2 + (6)^2 = (2.5 \times 2.5) + (6 \times 6) = 6.25 + 36 = 42.25$

Compare the square of the longest side with the sum of the squares of the other two sides:

$42.25 = 42.25$

Since the square of the longest side is equal to the sum of the squares of the other two sides, the Converse of the Pythagorean Property is satisfied.

Conclusion:

Yes, these can be the sides of a right triangle. The right angle is opposite the longest side, which has a length of 6.5 cm.

---

(ii) Sides: 2 cm, 2 cm, 5 cm

Given: The lengths of the sides are 2 cm, 2 cm, and 5 cm.

To Determine: If these sides form a right triangle and identify the right angle.

Solution:

The longest side is 5 cm. Let's check if the square of the longest side is equal to the sum of the squares of the other two sides.

Square of the longest side: $(5)^2 = 5 \times 5 = 25$

Sum of the squares of the other two sides:

$(2)^2 + (2)^2 = (2 \times 2) + (2 \times 2) = 4 + 4 = 8$

Compare the square of the longest side with the sum of the squares of the other two sides:

$25 \neq 8$

Since the square of the longest side is not equal to the sum of the squares of the other two sides, the Converse of the Pythagorean Property is not satisfied.

Also, let's check the Triangle Inequality Property: $2 + 2 > 5$? $4 > 5$? No. These lengths cannot even form a triangle.

Conclusion:

No, these cannot be the sides of a right triangle (or any triangle).

---

(iii) Sides: 1.5 cm, 2 cm, 2.5 cm

Given: The lengths of the sides are 1.5 cm, 2 cm, and 2.5 cm.

To Determine: If these sides form a right triangle and identify the right angle.

Solution:

The longest side is 2.5 cm. Let's check if the square of the longest side is equal to the sum of the squares of the other two sides.

Square of the longest side: $(2.5)^2 = 2.5 \times 2.5 = 6.25$

Sum of the squares of the other two sides:

$(1.5)^2 + (2)^2 = (1.5 \times 1.5) + (2 \times 2) = 2.25 + 4 = 6.25$

Compare the square of the longest side with the sum of the squares of the other two sides:

$6.25 = 6.25$

Since the square of the longest side is equal to the sum of the squares of the other two sides, the Converse of the Pythagorean Property is satisfied.

Conclusion:

Yes, these can be the sides of a right triangle. The right angle is opposite the longest side, which has a length of 2.5 cm.

Given:

A tree is broken at a height of 5 m from the ground.

The top of the tree touches the ground at a distance of 12 m from the base of the tree.

The standing part of the tree, the ground, and the broken part of the tree form a right-angled triangle.

---

To Find:

The original height of the tree.

---

Solution:

Let the height at which the tree is broken be the length of one leg of a right-angled triangle. Let this length be $a = 5$ m.

Let the distance from the base of the tree to the point where the top touches the ground be the length of the other leg. Let this length be $b = 12$ m.

The broken part of the tree, which is now the segment from the break point to the top on the ground, forms the hypotenuse of the right-angled triangle. Let the length of the broken part be $h$ m.

According to the Pythagorean Property in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Applying this to the triangle formed:

Substitute the given values for $a$ and $b$:

$h^2 = (5 \text{ m})^2 + (12 \text{ m})^2$

Calculate the squares:

$5^2 = 5 \times 5 = 25$

$12^2 = 12 \times 12 = 144$

Substitute these values into the equation:

$h^2 = 25 + 144$

Perform the addition:

$h^2 = 169$

To find the length $h$, take the square root of both sides of the equation:

$h = \sqrt{169}$

We know that $13 \times 13 = 169$.

So,

$h = 13$

The length of the broken part of the tree is 13 m.

The original height of the tree is the sum of the standing part and the broken part (the hypotenuse).

Original Height = Standing Part + Broken Part

Original Height = 5 m + 13 m

Original Height = 18 m

---

Answer:

The original height of the tree was 18 m.

Given:

In $\triangle$ PQR:

$\angle$Q = $25^\circ$

$\angle$R = $65^\circ$

---

To Determine:

Which of the given statements is true:

(i) PQ$^2$ + QR$^2$ = RP$^2$

(ii) PQ$^2$ + RP$^2$ = QR$^2$

(iii) RP$^2$ + QR$^2$ = PQ$^2$

---

Solution:

First, let's find the measure of the third angle, $\angle$P, in $\triangle$ PQR using the Angle Sum Property of a Triangle.

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

Substitute the given values of $\angle$Q and $\angle$R:

$\angle$P + $25^\circ + 65^\circ = 180^\circ$

Add the known angles:

$\angle$P + $90^\circ = 180^\circ$

Subtract $90^\circ$ from both sides to find $\angle$P:

$\angle$P = $180^\circ - 90^\circ$

$\angle$P = $90^\circ$

Since $\angle$P = $90^\circ$, $\triangle$ PQR is a right-angled triangle, with the right angle at vertex P.

In a right-angled triangle, the side opposite the right angle is called the hypotenuse, and the other two sides are called legs.

In $\triangle$ PQR, the angle at P is $90^\circ$. The side opposite to angle P is QR. Therefore, QR is the hypotenuse.

The other two sides, PQ and PR (or RP), are the legs.

According to the Pythagorean Property, in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

Applying this to $\triangle$ PQR:

Note that $\text{PR}^2$ is the same as $\text{RP}^2$ since it represents the square of the length of the side.

So the relationship is $\text{QR}^2 = \text{PQ}^2 + \text{RP}^2$, or $\text{PQ}^2 + \text{RP}^2 = \text{QR}^2$.

Now, let's compare this equation with the given options:

(i) PQ$^2$ + QR$^2$ = RP$^2$ (This would mean RP is the hypotenuse, and the right angle is at Q)

(ii) PQ$^2$ + RP$^2$ = QR$^2$ (This means QR is the hypotenuse, and the right angle is at P)

(iii) RP$^2$ + QR$^2$ = PQ$^2$ (This would mean PQ is the hypotenuse, and the right angle is at R)

Our derived equation, PQ$^2$ + RP$^2$ = QR$^2$, matches option (ii).

---

Answer:

The triangle PQR is right-angled at P.

According to the Pythagorean property, $\text{PQ}^2 + \text{RP}^2 = \text{QR}^2$.

Therefore, the true statement is (ii) PQ² + RP² = QR².

Solution:

---

Given:

Length of the rectangle $= 40$ cm.

Length of a diagonal $= 41$ cm.

---

To Find:

The perimeter of the rectangle.

---

Solution:

In a rectangle, the diagonal divides the rectangle into two right-angled triangles. The sides of the rectangle (length and width) are the legs of the right-angled triangle, and the diagonal is the hypotenuse.

Let the length of the rectangle be $l = 40$ cm.

Let the width of the rectangle be $w$ cm.

The length of the diagonal is $d = 41$ cm.

Using the Pythagorean Property in one of the right-angled triangles formed by the diagonal:

Substitute the given values:

Calculate the squares of the known values:

$40^2 = 40 \times 40 = 1600$

$41^2 = 41 \times 41 = 1681$

Substitute these values back into equation (i):

$1600 + \text{w}^2 = 1681$

Subtract 1600 from both sides to find $\text{w}^2$:

$\text{w}^2 = 1681 - 1600$

To find the width $w$, take the square root of both sides of equation (ii):

$\text{w} = \sqrt{81}$

Since width is a positive length, we take the positive square root.

$\text{w} = 9$ cm

---

Now that we have the length ($l = 40$ cm) and the width ($w = 9$ cm), we can calculate the perimeter of the rectangle.

The formula for the perimeter of a rectangle is:

Perimeter = 2 $\times$ $(40 \text{ cm} + 9 \text{ cm})$

Perimeter = 2 $\times$ $(49 \text{ cm})$

Perimeter = 98 cm

---

Answer:

The perimeter of the rectangle is 98 cm.

Solution:

---

Given:

The lengths of the diagonals of a rhombus are $d_1 = 16$ cm and $d_2 = 30$ cm.

---

To Find:

The perimeter of the rhombus.

---

Solution:

Let the side length of the rhombus be $s$ cm.

We know that the diagonals of a rhombus have two important properties:

1. They bisect each other (cut each other into two equal halves).
2. They intersect at right angles ($90^\circ$).

When the two diagonals intersect at the center of the rhombus, they divide the rhombus into four smaller triangles. Each of these four triangles is a right-angled triangle.

The lengths of the two shorter sides (legs) of each right-angled triangle are half the lengths of the diagonals. The longest side (hypotenuse) of each right-angled triangle is a side of the rhombus.

Length of the legs of each right triangle are:

Leg 1 = Half of diagonal $d_1 = \frac{16}{2} = 8$ cm

Leg 2 = Half of diagonal $d_2 = \frac{30}{2} = 15$ cm

The hypotenuse of these right triangles is the side $s$ of the rhombus.

Using the Pythagorean Property in one of the right-angled triangles:

Applying this to find the side $s$:

Calculate the squares of the leg lengths:

$8^2 = 8 \times 8 = 64$

$15^2 = 15 \times 15 = 225$

The equation becomes:

$s^2 = 64 + 225$

Add the values on the right side:

To find the side length $s$, take the square root of both sides of equation (i):

$\text{s} = \sqrt{289}$

We need to find a number that when multiplied by itself equals 289. We know that $17 \times 17 = 289$.

So,

$\text{s} = 17$

Since $s$ represents a length, it is a positive value. The unit is centimeters.

$s = 17$ cm

---

The perimeter of a rhombus is the sum of the lengths of its four sides. Since all sides of a rhombus are equal in length:

Perimeter = $4 \times s$

Perimeter = $4 \times 17$ cm

Perimeter = 68 cm

---

Answer:

The perimeter of the rhombus is 68 cm.