Congruence and Inequalities in Triangles Exercise 5.1 |

Congruence and Inequalities in Triangles Exercise 5.2 |

**Question 1:**

**In \(\Delta ABC\), if AB = AC and \(\angle\)A = 70**

^{o}, find \(\angle\)B and \(\angle\)C.

**Solution:**

AB = AC implies their opposite angles are equal

\(\Rightarrow \angle B=\angle C\)

But \(\angle A+\angle B+\angle C=180^{\circ}\)

\(\Rightarrow 70^{\circ}+\angle B+\angle B=180^{\circ}\)

\(\Rightarrow 70^{\circ}+2\angle B=180^{\circ}\)

\(\Rightarrow 2\angle B=180^{\circ}-70^{\circ}\)

\(\Rightarrow 2\angle B=110^{\circ}\)

\(\Rightarrow \angle B=\frac{110^{\circ}}{2}\)

\(\Rightarrow \angle B=55^{\circ}\)

\(\Rightarrow \angle B=\angle C=55^{\circ}\)

**Question 2:**

**The vertical angle of an isosceles triangle is 100 ^{o}. Find its base angles.**

**Solution:**

Consider the isosceles triangle \(\Delta ABC\)

Since the vertical angle of ABC is 100^{o}, we have, \(\angle A\)^{o}

By angle sum property of a triangle, we have,

\(\angle A+\angle B+\angle C=180^{\circ}\)

\(\Rightarrow 100^{\circ}+\angle B+\angle C=180^{\circ}\)

\(\Rightarrow 100^{\circ}+2\angle B=180^{\circ}\)

\(\Rightarrow 2\angle B=180^{\circ}-100^{\circ}=80^{\circ}\)

\(\Rightarrow \angle B=\frac{80^{\circ}}{2}\)

\(\Rightarrow \angle B=40^{\circ}\)

\(\Rightarrow \angle B=\angle C=40^{\circ}\)

**Question 3:**

**In a \(\Delta\)ABC, if AB = AC and \(\angle\)B = 65**

^{o}, find \(\angle\)C and \(\angle\)A.

**Solution:**

In \(\Delta ABC\)

\( \Rightarrow \Delta ABC\)

\(\Rightarrow\)

\(\Rightarrow \angle B=\angle C\)

\(\Rightarrow \angle C=65^{\circ}\)

Also by angle sum property, we have

\(\angle A+\angle B+\angle C=180^{\circ}\)

\(\Rightarrow \angle A+65^{\circ}+65^{\circ}=180^{\circ}\)

\(\Rightarrow \angle A=180^{\circ}-130^{\circ}=50^{\circ}\)

**Question 4:**

**In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.**

**Solution:**

Let ABC be an isosceles triangle in which AB = AC.

Then we have \(\angle B=\angle C\)

Let \(\angle B=\angle C=x\)

Then vertex angle A = 2(x + x) = 4x

Now, x + x + 4x = 180^{o}

\(\Rightarrow 6x=180^{\circ}\)

\(\Rightarrow x=\frac{180^{\circ}}{6}=30^{\circ}\)

Therefore, Vertex \(\angle A=4\times 30=120^{\circ}\)

And, \(\angle B=\angle C=30^{\circ}\)

**Question 5:**

**What is the measure of each of the equal angles of a right-angled isosceles triangle?**

**Solution:**

In a right angle isosceles triangle, the vertex angle is \(\angle A=90^{\circ}\)

Let x^{o} be the base angle and we have, \(\angle B=\angle C=x\)

By angle sum property of a triangle, we have

\(\angle A+\angle B+\angle C=180^{\circ}\)

\(\Rightarrow 90^{\circ}+x+x=180^{\circ}\)

\(\Rightarrow 90^{\circ}+2x=180^{\circ}\)

\(\Rightarrow 2x=180^{\circ}-90^{\circ}=90^{\circ}\)

\(\Rightarrow x=\frac{90^{\circ}}{2}=45^{\circ}\)

Thus, we have \(\angle B=\angle C=45^{\circ}\)

**Question 6:**

**If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.**

**Solution:**

Given: ABC is an isosceles triangle in which AB = AC and BC is produced both ways

To Prove: \(\angle\)

Proof: In \(\Delta\)

AB = AC

\(\Rightarrow\)

Now exterior \(\angle\)

and exterior \(\angle\)

\(\Rightarrow\)

**Question 7:**

**Find the measure of each exterior angle of an equilateral triangle.**

**Solution:**

Let \(\Delta ABC\)

Since it is an equilateral triangle, all the angle is 60^{o}

The exterior angle of \(\angle A\)

The exterior angle of \(\angle B\)

The exterior angle of \(\angle C\)

We can observe that angles \(\angle A\)

Therefore, we have

\(\angle A+\angle BAF=180^{\circ}\)

\(\Rightarrow 60^{\circ}+\angle BAF=180^{\circ}\)

\(\Rightarrow \angle BAF=180^{\circ}-60^{\circ}=120^{\circ}\)

Similarly, we have

\(\angle B+\angle ABD=180^{\circ}\)

\(\Rightarrow 60^{\circ}+\angle ABD=180^{\circ}\)

\(\Rightarrow \angle ABD=180^{\circ}-60^{\circ}=120^{\circ}\)

Also, we have

\(\angle C+\angle ACE=180^{\circ}\)

\(\Rightarrow 60^{\circ}+\angle ACE=180^{\circ}\)

\(\Rightarrow \angle ACE=180^{\circ}-60^{\circ}=120^{\circ}\)

Thus, we have, \(\angle BAF=120^{\circ}\)

So, the measure of each exterior angle of an equilateral triangle is 120^{o}.

**Question 8:**

**In the given figure, O is the mid-point of each of the line segments AB and CD. Prove that AC = BD and AC || BD.**

**Solution:**

Given: Two lines AB and CD intersect at O and O is the midpoint of AB and CD.

\(\Rightarrow\)

To prove: AC = BD and AC || BD

Proof:

In \(\Delta\)

AO = OB [Given: O is the midpoint of AB]

\(\angle\)

CO = OD [Given: O is the midpoint of CD]

So, by Side-Angle-Side congruence triangle are equal.

Therefore, we have, AC = BD.

Similarly, by c.p.c.t, we have, \(\angle\)

This implies that alternate angles formed by AC and BD with transversal CD are equal. This means that, AC || BD. Thus, AC = BD and AC || BD.’