RS Aggarwal Class 9 Solutions Congruence of Triangles Inequalities in Triangle

Question 1:

In \(\Delta ABC\), if AB = AC and \(\angle\)A = 70o, find \(\angle\)B and \(\angle\)C.

Solution:

AB = AC implies their opposite angles are equal

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\(\Rightarrow \angle B=\angle C\) [angles opposites to equal sides]

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 100o. Find its base angles.

Solution:

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Consider the isosceles triangle \(\Delta ABC\).

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

By angle sum property of a triangle, we have,

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

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

\(\Rightarrow 100^{\circ}+2\angle B=180^{\circ}\) [since in an isosceles triangle base angles are equal, \(\angle B=\angle C\)]

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

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

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

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

Question 3:

In a \(\Delta\)ABC, if AB = AC and \(\angle\)B = 65o, find \(\angle\)C and \(\angle\)A.

Solution:

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In \(\Delta ABC\), if AB = AC

\( \Rightarrow \Delta ABC\) is an isosceles triangle

\(\Rightarrow\) Base angles are equals

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

\(\Rightarrow \angle C=65^{\circ}\) [Since \(\angle B=65^{\circ}\)]

Also by angle sum property, we have

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

\(\Rightarrow \angle A+65^{\circ}+65^{\circ}=180^{\circ}\) [\(\angle B=\angle C=65^{\circ}\)]

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

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:

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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 = 180o

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

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

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:

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In a right angle isosceles triangle, the vertex angle is \(\angle A=90^{\circ}\) and the other two base angles are equal.

Let xo 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

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To Prove: \(\angle\)EBA = \(\angle\)DCA

Proof: In \(\Delta\)ABC we have,

AB = AC

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

Now exterior \(\angle\)EBA = \(\angle\)A + \(\angle\)C = \(\angle\)A + \(\angle\)B [\(\angle\)B = \(\angle\)C]

and exterior \(\angle\)DCA = \(\angle\)A + \(\angle\)B

\(\Rightarrow\) Exterior \(\angle\)EBA = Exterior \(\angle\)DCA.

Question 7:

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

Solution:

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Let \(\Delta ABC\) be an equilateral triangle.

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

The exterior angle of \(\angle A\) = \(\angle BAF\)

The exterior angle of \(\angle B\) = \(\angle ABD\)

The exterior angle of \(\angle C\) = \(\angle ACE\)

We can observe that angles \(\angle A\) and \(\angle BAF\), \(\angle B\) and \(\angle ABD\), \(\angle C\) and \(\angle ACE\) form linear pairs.

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, \(\angle BAF=120^{\circ}\), \(\angle ABD=120^{\circ}\), \(\angle ACE=120^{\circ}\)

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

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.

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Solution:

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

\(\Rightarrow\) AO = OB and CO = OD

To prove: AC = BD and AC || BD

Proof:

In \(\Delta\)AOC and \(\Delta\)BOD, we have,

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

\(\angle\)AOC = \(\angle\)BOD [Vertically opposite angles]

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, \(\angle\)ACO = \(\angle\)BDO and \(\angle\)CAO = \(\angle\)DDBO

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.’


Practise This Question

If 60 is divided in the ratio of 1:3, what is the value of the bigger part?