Chapter-5: Congruence of Triangles

Question 1:

In ΔABC, if AB = AC and A = 70o, find B and C.

Solution:

AB = AC implies their opposite angles are equal

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B=C [angles opposites to equal sides]

But A+B+C=180

70+B+B=180 70+2B=180 2B=18070 2B=110 B=1102 B=55 B=C=55

Question 2:

The vertical angle of an isosceles triangle is 100o. Find its base angles.

Solution:

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Consider the isosceles triangle ΔABC.

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

By angle sum property of a triangle, we have,

A+B+C=180 100+B+C=180

100+2B=180 [since in an isosceles triangle base angles are equal, B=C]

2B=180100=80 B=802 B=40 B=C=40

Question 3:

In a ΔABC, if AB = AC and B = 65o, find C and A.

Solution:

https://lh4.googleusercontent.com/V_jgpY8DL7on5_XYDRr7XCh-mM2L5X-qan8kDXc35Aw2SkWPxECljL3220dcBYH0VtW295tThG--Ka0v42YCGZVaXM8rto_bqxgia04iOlmU9R5CfnkIF-IWLIUL-xKRfdsX7QXfzH09iRhHdg

In ΔABC, if AB = AC

ΔABC is an isosceles triangle

Base angles are equals

B=C

C=65 [Since B=65]

Also by angle sum property, we have

A+B+C=180

A+65+65=180 [latex]\angle B=\angle C=65^{\circ}[/latex]

A=180130=50

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:

https://lh5.googleusercontent.com/cGd8dJM0NKsVpANYoIkj1Zm3X8emz5unXf9vNGO-SWkPeKqh621Y0Fyz-Q2u2vj1E8anz4GxOzjjlOPzHJC33l4BmXjH17ZtMvfn1szJ8hFoBzlLE4CTLAZySlz1rOM7PMKuIGlp8blJZwy_2A

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

Then we have B=C

Let B=C=x

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

Now, x + x + 4x = 180o

6x=180 x=1806=30

Therefore, Vertex A=4×30=120

And, B=C=30

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 A=90 and the other two base angles are equal.

Let xo be the base angle and we have, B=C=x.

By angle sum property of a triangle, we have

A+B+C=180 90+x+x=180 90+2x=180 2x=18090=90 x=902=45

Thus, we have B=C=45

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: EBA = DCA

Proof: In ΔABC we have,

AB = AC

B = C

Now exterior EBA = A + C = A + B [B = C]

and exterior DCA = A + B

Exterior EBA = Exterior DCA.

Question 7:

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

Solution:

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Let ΔABC be an equilateral triangle.

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

The exterior angle of A = BAF

The exterior angle of B = ABD

The exterior angle of C = ACE

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

Therefore, we have

A+BAF=180 60+BAF=180 BAF=18060=120

Similarly, we have

B+ABD=180 60+ABD=180 ABD=18060=120

Also, we have

C+ACE=180 60+ACE=180 ACE=18060=120

Thus, we have, BAF=120, ABD=120, ACE=120

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.

AO = OB and CO = OD

To prove: AC = BD and AC || BD

Proof:

In ΔAOC and ΔBOD, we have,

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

AOC = 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, ACO = BDO and CAO = DBO

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.

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