NCERT Solutions for Exercise 7.2 Class 9 Maths Chapter 7 - Triangles

Access Triangles Class 9 Chapter 7 Exercise: 7.2

Q1 (i) In an isosceles triangle ABC, with $AB=AC$ , the bisectors of $\mathrm{\angle }B$ and $\mathrm{\angle }C$ intersect each other at O. Join A to O. Show that : $OB=OC$

Answer:

In the triangle ABC,

Since AB = AC, thus $\mathrm{\angle }B=\mathrm{\angle }C$

or $\frac{1}{2}\mathrm{\angle }B=\frac{1}{2}\mathrm{\angle }C$

or $\mathrm{\angle }OBC=\mathrm{\angle }OCB$ (Angles bisectors are equal)

Thus $OB=OC$ as sides opposite to equal are angles are also equal.

Q1 (ii) In an isosceles triangle ABC, with $AB=AC$ , the bisectors of $\mathrm{\angle }B$ and $\mathrm{\angle }C$ intersect each other at O. Join A to O. Show that : AO bisects $\mathrm{\angle }A$

Answer:

Consider $\mathrm{\Delta }AOB$ and $\mathrm{\Delta }AOC$ ,

(i) $AB=AC$ (Given)

(ii) $AO=AO$ (Common in both the triangles)

(iii) $OB=OC$ (Proved in previous part)

Thus by SSS congruence rule, we can conclude that :

$\mathrm{\Delta }AOB\cong \mathrm{\Delta }AOC$

Now, by c.p.c.t.,

$\mathrm{\angle }BAO=\mathrm{\angle }CAO$

Hence AO bisects $\mathrm{\angle }A$ .

Q2 In $\mathrm{\Delta }ABC$ , AD is the perpendicular bisector of BC (see Fig). Show that $\mathrm{\Delta }ABC$ is an isosceles triangle in which $AB=AC$ .

Answer:

Consider $\mathrm{\Delta }$ ABD and $\mathrm{\Delta }$ ADC,

(i) $AD=AD$ (Common in both the triangles)

(ii) $\mathrm{\angle }ADB=\mathrm{\angle }ADC$ (Right angle)

(iii) <BD=<CD

Q3 ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig.). Show that these altitudes are equal.

Answer:

Consider $\mathrm{\Delta }AEB$ and $\mathrm{\Delta }AFC$ ,

(i) $\mathrm{\angle }A$ is common in both the triangles.

(ii) $\mathrm{\angle }AEB=\mathrm{\angle }AFC$ (Right angles)

(iii) $AB=AC$ (Given)

Thus by AAS congruence axiom, we can conclude that :

$\mathrm{\Delta }AEB\cong \mathrm{\Delta }AFC$

Now, by c.p.c.t. we can say : $BE=CF$

Hence these altitudes are equal.

Q4 (i) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig). Show that $\mathrm{\Delta }ABE\cong \mathrm{\Delta }ACF$

Answer:

Consider $\mathrm{\Delta }ABE$ and $\mathrm{\Delta }ACF$ ,

(i) $\mathrm{\angle }A$ is common in both the triangles.

(ii) $\mathrm{\angle }AEB=\mathrm{\angle }AFC$ (Right angles)

(iii) $BE=CF$ (Given)

Thus by AAS congruence, we can say that :

$\mathrm{\Delta }ABE\cong \mathrm{\Delta }ACF$

Q4 (ii) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig.). Show that $AB=AC$ , i.e., ABC is an isosceles triangle.

Answer:

From the prevoius part of the question we found out that : $\mathrm{\Delta }ABE\cong \mathrm{\Delta }ACF$

Now, by c.p.c.t. we can say that : $AB=AC$

Hence $\mathrm{\Delta }ABC$ is an isosceles triangle.

Q5 ABC and DBC are two isosceles triangles on the same base BC (see Fig.). Show that $\mathrm{\angle }ABD\cong \mathrm{\angle }ACD$ .

Answer:

Consider $\mathrm{\Delta }ABD$ and $\mathrm{\Delta }ACD$ ,

(i) $AD=AD$ (Common in both the triangles)

(ii) $AB=AC$ (Sides of isosceles triangle)

(iii) $BD=CD$ (Sides of isosceles triangle)

Thus by SSS congruency, we can conclude that :

$\mathrm{\angle }ABD\cong \mathrm{\angle }ACD$

Q6 $\mathrm{\Delta }ABC$ is an isosceles triangle in which $AB=AC$ . Side BA is produced to D such that $AD=AB$ (see Fig.). Show that $\mathrm{\angle }BCD$ is a right angle.

Answer:

Consider $\mathrm{\Delta }$ ABC,
It is given that AB = AC

So, $\mathrm{\angle }ACB=\mathrm{\angle }ABC$ (Since angles opposite to the equal sides are equal.)

Similarly in $\mathrm{\Delta }$ ACD,

We have AD = AB
and $\mathrm{\angle }ADC=\mathrm{\angle }ACD$
So,

$\mathrm{\angle }CAB+\mathrm{\angle }ACB+\mathrm{\angle }ABC={180}^{\circ }$

$\mathrm{\angle }CAB+2\mathrm{\angle }ACB={180}^{\circ }$
or $\mathrm{\angle }CAB={180}^{\circ }-2\mathrm{\angle }ACB$ ...........................(i)

And in $\mathrm{\Delta }$ ADC,
$\mathrm{\angle }CAD={180}^{\circ }-2\mathrm{\angle }ACD$ ..............................(ii)

Adding (i) and (ii), we get :
$\mathrm{\angle }CAB+\mathrm{\angle }CAD={360}^{\circ }-2\mathrm{\angle }ACD-2\mathrm{\angle }ACB$

or ${180}^{\circ }={360}^{\circ }-2\mathrm{\angle }ACD-2\mathrm{\angle }ACB$

and $\mathrm{\angle }BCD={90}^{\circ }$

Q7 ABC is a right angled triangle in which $\mathrm{\angle }A={90}^{\circ }$ and $AB=AC$ . Find $\mathrm{\angle }B$ and $\mathrm{\angle }C$ .

Answer:

In the triangle ABC, sides AB and AC are equal.

We know that angles opposite to equal sides are also equal.

Thus, $\mathrm{\angle }B=\mathrm{\angle }C$

Also, the sum of the interior angles of a triangle is ${180}^{\circ }$ .

So, we have :

$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$

or ${90}^{\circ }+2\mathrm{\angle }B={180}^{\circ }$

or $\mathrm{\angle }B={45}^{\circ }$

Hence $\mathrm{\angle }B=\mathrm{\angle }C={45}^{\circ }$

Q8 Show that the angles of an equilateral triangle are ${60}^{\circ }$ each.

Answer:

Consider a triangle ABC which has all sides equal.

We know that angles opposite to equal sides are equal.

Thus we can write : $\mathrm{\angle }A=\mathrm{\angle }B=\mathrm{\angle }C$

Also, the sum of the interior angles of a triangle is ${180}^{\circ }$ .

Hence, $\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$

or $3\mathrm{\angle }A={180}^{\circ }$

or $\mathrm{\angle }A={60}^{\circ }$

So, all the angles of the equilateral triangle are equal ( ${60}^{\circ }$ ).