9th Maths 7.2

NCERT Class 9th solution of Exercise 7.1

NCERT Class 9th Maths Projects

Exercise 7.2

Q1. In an isosceles triangle `ABC`, with `AB=AC`, the bisectors of `angleB` and `angleC` intersect each other at `O`. Join `A` to `O`. Show that :

i) `OB=OC`    ii) `AO` bisects `angleA`

Sol. :

i) In `triangleABC`m

`AB=AC` [Given]

`angleABC=angleACB` [Angles opposite to equal sides]

`angleOBC=angleOCB` [`OB` and `OC` bisect `angleB` and `angleC`]

`OB=OC` [Sides opposite to equal angles]

Proved.

ii) In `triangleAOB` and `triangleAOC`

`AB=AC` [Given]

`OB=OC` [Proved in (i)]

`OA=OA` [Common]

`triangleAOB=triangleAOC` [By `SSS` rule]

`angleOAB=angleOAC`  [CPCT]

So that, `OA` bisect `angleA`.

Proved.  

Q2. In `triangleABC`, `AD` is the perpendicular bisector of `BC` (see figure). Show that `triangleABC` is an isosceles triangle in which `AB=AC`.

Sol. :

In `triangleADB` and `triangleADC`

`AD=AD` [Common]

`angleADB=angleADC=90^circ` [Given `AD⊥BC`]

`BD=CD` [Given `AD` bisect `BC` at `D`]

`triangleADBcongtriangleADC` [By `SAS` rule]

`AB=AC` [CPCT]

Proved.

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

Sol. :

In `triangleABE` and `triangleACF`

`angleBAE=angleCAF` [Common]

`angleAEB=angleAFC=90^circ` [Given `BE⊥AC` and `CF⊥AB`]

`AB=AC` [Given]

`triangleABEcongtriangleACF`  [By `A``A``S` rule]

`BE=CF`  [CPCT]

So that, altitudes are equal.

Proved.

Q4. `ABC` is a triangle in which altitudes `BE` and `CF` to sides `AC` and `AB` are equal (see figure). Show that 

i) `triangleABEcongtriangleACF`

ii) `AB=AC`, i.e. `ABC` is an isosceles triangle.

Sol. :

i) `triangleABE` and `ACF`

`angleBAE=angleCAF` [Common]

`angleBEA=angleCFA` [`BE⊥AC` and `CF⊥AB`]

`BE=CF` [Given]

`triangleABEcongtriangleACF` [By `A``A``S` rule]

ii) `triangleABEcongtriangleACF` [Proved in (i)]

`AB=AC` [CPCT]

So that,`ABC` is an isosceles triangle.

Proved.

Q5. `ABC` and `DBC` are two isosceles triangles on the same base `BC` (see figure). Show that `angleABC=angleACD`.

Sol. :

In an isosceles `triangleABC`

`AB=AC` [Given]

`angleCBA=angleBCA`_______(1)

[angles opposite to equal sides]

In an isosceles `triangleDBC`

`DB=DC` [Given]

`angleCBD=angleBCD`_______(2)

[angles opposite to equal sides]

We add equation (1) and (2)

`angleCBA+angleCBD=angleBCA+angleBCD`

`angleABD=angleACD`

Proved.

Q6. `triangleABC` is an isosceles triangle in which `AB=AC`. Side `BA` is produced to `D` such that `AD=AB` (see figure). Show that `angleBCD` is a right angle.

Sol. :

In `triangleABC`,

`AB=AC`

`angleABC=angleACB`________(1)

[Angles oppostie to equal sides]

`AB=AD` and `AB=AC` [Given]

`AD=AC`

`angleADC=angleACD`________(2)

[Angles opposite to equal sides]

We add equation (1) and (2)

`angleABC+angleADC=angleACB+angleACD`

`angleDBC+angleBDC=angleBCD`_____(3)

By angle sum property of `triangle`

`angleDBC+angleBDC+angleBCD=180^circ`

`angleBCD+angleBCD=180^circ`[from equation (3)]

`2angleBCD=180^circ`

`angleBDC=180^circ/2`

`angleBDC=90^circ`

So that `angleBDC` is a right angle.

Proved.

Q7. `ABC` is a right angled triangle in which `angleA=90^circ` and `AB=AC`. Find `angleB` and `angleC`.

Sol. :

In `triangleABC`,

`AB=AC` [Given]

`angleB=angleC`_______(1) 

[Angles opposite to equal sides]

By angle sum property of `triangle`

`angleA+angleB+angleC=180^circ`

`90^circ+angleB+angleB=180^circ`

`2angleB=180^circ-90^circ`

`angleB=90^circ/2`

`angleB=45^circ`

So that,

Answer. 

`angleB=angleC=45^circ`

Q8. Show that the angles of an equilateral triangle are `60^circ` each.

Sol. :

Let each angle of an equilateral triangle is `x^circ`.

By angle sum property of `triangle`

`x^circ+x^circ+xcirc=180^circ`

`3x^circ=180^circ`

`x^circ=180^circ/3`

`x^circ=60^circ`

So that, angles of an equilateral triangle is equal to `60^circ` each.

Proved.