NCERT Solutions for Class 9 Maths Chapter 7 - Exercise

Exercise 7.2

1. In an isosceles triangle $\text{ABC}$, with $\text{AB}=\text{AC}$, the bisectors of $\angle \text{B}$ and $\angle \text{C}$ intersect each other at $O$. Join $\text{A}$ to $\text{O}$. Show that:

(i) $\text{OB}=\text{OC}$

(ii) $AO$ bisects $\angle \text{A}$

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Ans:

(i) It is given that in triangle $\text{ABC}$ is an isosceles triangle with $\text{AB}=\text{AC}$.

$\therefore \,\,\angle \text{ACB}=\angle \text{ABC}$ (Angles opposite to equal sides of a triangle are equal) 

Multiplying $\frac{1}{2}$ on both sides, we get

$\frac{1}{2}\angle \text{ACB}=\frac{1}{2}\angle \text{ABC}$

$\angle \text{OCB}=\angle \text{OBC}$

$\therefore \text{OB}=\text{OC}$ (Sides opposite to equal angles of a triangle are also equal)

(ii) In $\Delta \text{OAB}$ and $\Delta \text{OAC}$,

Side AO is present in both triangle. Therefore,

$\text{AO}=\text{AO}$ (Common)

Length of side AB is equal to AC, i.e., $\text{AB}=\text{AC}$ (Given)

$\text{OB}=\text{OC}$ (Proved above)

Therefore,

$\Delta \text{OAB}\cong \Delta \text{OAC}$ (By SSS axiom of congruency)

Therefore, $\angle \text{BAO}=\angle \text{CAO}$.

As they are corresponding parts of the corresponding triangle.

$\therefore $ AO bisects $\angle \text{A}$.     ($\angle \text{BAO}=\angle \text{CAO}$)

2. In $\Delta \text{ABC},\text{AD}$ is the perpendicular bisector of $\text{BC}$ (see the given figure). Show that $\vartriangle \text{ABC}$ is an isosceles triangle in which $\text{AB}=\text{AC}$.

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Ans:

In $\Delta \text{ADC}$ and $\Delta \text{ADB}$,

Side AD is present in both triangle. 

$\therefore \,\,\text{AD}=\text{AD}$ (Common)

Since AD is the perpendicular bisector of BC.

$\angle ADC=\angle ADB={{90}^{{}^\circ }}$ and

$\text{CD}=\text{BD}$

$\therefore \Delta \text{ADC}\cong \Delta \text{ADB}$ (By SAS axiom of congruency)

Therefore, $\text{AB}=\text{AC}$.

As they are corresponding parts of the corresponding triangle.

Hence, $\Delta ABC$ is an isosceles triangle as two of it’s sides are equal, i.e., $AB=AC$.

3. $\text{ABC}$ is an isosceles triangle in which altitudes $\text{BE}$ and $\text{CF}$ are drawn to equal sides $\text{AC}$ and $\text{AB}$ respectively (see the given figure). Show that these altitudes are equal.

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Ans:

In $\Delta \text{AEB}$ and $\Delta \text{AFC}$,

Since $\text{BE}$ and $\text{CF}$ are drawn to equal sides $\text{AC}$ and $\text{AB}$. Therefore,

$\angle \text{AEB}$ and $\angle \text{AFC}$ (Each \[{{90}^{{}^\circ }}\])

As $\angle A$ lies in both the triangles. Therefore,

$\angle \text{A}=\angle \text{A}$ (Common angle)

Length of side AB is equal to AC, i.e.,$\text{AB}=\text{AC}$ (Given)

$\therefore \Delta \text{AEB}\cong \Delta \text{AFC}$ (By AAS axiom of congruency) 

Therefore, $\text{BE}=\text{CF}$.

As they are corresponding parts of the corresponding triangle.

4.$\text{ABC}$ is a triangle in which altitudes $\text{BE}$ and $\text{CF}$ to sides $\text{AC}$ and $\text{AB}$ are equal (see the given figure). Show that

(i) $\Delta \text{ABE}\cong \Delta \text{ACF}$

(ii) $\text{AB}=\text{AC}$, ie., $\text{ABC}$ is an isosceles triangle.

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Ans:

(i) In $\Delta \text{ABE}$ and $\Delta \text{ACF}$,

$\text{ABC}$ is a triangle in which altitudes $\text{BE}$ and $\text{CF}$ to sides $\text{AC}$ and $\text{AB}$.

Therefore, $\angle \text{ABE}$ and $\angle ACF={{90}^{{}^\circ }}$

Since $\angle A$ is present in both triangles,

$\therefore \,\,\angle \text{A}=\angle \text{A}$ (Common angle)

Length of side BE is equal to CF, i.e., $\text{BE}=\text{CF}$ (Given)

$\therefore \Delta \text{ABE}\cong \Delta \text{ACF}$ (By AAS axiom of congruency)

(ii) Since $\Delta \text{ABE}\cong \Delta \text{ACF}$   (proved above)

Therefore, $\text{AB}=\text{AC}$.

As they are corresponding parts of the corresponding triangle.

5. $\text{ABC}$ and $\text{DBC}$ are two isosceles triangles on the same base $\text{BC}$ (see the given figure). Show that $\angle \text{ABD}=\angle \text{ACD}$.

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Ans:

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Construction: join $\text{AD}$.

In $\Delta \text{ABD}$ and $\Delta \text{ACD}$,

Length of side AB is equal to AC, i.e., $\text{AB}=\text{AC}$ (Given)

Length of side BD is equal to CD, i.e., $\text{BD}=\text{CD}$ (Given)

As side AD is present in both triangles. Therefore,

$\text{AD}=\text{AD}$ (Common side)

$\therefore \Delta \text{ABD}\cong \Delta \text{ACD}$ (By SSS axiom of congruency)

Therefore, $\angle \text{ABD}=\angle \text{ACD}$.

As they are corresponding parts of the corresponding triangle.

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

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Ans:

In $\Delta \text{ABC}$,

Length of side AB is equal to AC, i.e., $\text{AB}=\text{AC}$ (Given)

$\therefore \angle \text{ACB}=\angle \text{ABC}$ (Angles opposite to equal sides of a triangle are also equal)

In $\Delta \text{ACD}$,

Length of side AC is equal to AD, i.e., $\text{AC}=\text{AD}$  (Given)

$\therefore \angle \text{ADC}=\angle \text{ACD}$ (Angles opposite to equal sides of a triangle are also equal)

In $\Delta \text{BCD}$,

$\angle \text{ABC}+\angle \text{BCD}+\angle \text{ADC}={{180}^{{}^\circ }}$ (Sum of angles in a triangle is ${{180}^{{}^\circ }}$)

$\angle \text{ACB}+\angle \text{ACB}+\angle \text{ACD}+\angle \text{ACD}={{180}^{{}^\circ }}$  ($\angle BCD=\angle BCA+\angle ACD$)

$2(\angle \text{ACB}+\angle \text{ACD})={{180}^{{}^\circ }}$

$2(\angle \text{BCD})={{180}^{{}^\circ }}$

$\therefore \angle \text{BCD}={{90}^{{}^\circ }}$

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

Ans:

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It is given that the side $\text{AB}=\text{AC}$

Angles opposite to equal sides are also equal

$\therefore \angle \text{C}=\angle \text{B}$  …(i)

In $\Delta \text{ABC}$,

$\angle \text{A}+\angle \text{B}+\angle \text{C}={{180}^{{}^\circ }}$ (Sum of angles in a triangle is ${{180}^{{}^\circ }}$)${{90}^{{}^\circ }}+\angle \text{B}+\angle \text{C}={{180}^{{}^\circ }}$  …from equation (i)

${{90}^{{}^\circ }}+\angle \text{B}+\angle \text{B}={{180}^{{}^\circ }}$

$2\angle \text{B}={{90}^{{}^\circ }}$

$\angle \text{B}={{45}^{{}^\circ }}$

$\begin{align} & \because \angle \text{B}=\angle \text{C}\,\,\,\,\,\,\,\,\,\,\,...\text{from equation}\,\left( i \right) \\  & \therefore \angle \text{C}={{45}^{{}^\circ }} \\ \end{align}$

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

Ans:

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Consider that $\text{ABC}$ is an equilateral triangle in which all sides are equal.

Therefore, $\text{AB}=\text{BC}=\text{AC}$

$\text{AB}=\text{AC}$

Angles opposite to equal sides of a triangle are equal

$\therefore \angle \text{C}=\angle \text{B}\,\,\,\,\,\,...\left( i \right)$ 

Also, $\text{AC}=\text{BC}$

Angles opposite to equal sides of a triangle are equal

$\therefore \angle \text{B}=\angle \text{A}\,\,\,\,\,\,\,\,\,...\left( ii \right)$ 

Therefore, from equation (i) and (ii), we get

$\angle \text{A}=\angle \text{B}=\angle \text{C}\,\,\,\,\,\,\,\,...\left( iii \right)$

In $\Delta \text{ABC}$,

$\angle \text{A}+\angle \text{B}+\angle \text{C}={{180}^{{}^\circ }}$   (Sum of angles in a triangle is ${{180}^{{}^\circ }}$)

$\angle \text{A}+\angle \text{A}+\angle \text{A}={{180}^{{}^\circ }}$     …from equation (iii)

$3\angle \text{A}={{180}^{{}^\circ }}$

$\angle \text{A}={{60}^{{}^\circ }}$

$\therefore \angle \text{A}=\angle \text{B}=\angle \text{C}={{60}^{{}^\circ }}$

From above we can conclude that all the interior angles of an equilateral triangle has an angle of ${{60}^{{}^\circ }}$.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2

Opting for the NCERT solutions for Ex 7.2 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 7.2 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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