Exercise 7.2 (Revised) - Chapter 7 - Triangles - Ncert Solutions class 9 - Maths

Share this to Friend on WhatsApp

Chapter 7 - Triangles | NCERT Solutions for Class 9 Maths

Ex 7.2 Question 1.

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

Answer.

(i) $A B C$ is an isosceles triangle in which $A B=A C$.

$\therefore \angle \mathrm{C}=\angle \mathrm{B}$ [Angles opposite to equal sides]
$\Rightarrow \angle \mathrm{OCA}+\angle \mathrm{OCB}=\angle \mathrm{OBA}+\angle \mathrm{OBC}$
$\because \mathrm{OB}$ bisects $\angle \mathrm{B}$ and $\mathrm{OC}$ bisects $\angle \mathrm{C}$
$\therefore \angle \mathrm{OBA}=\angle \mathrm{OBC}$ and $\angle \mathrm{OCA}=\angle \mathrm{OCB}$
$\Rightarrow \angle \mathrm{OCB}+\angle \mathrm{OCB}=\angle \mathrm{OBC}+\angle \mathrm{OBC}$
$\Rightarrow 2 \angle \mathrm{OCB}=2 \angle \mathrm{OBC}$
$\Rightarrow \angle \mathrm{OCB}=\angle \mathrm{OBC}$

Now in $\triangle \mathrm{OBC}$,
$\angle \mathrm{OCB}=\angle \mathrm{OBC}$ [Prove above]
$\therefore \mathrm{OB}=\mathrm{OC}$ [Sides opposite to equal sides]
(ii) In $\triangle \mathrm{AOB}$ and $\triangle \mathrm{AOC}$,
$\mathrm{AB}=\mathrm{AC}[$ Given $]$
$\angle \mathrm{OBA}=\angle \mathrm{OCA}$ [Given]

And $\angle \mathrm{B}=\angle \mathrm{C}$

$
\begin{aligned}
& \Rightarrow \frac{1}{2} \angle \mathrm{B}=\frac{1}{2} \angle \mathrm{C} \\
& \Rightarrow \angle \mathrm{OBA}=\angle \mathrm{OCA} \\
& \Rightarrow \mathrm{OB}=\mathrm{OC} \text { [Prove above] } \\
& \therefore \triangle \mathrm{AOB} \cong \triangle \mathrm{AOC} \text { [By SAS congruency] } \\
& \Rightarrow \angle \mathrm{OAB}=\angle \mathrm{OAC} \text { [By C.P.C.T.] }
\end{aligned}
$

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

Ex 7.2 Question 2.

In $\triangle A B C, A D$ is the perpendicular bisector of $B C$ (See figure). Show that $\triangle A B C$ is an isosceles triangle in which $A B=A C$.

Answer.

In $\triangle \mathrm{AOB}$ and $\triangle \mathrm{AOC}$,
$B D=C D[A D$ bisects $B C]$
$
\angle \mathrm{ADB}=\angle \mathrm{ADC}=90^{\circ}[\mathrm{AD} \perp \mathrm{BC}]
$
$
\begin{aligned}
& A D=A D[C o m m o n] \\
& \therefore \triangle A B D \cong \triangle A C D \text { [By SAS congruency] } \\
& \Rightarrow A B=A C \text { [By C.P.C.T.] }
\end{aligned}
$

Therefore, $\mathrm{ABC}$ is an isosceles triangle.

Ex 7.2 Question 3.

$A B C$ is an isosceles triangle in which altitudes $B E$ and $C F$ are drawn to sides $A C$ and $A B$ respectively (See figure). Show that these altitudes are equal.

Answer.

In $\triangle \mathrm{ABE}$ and $\triangle \mathrm{ACF}$,
$
\begin{aligned}
& \angle \mathrm{A}=\angle \mathrm{A} \text { [Common] } \\
& \angle \mathrm{AEB}=\angle \mathrm{AFC}=90^{\circ} \text { [Given] } \\
& \mathrm{AB}=\mathrm{AC} \text { [Given] } \\
& \therefore \triangle \mathrm{ABE} \cong \triangle \mathrm{ACF} \text { [By ASA congruency] } \\
& \Rightarrow \mathrm{BE}=\mathrm{CF} \text { [By C.P.C.T.] }
\end{aligned}
$
$\Rightarrow$ Altitudes are equal.

Ex 7.2 Question 4.

$A B C$ is a triangle in which altitudes $B E$ and CF to sides $A C$ and $A B$ are equal (See figure). Show that:
(i) $\triangle \mathrm{ABE} \cong \triangle A C F$
(ii) $A B=A C$ or $\triangle A B C$ is an isosceles triangle.

Answer.

(i) In $\triangle \mathrm{ABE}$ and $\triangle A C F$,
$
\begin{aligned}
& \angle \mathrm{A}=\angle \mathrm{A} \text { [Common] } \\
& \angle \mathrm{AEB}=\angle \mathrm{AFC}=90^{\circ} \text { [Given] } \\
& \mathrm{BE}=\mathrm{CF} \text { [Given] } \\
& \therefore \triangle \mathrm{ABE} \cong \triangle \mathrm{ACF} \text { [By ASA congruency] }
\end{aligned}
$
(ii) Since $\triangle \mathrm{ABE} \cong \triangle \mathrm{ACF}$
$
\Rightarrow \mathrm{BE}=\mathrm{CF} \text { [By C.P.C.T.] }
$
$\Rightarrow \mathrm{ABC}$ is an isosceles triangle.

Ex 7.2 Question 5.

ABC and DBC are two isosceles triangles on the same base BC (See figure). Show that $\angle$ $\mathbf{A B D}=\angle \mathrm{ACD}$

Answer.

In isosceles triangle $\mathrm{ABC}$,
$\mathrm{AB}=\mathrm{AC}[$ Given $]$
$\angle \mathrm{ACB}=\angle \mathrm{ABC}$ $\qquad$ (i) [Angles opposite to equal sides]

Also in Isosceles triangle BCD.
$B D=D C$
$\therefore \angle B C D=\angle C B D$ $\qquad$ (ii) [Angles opposite to equal sides]

Adding eq. (i) and (ii),
$\angle \mathrm{ACB}+\angle \mathrm{BCD}=\angle \mathrm{ABC}+\angle \mathrm{CBD}$
$\Rightarrow \angle \mathrm{ACD}=\angle \mathrm{ABD}$
or $\angle \mathrm{ABD}=\angle \mathrm{ACD}$

Ex 7.2 Question 6.

$\triangle A B C$ is an isosceles triangle in which $A B=A C$. Side $B A$ is produced to $D$ such that $A D=$ $A B$. Show that $\angle B C D$ is a right angle (See figure).

Answer.

In isosceles triangle $\mathrm{ABC}$,
$A B=A C[$ Given $]$
$\angle \mathrm{ACB}=\angle \mathrm{ABC}$ $\qquad$ .(i) [Angles opposite to equal sides]

Now $A D=A B[$ By construction]

But $\mathrm{AB}=\mathrm{AC}[$ Given $]$
$\therefore A D=A B=A C$
$\Rightarrow \mathrm{AD}=\mathrm{AC}$

Now in triangle ADC,
$A D=A C$
$\Rightarrow \angle \mathrm{ADC}=\angle \mathrm{ACD}$ $\qquad$ (ii) [Angles opposite to equal sides]

Since $\angle \mathrm{BAC}+\angle \mathrm{CAD}=180^{\circ}$ $\qquad$ (iii) [Linear pair]

And Exterior angle of a triangle is equal to the sum of its interior opposite angles.
$\therefore$ In $\triangle \mathrm{ABC}$,
$\angle \mathrm{CAD}=\angle \mathrm{ABC}+\angle \mathrm{ACB}=\angle \mathrm{ACB}+\angle \mathrm{ACB}$ [Using (i)]

$
\Rightarrow \angle \mathrm{CAD}=2 \angle \mathrm{ACB}
$

Similarly, for $\triangle \mathrm{ADC}$,
$
\begin{aligned}
& \angle \mathrm{BAC}=\angle \mathrm{ACD}+\angle \mathrm{ADC} \\
& =\angle \mathrm{ACD}+\angle \mathrm{ACD}=2 \angle \mathrm{ACD}
\end{aligned}
$

From eq. (iii), (iv) and (v),
$
\begin{aligned}
& 2 \angle \mathrm{ACB}+2 \angle \mathrm{ACD}=180^{\circ} \\
& \Rightarrow 2(\angle \mathrm{ACB}+\angle \mathrm{ACD})=180^{\circ} \\
& \Rightarrow \angle \mathrm{ACB}+\angle \mathrm{ACD}=90^{\circ} \\
& \Rightarrow \angle \mathrm{BCD}=90^{\circ}
\end{aligned}
$

Hence $\angle \mathrm{BCD}$ is a right angle.

Ex 7.2 Question 7.

$\mathbf{A B C}$ is a right angled triangle in which $\angle \mathbf{A}=90^{\circ}$ and $\mathbf{A B}=\mathbf{A C}$. Find $\angle \mathbf{B}$ and $\angle \mathbf{C}$.

Answer.

$\mathrm{ABC}$ is a right triangle in which,

$
\begin{aligned}
& \angle \mathrm{A}=90^{\circ} \text { And } \mathrm{AB}=\mathrm{AC} \\
& \text { In } \triangle \mathrm{ABC} \\
& \mathrm{AB}=\mathrm{AC} \\
& \Rightarrow \angle \mathrm{C}=\angle \mathrm{B} \ldots \ldots \ldots . . \text { (i) }
\end{aligned}
$

We know that, in $\triangle \mathrm{ABC}$,
$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}$ [Angle sum property]
$
\Rightarrow 90^{\circ}+\angle \mathrm{B}+\angle \mathrm{B}=180^{\circ}
$
$\left[\angle \mathrm{A}=90^{\circ}\right.$ (given) and $\angle \mathrm{B}=\angle \mathrm{C}$ (from eq. (i)]
$
\begin{aligned}
& \Rightarrow 2 \angle \mathrm{B}=90^{\circ} \\
& \Rightarrow \angle \mathrm{B}=45^{\circ}
\end{aligned}
$

Also $\angle \mathrm{C}=45^{\circ}[\angle \mathrm{B}=\angle \mathrm{C}]$

Ex 7.2 Question 8.

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

Answer.

Let $A B C$ be an equilateral triangle.

$
\begin{aligned}
& \therefore A B=B C=A C \\
& \Rightarrow A B=B C \\
& \Rightarrow \angle C=\angle A \ldots
\end{aligned}
$
Similarly, $A B=A C$
$
\Rightarrow \angle \mathrm{C}=\angle \mathrm{B}
$
From eq. (i) and (ii),
$
\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}
$
Now in $\triangle \mathrm{ABC}$
$
\begin{aligned}
& \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ} \ldots \ldots . . . \text { (iv) } \\
& \Rightarrow \angle \mathrm{A}+\angle \mathrm{A}+\angle \mathrm{A}=180^{\circ} \\
& \Rightarrow 3 \angle \mathrm{A}=180^{\circ} \\
& \Rightarrow \angle \mathrm{A}=60^{\circ} \\
& \text { Since } \angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}[\text { From eq. (iii) }] \\
& \therefore \angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=60^{\circ}
\end{aligned}
$

Since $\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}$ [From eq. (iii)]

Hence each angle of equilateral triangle is $60^{\circ}$.