Exercise 9.2 (Revised) - Chapter 10 - Circles - Ncert Solutions class 9 - Maths

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Chapter 9: Circles - NCERT Solutions for Class 9 Maths | Expert Answers & PDF

Ex 9.2 Question 1.

Two circles of radii $5 \mathrm{~cm}$ and $3 \mathrm{~cm}$ intersect at two points and the distance between their centers is $4 \mathrm{~cm}$. Find the length of the common chord.

Answer.

Let two circles with centres $O$ and $O$ ' intersect each other at points $A$ and $B$. On joining $A$ and $B$, $A B$ is a common chord.

Radius $\mathrm{OA}=5 \mathrm{~cm}$, Radius $\mathrm{O}^{\prime} \mathrm{A}=3 \mathrm{~cm}$,

Distance between their centers $O \mathrm{O}^{\prime}=4 \mathrm{~cm}$

In triangle $\mathrm{AOO}$,
$
\begin{aligned}
& 5^2=4^2+3^2 \\
& \Rightarrow 25=16+9 \\
& \Rightarrow 25=25
\end{aligned}
$

Hence $A O O^{\prime}$ is a right triangle, right angled at $O^{\prime}$.

Since, perpendicular drawn from the center of the circle bisects the chord.

Hence $\mathrm{O}^{\prime}$ is the mid-point of the chord $\mathrm{AB}$. Also $\mathrm{O}^{\prime}$ is the centre of the circle II.

Therefore length of chord $A B=$ Diameter of circle II
$\therefore$ Length of chord $\mathrm{AB}=2 \times 3=6 \mathrm{~cm}$.

To prove: (a) $A E=C E$
(b) $B E=D E$

Construction: Draw $\mathrm{OM} \perp \mathrm{AB}$, $\mathrm{ON} \perp \mathrm{CD}$. Also join $\mathrm{OE}$.

Proof: In right triangles OME and ONE,
$\angle \mathrm{OME}=\angle \mathrm{ONE}=90^{\circ}$
$\mathrm{OM}=\mathrm{ON}$
[Equal chords are equidistance from the centre]
$\mathrm{OE}=\mathrm{OE}[$ Common]
$\therefore \triangle \mathrm{OME} \cong \triangle$ ONE [RHS rule of congruency]
$\therefore M E=N E[B y C P C T]$

Now, $\mathrm{O}$ is the centre of circle and $\mathrm{OM} \perp \mathrm{AB}$
$\therefore A M=\frac{1}{2} A B$
[Perpendicular from the centre bisects the chord] $\qquad$
Similarly, $\mathrm{NC}=\frac{1}{2} \mathrm{CD}$ $\qquad$ .(iii)

But $A B=C D$ [Given]

From eq. (ii) and (iii), $\mathrm{AM}=\mathrm{NC}$ $\qquad$ (iv)

Also $\mathrm{MB}=\mathrm{DN}$ $\qquad$ (V)

Adding (i) and (iv), we get,
$A M+M E=N C+N E$
$\Rightarrow \mathrm{AE}=\mathrm{CE}$ [Proved part (a)]

Now $A B=C D$ [Given]
$\mathrm{AE}=\mathrm{CE}[$ Proved $]$
$\Rightarrow A B-A E=C D-C E$
$\Rightarrow B E=D E$ [Proved part (b)]

Ex 9.2 Question 3.

If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chord.

Answer.

Given: $A B$ and $C D$ be two equal chords of a circle with centre $O$ intersecting each other with in the circle at point $E$. OE is joined.

To prove: $\angle \mathrm{OEM}=\angle \mathrm{OEN}$

Construction: Draw $\mathrm{OM} \perp \mathrm{AB}$ and
$O N \perp C D$.

Proof: In right angled triangles OME and ONE,
$\angle \mathrm{OME}=\angle \mathrm{ONE}\left[\right.$ Each $90^{\circ}$ ]
$\mathrm{OM}=\mathrm{ON}$ [Equal chords are equidistant from the centre]
$O E=O E[$ Common $]$
$\therefore \triangle \mathrm{OME} \cong \triangle \mathrm{ONE}[R H S$ rule of congruency]
$\therefore \angle \mathrm{OEM}=\angle \mathrm{OEN}$ [By CPCT]

Ex 9.2 Question 4.

If a line intersects two concentric circles (circles with the same centre) with centre $O$ at $A$, $B, C$ and $D$, prove that $A B=C D$. (See figure)

Answer.
Given: Line $l$ intersects two concentric circles with centre $O$ at points $A, B, C$ and $D$.

To prove: $A B=C D$

Construction: Draw $\mathrm{OL} \perp l$

Proof: AD is a chord of outer circle and $\mathrm{OL} \perp \mathrm{AD}$.
$\therefore A L=L D$ $\qquad$ .(i) [Perpendicular drawn from the centre bisects the chord]

Now, $\mathrm{BC}$ is a chord of inner circle and
$\mathrm{OL} \perp \mathrm{BC}$
$\therefore B L=L C$...(ii) [Perpendicular drawn from the centre bisects the chord]

Subtracting (ii) from (i), we get,
$A L-B L=L D-L C$
$\Rightarrow A B=C D$

Ex 9.2 Question 5.

Three girls Reshma, Salma and Mandip are standing on a circle of radius $\mathbf{5} \mathbf{~ m}$ drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is $6 \mathrm{~m}$ each, what is the distance between Reshma and Mandip?

Answer.

Let Reshma, Salma and Mandip takes the position $C, A$ and $B$ on the circle. Since $A B=A C$

The centre lies on the bisector of $\angle \mathrm{BAC}$.

Let $M$ be the point of intersection of $B C$ and $O A$.

Again, since $A B=A C$ and $A M$ bisects
$\angle \mathrm{CAB}$.
$\therefore A M \perp C B$ and $M$ is the mid-point of $C B$.

Let $O M=x$, then $M A=5-x$

From right angled triangle $\mathrm{OMB}$,
$O B^2=O M^2+M B^2$
$
\Rightarrow 5^2=x^2+\mathrm{MB}^2
$

Again, in right angled triangle AMB,
$
\begin{aligned}
& A B^2=A M^2+M B^2 \\
& \Rightarrow 6^2=(5-x)^2+M B^2 .
\end{aligned}
$

Equating the value of $M B^2$ from eq. (i) and (ii),
$
\begin{aligned}
& 5^2-x^2=6^2-(5-x)^2 \\
& \Rightarrow(5-x)^2-x^2=6^2-5^2
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow 25-10 x+x^2-x^2=36-25 \\
& \Rightarrow 10 x=25-11 \\
& \Rightarrow 10 x=14 \\
& \Rightarrow x=\frac{14}{10}
\end{aligned}
$

Hence, from eq. (i),
$
\begin{aligned}
& M B^2=5^2-x^2=5^2-\left(\frac{14}{10}\right)^2 \\
& =\left(5+\frac{4}{10}\right)\left(5-\frac{14}{10}\right)=\frac{64}{10} \times \frac{36}{10}
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow M B=\frac{8 \times 6}{10}=4.8 \mathrm{~cm} \\
& \therefore B C=2 M B=2 \times 4.8=9.6 \mathrm{~cm}
\end{aligned}
$

Ex 9.2 Question 6.

A circular park of radius $\mathbf{2 0} \mathrm{m}$ is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Answer.

Let position of three boys Ankur, Syed and David are denoted by the points A, B and C respectively.

$
\mathrm{A}=\mathrm{B}=\mathrm{C}=a[\text { say }]
$

Since equal sides of equilateral triangle are as equal chords and perpendicular distances of equal chords of a circle are equidistant from the centre.
$
\therefore O D=O E=O F=x \mathrm{~cm}[\mathrm{say}]
$

Join $\mathrm{OA}, \mathrm{OB}$ and $\mathrm{OC}$.
$\Rightarrow$ Area of $\triangle \mathrm{AOB}$
$=$ Area of $\triangle \mathrm{BOC}=$ Area of $\triangle \mathrm{AOC}$

And Area of $\triangle \mathrm{ABC}$
$=$ Area of $\triangle \mathrm{AOB}+$ Area of $\triangle \mathrm{BOC}+$ Area of $\triangle \mathrm{AOC}$
$\Rightarrow$ And Area of $\triangle \mathrm{ABC}=3 \times$ Area of $\triangle \mathrm{BOC}$

$
\begin{aligned}
& \Rightarrow \frac{\sqrt{3}}{4} a^2=3\left(\frac{1}{2} \mathrm{BC} \times \mathrm{OE}\right) \\
& \Rightarrow \frac{\sqrt{3}}{4} a^2=3\left(\frac{1}{2} \times a \times x\right) \\
& \Rightarrow \frac{a^2}{a}=3 \times \frac{1}{2} \times \frac{4}{\sqrt{3}} \times x \\
& \Rightarrow a=2 \sqrt{3} x \ldots \ldots . . \text { (i) }
\end{aligned}
$

Now, $C E \perp B C$
$\therefore B E=E C=\frac{1}{2} B C \quad[\because$ Perpendicular drawn from the centre bisects the chord $]$
$
\begin{aligned}
& \Rightarrow \mathrm{BE}=\mathrm{EC}=\sqrt{3} x \\
&
\end{aligned}
$

Now in right angled triangle BEO,
$\mathrm{OE}^2+\mathrm{BE}^2=\mathrm{OB}^2$ [Using Pythagoras theorem $]$
$
\Rightarrow x^2+(\sqrt{3} x)^2=(20)^2
$

$
\begin{aligned}
& \Rightarrow x^2+3 x^2=400 \\
& \Rightarrow 4 x^2=400 \\
& \Rightarrow x^2=100 \\
& \Rightarrow x=10 \mathrm{~m}
\end{aligned}
$

And $a=2 \sqrt{3} x=2 \sqrt{3} \times 10=20 \sqrt{3} \mathrm{~m}$

Thus distance between any two boys is $20 \sqrt{3} \mathrm{~m}$.