Exercise 9.3 (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 9.3 Question 1.

In figure, $\mathbf{A}, \mathbf{B}, \mathbf{C}$ are three points on a circle with centre $\mathrm{O}$ such that $\angle \mathbf{B O C}=30^{\circ}, \angle \mathbf{A O B}=$ $60^{\circ}$. If $\mathrm{D}$ is a point on the circle other than the arc $\mathrm{ABC}$, find $\angle \mathrm{ADC}$.

Answer.

$\angle \mathrm{AOC}=\angle \mathrm{AOB}+\angle \mathrm{BOC}$
$
\Rightarrow \angle \mathrm{AOC}=60^{\circ}+30^{\circ}=90^{\circ}
$

Now $\angle \mathrm{AOC}=2 \angle \mathrm{ADC}$
$[\because$ Angled subtended by an arc, at the centre of the circle is double the angle subtended by the same arc at any point in the remaining part of the circle]
$\Rightarrow \angle \mathrm{ADC}=\frac{1}{2} \angle \mathrm{AOC}$
$\Rightarrow \angle \mathrm{ADC}=\frac{1}{2} \times 90^{\circ}=45^{\circ}$
Ex 9.3 Question 2.

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord on a point on the minor arc and also at a point on the major arc.

Answer.

Let $A B$ be the minor arc of circle.

$\therefore$ Chord $\mathrm{AB}=$ Radius $\mathrm{OA}=$ Radius $\mathrm{OB}$
$\therefore \triangle \mathrm{AOB}$ is an equilateral triangle.
$
\Rightarrow \angle \mathrm{AOB}=60^{\circ}
$

Now $m \overparen{\mathrm{AB}}+m \overparen{\mathrm{BA}}=360^{\circ}$
$
\begin{aligned}
& \Rightarrow \angle \mathrm{AOB}+\angle \mathrm{BOA}=360^{\circ} \\
& \Rightarrow 60^{\circ}+\angle \mathrm{BOA}=360^{\circ}
\end{aligned}
$

$
\Rightarrow \angle \mathrm{BOA}=360^{\circ}-60^{\circ}=300^{\circ}
$
$\mathrm{D}$ is a point in the minor arc.
$
\begin{aligned}
& \therefore m \overparen{\mathrm{BA}}=2 \angle \mathrm{BDA} \\
& \Rightarrow \angle \mathrm{BOA}=2 \angle \mathrm{BDA} \\
& \Rightarrow \angle \mathrm{BDA}=\frac{1}{2} \angle \mathrm{BOA}=\frac{1}{2} \times 300^{\circ} \\
& \Rightarrow \angle \mathrm{BDA}=150^{\circ}
\end{aligned}
$

Thus angle subtended by major arc, $\overparen{\mathrm{BA}}$ at any point $\mathrm{D}$ in the minor arc is $150^{\circ}$.

Let $E$ be a point in the major arc $\overparen{\mathrm{BA}}$.
$
\begin{aligned}
& \therefore m \widetilde{\mathrm{AB}}=2 \angle \mathrm{AEB} \\
& \Rightarrow \angle \mathrm{AOB}=2 \angle \mathrm{AEB} \\
& \Rightarrow \angle \mathrm{AEB}=\frac{1}{2} \angle \mathrm{AOB} \\
& \Rightarrow \angle \mathrm{AEB}=\frac{1}{2} \times 60^{\circ}=30^{\circ}
\end{aligned}
$
Ex 9.3 Question 3.

In figure, $\angle \mathbf{P Q R}=100^{\circ}$ : where $\mathbf{P}, \mathbf{Q}, \mathbf{R}$ are points on a circle with centre $\mathbf{O}$. Find $\angle \mathbf{O P R}$.

Answer.

In the figure, $\mathrm{Q}$ is a point in the minor arc $\overline{\mathrm{PQR}}$.
$
\begin{aligned}
& \therefore m \overparen{R P}=2 \angle \mathrm{PQR} \\
& \Rightarrow \angle \mathrm{ROP}=2 \angle \mathrm{PQR} \\
& \Rightarrow \angle \mathrm{ROP}=2 \times 100^{\circ}=200^{\circ}
\end{aligned}
$

Now $m \overparen{\mathrm{PR}}+m \overparen{\mathrm{RP}}=360^{\circ}$
$
\begin{aligned}
& \Rightarrow \angle \mathrm{POR}+\angle \mathrm{ROP}=360^{\circ} \\
& \Rightarrow \angle \mathrm{POR}+200^{\circ}=360^{\circ} \\
& \Rightarrow \angle \mathrm{POR}=360^{\circ}-200^{\circ}=160^{\circ} \ldots . . \text { (i) }
\end{aligned}
$

Now $\triangle \mathrm{OPR}$ is an isosceles triangle.
$\therefore \mathrm{OP}=\mathrm{OR}$ [radii of the circle]
$\Rightarrow \angle \mathrm{OPR}=\angle \mathrm{ORP}$ [angles opposite to equal sides are equal]

Now in isosceles triangle OPR,
$
\begin{aligned}
& \angle \mathrm{OPR}+\angle \mathrm{ORP}+\angle \mathrm{POR}=180^{\circ} \\
& \Rightarrow \angle \mathrm{OPR}+\angle \mathrm{ORP}+160^{\circ}=180^{\circ} \\
& \Rightarrow 2 \angle \mathrm{OPR}=180^{\circ}-160^{\circ} \text { [Using (i) \& (ii)] } \\
& \Rightarrow 2 \angle \mathrm{OPR}=20^{\circ} \\
& \Rightarrow \angle \mathrm{OPR}=10^{\circ}
\end{aligned}
$
Ex 9.3 Question 4.

In figure, $\angle \mathbf{A B C}=69^{\circ}, \angle \mathbf{A C B}=31^{\circ}$ : find $\angle \mathbf{B D C}$.

Answer.

In triangle $A B C$,
$
\begin{aligned}
& \angle \mathrm{BAC}+\angle \mathrm{ABC}+\angle \mathrm{ACB}=180^{\circ} \\
& \Rightarrow \angle \mathrm{BAC}+69^{\circ}+31^{\circ}=180^{\circ} \\
& \Rightarrow \angle \mathrm{BAC}=180^{\circ}-69^{\circ}-31^{\circ} \\
& \Rightarrow \angle \mathrm{BAC}=80^{\circ} \ldots \ldots . \text { (i) }
\end{aligned}
$

Since, $A$ and $D$ are the points in the same segment of the circle.
$
\therefore \angle \mathrm{BDC}=\angle \mathrm{BAC}
$
[Angles subtended by the same arc at any points in the alternate segment of a circle are equal] $\Rightarrow \angle \mathrm{BDC}=80^{\circ}[$ Using (i) $]$

Ex 9.3 Question 5.

In figure, A, B, C, D are four points on a circle. AC and BD intersect at a point $E$ such that $\angle$ BEC $=130^{\circ}$ and $\angle E C D=20^{\circ}$. Find $\angle$ BAC.

Answer.

Given: $\angle \mathrm{BEC}=130^{\circ}$ and $\angle \mathrm{ECD}=20^{\circ}$
$\angle \mathrm{DEC}=180^{\circ}-\angle \mathrm{BEC}=180^{\circ}-130^{\circ}=50^{\circ}$ [Linear pair]

Now in $\triangle \mathrm{DEC}$,
$\angle \mathrm{DEC}+\angle \mathrm{DCE}+\angle \mathrm{EDC}=180^{\circ}$ [Angle sum property]
$\Rightarrow 50^{\circ}+20^{\circ}+\angle E D C=180^{\circ}$
$\Rightarrow \angle \mathrm{EDC}=110^{\circ}$
$\Rightarrow \angle \mathrm{BAC}=\angle \mathrm{EDC}=110^{\circ}$ [Angles in same segment]

Ex 9.3 Question 6.

$\mathbf{A B C D}$ is a cyclic quadrilateral whose diagonals intersect at a point $\mathrm{E} . \angle \mathrm{DBC}={ }^{70^{\circ}}=\angle \mathrm{BAC}$ is $30^{\circ}$ find $\angle B C D$. Further if $A B=B C$, find $\angle E C D$.

Answer.

For chord CD

$\begin{aligned}
& \angle B C D=80^{\circ} \angle C B D=\angle C A D \text { (Angles in same segment) } \\
& \angle C A D=70^{\circ} \\
& \angle B A D=\angle B A C+\angle C A D \\
& =30^{\circ}+70^{\circ}=100^{\circ} \\
& \angle B C D+\angle B A D=180^{\circ} \text { (Opposite angles of a cyclic quadrilateral) } \\
& \angle B C D+100=180^{\circ} \\
& \angle B C D=80^{\circ} \\
& \text { In } \triangle A B C \\
& \mathrm{AB}=\mathrm{BC} \text { (given) } \\
& \therefore \angle B C A=\angle C A B \text { (Angles opposite to equal sides of a triangle) } \\
& \angle B C A=30^{\circ} \\
& \text { We have } \angle B C D=80^{\circ} \angle B C A+\angle A C D=80^{\circ} \\
& 30^{\circ}+\angle A C D=80^{\circ} \\
& \angle A C D=50^{\circ} \\
& \angle E C D=50^{\circ}
\end{aligned}$

Ex 9.3 Question 7.

If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Answer.

Let $\mathrm{ABCD}$ a cyclic quadrilateral having diagonals as $\mathrm{BD}$ and $\mathrm{AC}$ intersecting each other at point $\mathrm{O}$.

$
\angle B A D=\frac{1}{2} \angle B O D=\frac{180^{\circ}}{2}=90^{\circ}
$
(Consider $\mathrm{BD}$ as a chord) $\angle B C D+\angle B A D=180^{\circ}$ (Cyclic quadrilateral)
$
\angle B C D=180^{\circ}-90^{\circ}=90^{\circ} \angle A D C=\frac{1}{2} \angle A O C=\frac{180^{\circ}}{2}=90^{\circ}
$
(Considering AC as a chord)
$
\begin{aligned}
& \angle A D C+\angle A B C=180^{\circ} \text { (Cyclic quadrilateral) } \\
& 90^{\circ}+\angle A B C=180^{\circ} \\
& \angle A B C=90^{\circ}
\end{aligned}
$

Here, each interior angle of cyclic quadrilateral is of $90^{\circ}$. Hence it is a rectangle.

Ex 9.3 Question 8.

If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Answer.

Given: $A$ trapezium $A B C D$ in which $A B{ }^{\|} C D$ and $A D=B C$.

To prove: The points $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ are concyclic.

Construction: Draw DE ${ }^{\|} \mathrm{CB}$.

Proof: Since $\mathrm{DE}^{\|}{ }_{\mathrm{CB}}$ and $\mathrm{EB} \|_{\mathrm{DC}}$.

$\therefore E B C D$ is a parallelogram.
$\therefore \mathrm{DE}=\mathrm{CB}$ and $\angle \mathrm{DEB}=\angle \mathrm{DCB}$

Now $\mathrm{AD}=\mathrm{BC}$ and $\mathrm{DA}=\mathrm{DE}$
$\Rightarrow \angle \mathrm{DAE}=\angle \mathrm{DEB}$

But $\angle \mathrm{DEA}+\angle \mathrm{DEB}=180^{\circ}$
$\Rightarrow \angle \mathrm{DAE}+\angle \mathrm{DCB}=180^{\circ}$
$[\because \angle \mathrm{DEA}=\angle \mathrm{DAE}$ and $\angle \mathrm{DEB}=\angle \mathrm{DCB}]$
$\Rightarrow \angle \mathrm{DAB}+\angle \mathrm{DCB}=180^{\circ}$
$\Rightarrow \angle \mathrm{A}+\angle \mathrm{C}=180^{\circ}$

Hence, $A B C D$ is a cyclic trapezium.

Ex 9.3 Question 9.

Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D, P, Q respectively (see figure). Prove that $\angle \mathrm{ACP}=$ $\angle$ QCD.

Answer.

In triangles ACD and QCP,
$
\begin{aligned}
& \angle \mathrm{A}=\angle \mathrm{P} \text { and } \angle \mathrm{Q}=\angle \mathrm{D} \text { [Angles in same segment] } \\
& \therefore \angle \mathrm{ACD}=\angle \mathrm{QCP} \text { [Third angles] ..........(i) }
\end{aligned}
$

Subtracting $\angle \mathrm{PCD}$ from both the sides of eq. (i), we get,
$
\begin{aligned}
& \angle \mathrm{ACD}-\angle \mathrm{PCD}=\angle \mathrm{QCP}-\angle \mathrm{PCD} \\
& \Rightarrow \angle \mathrm{ACPO}=\angle \mathrm{QCD}
\end{aligned}
$

Hence proved.

Ex 9.3 Question 10.

If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Answer.

Given: Two circles intersect each other at points $A$ and $B$. AP and $A Q$ be their respective diameters.

To prove: Point $\mathrm{B}$ lies on the third side $\mathrm{PQ}$.

Construction: Join A and B.

Proof: AP is a diameter.

$
\therefore \angle 1=90^{\circ}
$
[Angle in semicircle]
AlsoAQ is a diameter.
$
\therefore \angle 2=90^{\circ}
$
[Angle in semicircle]
$
\begin{aligned}
& \angle 1+\angle 2=90^{\circ}+90^{\circ} \\
& \Rightarrow \angle \mathrm{PBQ}=180^{\circ}
\end{aligned}
$
$\Rightarrow P B Q$ is a line.

Thus point B. i.e. point of intersection of these circles lies on the third side i.e., on $\mathrm{PQ}$.

Ex 9.3 Question 11.

ABC and ADC are two right triangles with common hypotenuse AC. Prove that $\angle$ CAD $=\angle$ ABD.

Answer.

We have $A B C$ and $A D C$ two right triangles, right angled at $B$ and $D$ respectively.

$
\Rightarrow \angle \mathrm{ABC}=\mathrm{ADC}\left[\text { Each } 90^{\circ}\right]
$

If we draw a circle with $\mathrm{AC}$ (the common hypotenuse) as diameter, this circle will definitely passes through of an arc $A C$, Because $B$ and $D$ are the points in the alternate segment of an arc $A C$.

Now we have $\overparen{\mathrm{CD}}$ subtending $\angle \mathrm{CBD}$ and $\angle \mathrm{CAD}$ in the same segment.
$
\therefore \angle C A D=\angle C B D
$

Hence proved.