2 Marks Questions

Question 12 Marks

In given figure, two circles intersect at two points $B$ and $C$. Through $B$, two line segments $A B D$ and $P B Q$ are drawn to intersect the circles at $A, D$ and $P, Q$ respectively. Prove that $\angle A C P=\angle Q C D$.

Answer

As angles in the same segment of circle are equal
$\angle ABP = \angle ACP ..........................(i)$
$\angle ABP =\angle QBD$ (Vertically opposite angles)
Also, $\angle QCD =\angle QBD$ (Angles in the same segment)
$\therefore \angle ABP=\angle QCD..........(ii)$
From $(i)$ and $(ii),$ we have
$\angle ACP =\angle QCD$.

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Question 22 Marks

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

Answer

From the given figure, in $\triangle A B C$, we can write
$\angle A B C+\angle A C B+\angle B A C=180^{\circ}(\text { by angle sum property })$
$69^{\circ}+31^{\circ}+\angle B A C=180^{\circ}$
$\Rightarrow \angle B A C=180^{\circ}-100^{\circ}=80^{\circ}$
$\angle B D C=\angle B A C \text { (Angles in the same segment) }$
$\therefore \angle B D C=80^{\circ}$

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Question 32 Marks

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

Answer

$OA = OB = AB$ |Given
$\therefore$ $\triangle OAB$ is equilateral
$\therefore$ $\angle AOB = 60^o$

$\angle A C B=\frac{1}{2} \angle A O B$
[The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle]
$=\frac{1}{2} \times 60^{\circ}=30^{\circ}$
Now, $\because$ ADBC is a cyclic quadrilateral.
$\therefore \angle ADB +$ $\angle ACB$ [The sum of either pair of opposite angles of a cyclic quadrilateral is $180^o$]
$\Rightarrow \angle A D B+30^{\circ}=180^{\circ}$
$\Rightarrow \angle A D B=180^{\circ}-30^{\circ}$
$\Rightarrow \angle A D B=150^{\circ}$

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Question 42 Marks

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

Answer

Given: Circles are described with sides $A B$ and $A C$ of a triangle $A B C$ as diameters. They intersect in a point $D$. To prove: $D$ lies on the third side $B C$ of $\triangle A B C$.

Construction: Join $AD.$
Proof: Circle described on $A B$ as diameter intersects $B C$ in $D$.
$\angle A D B=90^{\circ}$ [Angle in a semi-circle]
But $\angle ADB +\angle ADC =180^{\circ}$ [Linear Pair Axiom]
$\therefore \angle A D C=90^{\circ}$
Hence, the circle described on $AC$ as diameter must pass through $D.$
Thus, the two circles intersect in $D .$
Now $\angle A D B+\angle A D C=180^{\circ}$
$\therefore$ Points B, D, C are collinear
$\therefore$ D lies on $B C$.

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Question 52 Marks

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

Answer

$\angle A O C=\angle A O B+\angle B O C$
$\Rightarrow \angle A O C=60^{\circ}+30^{\circ}=90^{\circ}$
Now $\angle A O C=2 \angle A D C$
[ $\because$ 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]
$\Rightarrow \angle A D C=\frac{1}{2} \angle A O C$
$\Rightarrow \angle A D C=\frac{1}{2} \times 90^{\circ}=45^{\circ}$

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Question 62 Marks

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 chords.

Answer

Given: Two equal chords $A B$ and $C D$ of a circle with centre $O$ intersect within the circle. Their point of intersection is $E.$
To prove $\angle O E A=\angle O E D$
Construction: Join $OA$ and $OD$
Proof: In $\triangle O E A$ and $\triangle O E D$
$OE = OE \mid $ Common
$O A=O D \mid$ Radii of a circle
$A E=D E$
$\therefore \triangle O E A \cong \triangle O E D[$ SSS Rule]
$\therefore \angle O E A=\angle O E D$ [c.p.c.t]

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Question 72 Marks

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

Answer

We know that if two circles intersect each other at two points, then the line joining their centres is the perpendicular bisector of their common chord.
$\therefore$Length of the common chord
$= PQ =2O'P$
$= 2 \times 3cm = 6cm$

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Question 82 Marks

Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Answer

Given: In a circle ( $O$ ), AB and CD subtend two angles at the centre such that $\angle A O B=\angle C O D$

To Prove: $A B=C D$
Proof: In $\triangle A O B$ and $\triangle C O D$,
$A O=C O$ [Radii of the same circle]
$B O=D O$ [Radii of the same circle]
$\angle A O B=\angle C O D$ [Given]
$\therefore \triangle AOB \cong \triangle COD [ By$ SAS congruency]
$\Rightarrow AB=CD[By CPCT]$
Hence proved.

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Question 92 Marks

Given an arc of a circle, complete the circle.

Answer

Let arc $PQ$ of a circle be given. We have to complete the circle, which means that we have to find its centre and radius. Take a point $R$ on the arc. Join $PR$ and $RQ$. Use the construction that has been used in proving There is one and only one circle passing through three given non-collinear points., to find the centre and radius.
Taking the centre and the radius so obtained, we can complete the circle (See Figure).

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