2 Marks Questions

Question 12 Marks

In given figure, 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 and P, Q respectively. Prove that $\angle$ACP = $\angle$QCD.

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, $\angle$ABC = 69°, $\angle$ACB = 31°, find $\angle$BDC.

Answer

From the given figure, in $\triangle$ABC, we can write
$\angle$ABC + $\angle$ACB + $\angle$BAC = 180^o ( by angle sum property)
69^o + 31^o + $\angle$BAC = 180^o
$\Rightarrow$ $\angle$BAC = 180° -100° = 80°
$\angle$BDC = $\angle$BAC (Angles in the same segment)
$\therefore$ $\angle$BDC = 80°

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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

$\angle$ACB = $\frac{1}{2}$$\angle$AOB
[The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle]
=$\frac{1}{2}$ $\times$ 60^o = 30^o
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]
$\Rightarrow$ $\angle$ADB + 30^o = 180^o
$\Rightarrow$ $\angle$ADB = 180^o - 30^o
$\Rightarrow$ $\angle$ADB = 150^o

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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 AB and AC of a triangle ABC as diameters. They intersect in a point D.
To prove: D lies on the third side BC of $\triangle$ABC.

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

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Question 52 Marks

In figure, A, B, C are three points on a circle with centre O such that $\angle$BOC = 30°, $\angle$AOB = 60°. If D is a point on the circle other than the arc ABC, find $\angle$ADC.

Answer

$\angle$AOC = $\angle$AOB + $\angle$BOC
$\Rightarrow$$\angle$AOC = 60° + 30° = 90°
Now $\angle$AOC = 2$\angle$ADC
[$\because$ Angled subtended by an arc, at the centre of the circle is double the angle subtended by the same arc at any point in the remaining part of the circle]
$\Rightarrow$ $\angle$ADC = $\frac{1}{2}$$\angle$AOC
$\Rightarrow$$\angle$ADC = $\frac{1}{2}$ x 90°= 45°

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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 AB and CD of a circle with centre O intersect within the circle. Their point of intersection is E.
To prove $\angle$OEA = $\angle$OED
Construction: Join OA and OD
Proof: In $\triangle$OEA and $\triangle$OED
OE = OE |Common
OA = OD |Radii of a circle
AE = DE
$\therefore$ $\triangle$OEA $\cong$ $\triangle$OED [SSS Rule]
$\therefore$ $\angle$OEA = $\angle$OED [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 $AOB =$ \angle $COD

To Prove: AB = CD
Proof: In $\triangle $AOB and $\triangle $COD,
AO = CO [Radii of the same circle]
BO = DO [Radii of the same circle]
$\angle $AOB = $\angle $ COD [Given]
$\therefore $ $\triangle$AOB $ \cong$ $\triangle $COD[By SAS congruency]
$ \Rightarrow $ AB = CD [By CPCT]
Hence proved.

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Maths 2 Marks Questions

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