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Class 10: Circles – Sample Problems 18(b-4)

c591Question 31: AB is a diameter of a circle, APBR as shown in the figure. APQ and RBQ are straight lines. Find (i) \angle PRB (ii) \angle PBR (iii) \angle BPR


(i) \angle BAP = \angle BRP = 35^o (angles in the same segment of the circle subtended by the same chord)

(ii) \angle ABQ = 180^o-35^o-25^o=120^o

\angle APB = 90^o (angle in a semi circle subtended by the diameter)

\angle PBA = 180^o-35^o-90^o = 55^o

\therefore \angle PBR = \angle PBA + \angle ABR = 55^o+60^o = 115^o

(iii) \angle BPR = 180^o-35^o-115^o = 30^o


c59Question 32: In the given figure, SP is bisector of \angle RPT and PQRS is a cyclic quadrilateral. Prove that: SQ=SR .


PQRS is a cyclic quadrilateral

\angle SPR = \angle SPT = x

\angle QPR = 180^o-2x

\angle SQR + \angle QPS = 180 (cyclic quadrilateral)

\angle SRQ = 180^o-(180^o-x)= x

Also \angle RQS = \angle RPS = x (angles in the same segment of the circle subtended by the same chord)

\therefore \angle RQS = \angle QRS

\therefore RS=QS


c58Question 33: In the figure, O is the center of the circle, \angle AOE=150^o, \angle DAO=51^o . Calculate \angle CEB and \angle OCE .


\angle ADE = \frac{1}{2} Reflex (\angle AOE) = \frac{1}{2} (360^o-150^o) = 105^o

\angle DAB + \angle DEB = 180^o

\angle DEB = 180^o-51^o=129^o

\angle CEB=180^o-129^o=51^o

\angle OCE = 180^o-51^o-105^o=24^o


c57Question 34: In the given figure, P and Q are the centers of two intersecting circles intersecting at B and C . ACD is a straight line. Calculate the numerical value of x .


\angle ACB = \frac{1}{2} \angle APB =75^o

ACD is a straight line

\angle BCD = 180^o-75^o = 105^o

\angle BCD = \frac{1}{2} (360^o-x)

\Rightarrow x = 360^o-210^o = 150^o


c56Question 35: In the figure given below, two circles intersect at A and B . The center of the smaller circle is O and lines on the circumference of the larger circle. Given \angle APB = a . Find in terms of a the value of (i) Obtuse \angle AOB (ii) \angle ACB (iii) \angle ADB . Give reasons.


(i) \angle APB = \frac{1}{2} (\angle AOB)

\angle AOB = 2a

(ii) \angle BOA + \angle ACB = 180^o \  (AOBC is a cyclic quadrilateral)

\therefore \angle ACB = 180^o-2a

(iii) \angle ADB = \angle ACB (angles in the same segment of the circle subtended by the same chord)

\therefore \angle ADB = 180^o-2a


c55Question 36: In the given figure O is the cent of the circle and \angle ABC=55^o . Calculate x and y .


\angle AOC = 2 \angle ABC \Rightarrow \angle AOC = 110^o = x

ABCD is a cyclic quadrilateral

\angle ADC + \angle ABC = 180^o

\Rightarrow y = 180^o-55^o=125^o


c54Question 37: In the given figure, A is the center of the circle, ABCD is a parallelogram and CDE is a straight line. Prove that \angle BCD = 2 \angle ABE .


 \angle BAD = 2 \angle BED (angle subtended at the center is twice subtended on the circumference by the same chord)

 \angle BED = \angle ABE (alternate angles)

\therefore \angle BAD = 2 \angle ABE

ABCD is a parallelogram

\therefore \angle BAD = \angle BCD (opposite angles in a parallelogram are equal)

 \angle BCD = 2 \angle ABE


c53Question 38: ABCD is a cyclic quadrilateral in which AB is parallel to DC and AB is a diameter of the circle. Given \angle BED = 65^o ; calculate: (i) \angle DAB (ii) \angle BDC .


\angle BED = 65^o

\angle DAB = \angle DEB = 65^o (angles in the same segment of the circle subtended by the same chord)

\angle ADB = 90^o (angle subtended by the diameter on a semi circle)

\angle DBA = 180^o-65^o-90^o = 25^o

\therefore \angle BDC = 25^o (alternate angles)


c52Question 39: In the given figure AB is the diameter of the circle.  Chord ED \parallel AB and \angle EAB=63^o . Calculate (i) \angle EBA (ii) \angle BCD .


(i) \angle AEB = 90^o (angle in the semi circle)

\angle EBA = 180^o-90^o-63^o = 27^o

(ii) ED \parallel AB

\angle DEB = \angle EBA = 27^o (alternate angles)

EBCD is a cyclic quadrilateral

\angle DEB + \angle DCB = 180^o

\angle DCB = 180^o-27^o=153^o


c51.jpgQuestion 40: The sides AB and CD of a cyclic quadrilateral ABCD are produced to meet at E , the sides DA and CB are produced to meet at F . If  \angle BEC = 42^o and \angle BAD=98^o find (i) \angle AFB (ii) \angle ADC


\angle BAF = 180^o-98^o=82^o

\angle DCB = 180^o-98^o = 82^o

\angle BCE = 180^o-82^o = 98^o

\angle CBE = 180^o-98^o-42^o = 40^o

\angle ABF = 40^o (vertically opposite angles)

\angle AFB = 180^o-82^o-40^o=58^o

\angle CBA = 180^o-40^o = 140^o

\angle ADC = 180^o-140^o=40^o (cyclic quadrilateral)


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Class 10: Circles – Sample Problems 18(b-4)


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