Chapter 10 Circles Ex 10.5 Solutions
Question - 1 : - In figure A,B and C are three points on a circlewith centre 0 such that ∠BOC= 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ ADC.
Answer - 1 : -
_We have a circle with centre O, such that
_∠_AOB = 60° and _∠_BOC = 30°
_∵∠_AOB + _∠_BOC = _∠_AOC
_∴_ _∠_AOC = 60° + 30° = 90°
_The angle subtended by an arc at the circle ishalf the angle subtended by it at the centre.
_∴_ _∠_ ADC = 
(∠AOC) = (90°)= 45°
Question - 2 : - A chord of a circle is equal to the radius of the circle, find the anglesubtended by the chord at a point on the minor arc and also at a point on themajor arc.
Answer - 2 : -
Wehave a circle having a chord AB equal to radius of the circle.
∴ AO = BO = AB
⇒ ∆AOB is an equilateraltriangle.
Since, each angle of an equilateral triangle is 60°.
⇒ ∠AOB = 60°
Since, the arc ACB makes reflex ∠AOB= 360° – 60° = 300° at the centre of the circle and ∠ACB at a point on the minorarc of the circle.
Hence,the angle subtended by the chord on the minor arc = 150°.
Similarly,, ∠ADB= 1 [∠AOB] = 1 x60° = 30°
Hence, the angle subtended by the chord on the major arc = 30°
Question - 3 : - In figure, ∠PQR= 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
Answer - 3 : -
The angle subtended by an arc of a circle at its centre is twicethe angle subtended by the same arc at a point pn the circumference.
∴ reflex∠POR = 2∠PQR
But ∠PQR =100°
∴ reflex∠POR = 2x 100° = 200°
Since, ∠POR +reflex ∠POR =360°
⇒ ∠POR = 360° – 200°
⇒ ∠POR = 160°
Since, OP = OR [Radii of the same circle]
∴ In∆POR, ∠OPR = ∠ORP
[Angles opposite to equal sides of a triangle are equal]
Also, ∠OPR + ∠ORP + ∠POR = 180°
[Sum of the angles of a triangle is 180°]
⇒ ∠OPR + ∠ORP + 160° = 180°
⇒ 2∠OPR = 180° -160° = 20° [∠OPR = ∠ORP]
Question - 4 : - In figure, ∠ABC= 69°,∠ACB = 31°, find ∠BDC.
Answer - 4 : - In ∆ABC, ∠ABC+ ∠ACB + ∠BAC = 180°
⇒ 69° + 31° + ∠BAC = 180°
⇒ ∠BAC = 180° – 100° = 80°
Since, angles in the same segment are equal.
∴∠BDC = ∠BAC ⇒ ∠BDC = 80
Question - 5 : - In figure, A, B and C are four points on a circle. ACand BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.
Answer - 5 : -
∠BEC = ∠EDC + ∠ECD
[Sum of interior opposite angles is equal to exterior angle]
⇒ 130° =∠EDC +20°
⇒ ∠EDC = 130° – 20° = 110°
⇒ ∠BDC = 110°
Since, angles in the same segment are equal.
∴ ∠BAC = ∠BDC
⇒ ∠BAC = 110°
Question - 6 : - ABCD is a cyclic quadrilateral whose diagonalsintersect at a point E. If ∠DBC= 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC,find ∠ECD.
Answer - 6 : - Since angles in the same segment of a circle are equal.
∴ ∠BAC = ∠BDC
⇒ ∠BDC = 30°
llso, ∠DBC= 70° [Given]
In ∆BCD, we have
∠BCD + ∠DBC + ∠CDB = 180° [Sum of anglesof a triangle is 180°]
⇒ ∠BCD + 70° + 30° = 180°
⇒ ∠BCD = 180° -100° = 80°
Now, in ∆ABC,
AB = BC [Given]
∴ ∠BCA = ∠BAC [Angles opposite toequal sides of a triangle are equal]
⇒ ∠BCA = 30° [∵ ∠B AC = 30°]
Now, ∠BCA + ∠BCD = ∠BCD
⇒ 30° + ∠ECD = 80°
⇒ ∠BCD = 80° – 30° = 50°
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 - 7 : - 
Let ABCD be a cyclicquadrilateral having diagonals BD and AC, intersecting each other at point O.
(Consider BD as achord)
∠BCD+ ∠BAD = 180° (Cyclic quadrilateral)
∠BCD= 180° − 90° = 90°
(Considering AC as achord)
∠ADC+ ∠ABC = 180° (Cyclic quadrilateral)
90° + ∠ABC= 180°
∠ABC= 90°
Eachinterior angle of a cyclic quadrilateral is of 90°. Hence, it is a rectangle.
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Answer - 8 : -
Consider a trapezium ABCD with AB | |CD and BC = AD.
Draw AM ⊥ CD and BN ⊥ CD.
In ΔAMD and ΔBNC,
AD = BC (Given)
∠AMD = ∠BNC (By construction, each is 90°)
AM = BN (Perpendicular distance between two parallel lines is same)
∴ ΔAMD ≅ ΔBNC (RHS congruence rule)
∴ ∠ADC = ∠BCD (CPCT) … (1)
∠BAD and ∠ADC are on the same side of transversal AD.
∠BAD + ∠ADC = 180° … (2)
∠BAD + ∠BCD = 180° [Using equation (1)]
This equation shows that the opposite angles are supplementary.
Therefore, ABCD is a cyclic quadrilateral.
Two circlesintersect at two points B and C. Through B, two line segments ABD and PBQ aredrawn to intersect the circles at A, D and P, Q respectively (see the givenfigure). Prove that ∠ACP = ∠QCD.
Answer - 9 : - 
Join chordsAP and DQ.
For chord AP,
∠PBA = ∠ACP (Angles in the same segment) … (1)
For chord DQ,
∠DBQ = ∠QCD (Angles in the same segment) … (2)
ABD and PBQ are line segments intersecting at B.
∴ ∠PBA = ∠DBQ (Vertically opposite angles) … (3)
From equations (1), (2), and (3), we obtain
∠ACP = ∠QCD
If circles are drawn taking two sides of a triangle asdiameters, prove that the point of intersection of these circles lie on thethird side.
Answer - 10 : - 
Consider aΔABC.
Two circles are drawn while taking AB and AC as thediameter.
Let they intersect each other at D and let D not lie onBC.
Join AD.
∠ADB = 90° (Angle subtended by semi-circle)
∠ADC = 90° (Angle subtended by semi-circle)
∠BDC = ∠ADB + ∠ADC = 90° +90° = 180°
Therefore, BDC is a straight line and hence, ourassumption was wrong.
Thus, Point D lies on third side BC of ΔABC.