Class IX Math
NCERT Solution for Circles
NCERT Solution for Circles
EXERCISE 10.5
Q1. In the figure, A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOS = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.
Solution: We have a circle with centre O, such that ∠AOB 60° and ∠BOC = 30°
∵ ∠AOB + ∠BOC = ∠AOC
∴ ∠AOC = 60° + 30° = 90°
Now, the arc ABC subtends ∠AOC = 90° at the centre and ∠ADC at a point D on the circle other than the arc ABC.
Q2. 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.
Solution: We have a circle having a chord AB equal to radius of the circle.
∴ AO = BO = AB
∴ ΔAOB is an equilateral triangle.
Since, each angle of an equilateral = 60°.
⇒ ∠AOB = 60°
Since, the arc ACB makes reflex ∠AOB = 360° – 60° = 300° at the centre of the circle and ∠ABC at a point on the minor arc of the circle. 
Thus, the angle subtended by the chord on the minor arc = 150° and on the major arc = 30°.
Q3. In the figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
Solution: The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the p circumference.
∴ reflex ∠POR = 2∠PQR
But ∠POR = 100°
∴ reflex ∠POR = 2 × 100° = 200°
Since, ∠POR + reflex = ∠POR = 360°
⇒ ∠POR + 200° = 360°
⇒ ∠POR = 360°
⇒ ∠POR = 360° – 200°
⇒ ∠POR = 160°
Since, OP = OR
∴ In ∠POR, ∠OPR = ∠OPR
Also, ∠OPR + ∠ORP + ∠POR = 180°
∠OPR + ∠ORP + 160° = 180°
2∠OPR = 180° – 160° = 20° 
Q4. In the figure, ∠ABC = 69° ∠ACB 31°, find ∠BDC.
Solution: We have, in ΔABC,
∠ABC = 69° and ∠ACB = 31°
But ∠ABC + ∠ACB + ∠BAC = 180°
∴ 69° + 31° + ∠BAC = 180°
⇒ ∠BAC = 180° – 69° – 31° = 80°
Since, angles in the same segment are equal.
∴ ∠BAC = ∠BAC
⇒ ∠BDC = 80°
Q5. In the figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°.Find ∠BAC.
Solution: In ΔCDE,
Exterior ∠BEC = {Sum of interior opposite angles}
[BD is a straight line.]
130° = ∠EDC + ∠ECD
130° = ∠EDC + 20°
⇒ ∠EDC = 130° – 20° = 110°
⇒ ∠BDC = 110°
Since, angles in the same segment are equal.
∠BAC = ∠BDC
⇒ ∠BAC = 110°
Q6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
Solution: ∵ Angles in teh same segment of a circle are equal.
∴ ∠BAC = ∠BDC
⇒ 30° = ∠BDC
Also ∠DBC=70° [Given]
∴In ΔBCD, we have
∠BCD + ∠DBC + ∠CDB = 180° [Sum of angles of a triangle is 180°]
⇒ ∠BCD + 70° + 30° = 180°
⇒ ∠BCD = 180° – 70° – 30° = 80°
Now, in ΔABC,
AB = BC
⇒ ∠BCA = ∠BAC [Angles opposite to equal sides of a triangle are equal]
⇒∠BCA = 30°
Now, ∠BCA = ∠ECD = ∠BCD
⇒ 30° + ∠ECD = 80°
⇒ ∠ECD = 80° – 30° = 50°
Q7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Solution: ∵AC and BD are diameters.
∴ AC = BD [∵All diameters of a circle are equal]
Since a diameter divides a circle into equal parts.
∠ABC = 90°
∠BCD = 90°and ∠CDA = 90°
Now, in right ΔABC and right ΔBAD,
AC = BD [From (1)]
AB = AB [Common]
∴ ABC ≌ ΔBAD [RHS criterion]
⇒ BC = AD [c.p.c.t.]
Similarly, AB = CD
Thus, the cyclic quadrilateral ABCD is such that its opposite sides are equal and each of its angle is right angle.
Q8. If the nonparallel sides of a trapezium are equal, prove that it is cyclic. Solution: We have a trapezium ABCD such that AB || CD and AD = BC.
Solution : Let us draw BE || AD such that ABED is a parallelogram.
∵The opposite angles of a parallelogram are equal.
∴ ∠BAD =∠BED …(1)
and AD = BE [Opposite sides of a parallelogram] …(2)
But AD = BC [Given] …(3)
∴ From (2) and (3), we have
BE = BC
⇒ ∠BEC = ∠BCE [Angles opposite to equal sides of a triangle Δ are equal] …(4)
Now, ∠BED + ∠BEC = 180° [Linear pairs]
⇒ ∠BAD + ∠BCE = 180° [Using (1) and (4)]
i.e. A pair of opposite angles of quadrilateral ABCD is 180°.
∴ ABCD is cyclic.
⇒ The trapezium ABCD is a cyclic.
Q9. 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 (see the figure). Prove that ∠ACP = ∠QCD.
Solution: Since angles in the same segment of a circle are equal.
∴ ∠ACP = ∠ABP …(1)
Similarly, ∠QCD = ∠QBD …(2)
Since ∠ABP = ∠QBD [Vertically opposite angles are equal]
From (1) and (2), we have
∠ACP = ∠QCD
EXERCISE 10.5 Class IX
Q10. 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.
Solution: We have a ΔABC, and two circles described with diameters as AB and AC respectively. They intersect in a point D.
Let us join A and D.
AB is a diameter.
∴ ∠ADE is an angle formed in a semicircle.
⇒ ∠ADB = 90° …(1)
Similarly, ∠ADC = 90° …(2)
Adding (1) and (2), we have
∠ADB + ∠ADC = 90° + 90° = 180°
i.e. B, D and C are collinear points.
⇒ BC is a straight line.
Thus, D lies on BC.
EXERCISE 10.5 Class IX
Q11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Solution: We have right ΔABC and right ΔADC such that they are having AC as their common hypotenuse.
∵ AC is a hypotenuse.
∵ ∠ADC = 90° = ∠ABC
∴ Both the triangles are in the same semicircle.
⇒ A, B, C and D are concyclic.
Let us join B and D.
∵ DC is a chord.
∴ ∠CAD and ∠CBD are formed in the same segment.
⇒ ∠CAD = ∠CBD.
Q12. Prove that a cyclic parallelogram is a rectangle.
Solution: We have a cyclic parallelogram ABCD.
Since, ABCD is a cyclic quadrilateral.
∴ Sum of its opposite angles = 180°
⇒∠A + ∠C = 180° …(1)
But∠A = ∠C …(2)
[Opposite angles of parallelogram are equal]
From (1) and (2), we have
∠A = ∠C = 90°
Similarly, ∠B = ∠D = 90°
⇒ Each angle of the parallelogram ABCD is of 90°.