NCERT Solutions Class 9th Maths Chapter – 9 Circles Exercise 9.3
| Textbook | NCERT |
| Class | 9th |
| Subject | Mathematics |
| Chapter | 9th |
| Chapter Name | Circles |
| Category | Class 9th Math Solutions |
| Medium | English |
| Source | Last Doubt |
NCERT Solutions Class 9th Maths Chapter – 9 Circles Exercise 9.3
Chapter – 9
Circles
Exercise 9.3
Question 1. In figure A,B and C are three points on a circle with centre 0 such that ∠BOC = 30° and ∠ AOB = 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° The angle subtended by an arc at the circle is half the angle subtended by it at the Centre. ∴ ∠ ADC = 1/2 (∠AOC) = 1/2(90°) = 45° |
| Question 2. 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 ⇒ ∆AOB is an equilateral triangle. 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 minor arc of the circle.  Hence, the angle subtended by the chord on the minor arc = 150°. Similarly, ∠ADB = 1/2 [∠AOB] = 12 x 60° = 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. 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 pn the circumference. ∴ reflex ∠POR = 2∠PQR But ∠PQR = 100° ∴ reflex ∠POR = 2 x 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] ⇒ ∠OPR = 20∘/2 = 10° |
Question 4. In figure, ∠ABC = 69°,∠ACB = 31°, find ∠BDC. Solution 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. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠BAC. Solution ∠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 diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD. Solution – Since angles in the same segment of a circle are equal. ∴ ∠BAC = ∠BDC ⇒ ∠BDC = 30°  lso, ∠DBC = 70° [Given] In ∆BCD, we have ∠BCD + ∠DBC + ∠CDB = 180° [Sum of angles of a triangle is 180°] ⇒ ∠BCD + 70° + 30° = 180° ⇒ ∠BCD = 180° -100° = 80° Now, in ∆ABC, AB = BC [Given] ∴ ∠BCA = ∠BAC [Angles opposite to equal 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. Solution – Since AC and BD are diameters. ⇒ AC = BD …(i) [All diameters of a circle are equal] Also, ∠BAD = 90° [Angle formed in a semicircle is 90°] Similarly, ∠ABC = 90°, ∠BCD = 90° and ∠CDA = 90°  Now, in ∆ABC and ∆BAD, we have AC = BD [From (i)] AB = BA [Common hypotenuse] ∠ABC = ∠BAD [Each equal to 90°] ∴ ∆ABC ≅ ∆BAD [By RHS congruence criteria] ⇒ BC = AD [C.P.C.T.] Similarly, AB = DC Thus, the cyclic quadrilateral ABCD is such that its opposite sides are equal and each of its angle is a right angle. ∴ ABCD is a rectangle. |
| Question 8. If the non – parallel sides of a trapezium are equal, prove that it is cyclic. Solution – We have a trapezium ABCD such that AB ॥ CD and AD = BC. Let us draw BE ॥ AD such that ABED is a parallelogram. ∵ The opposite angles and opposite sides of a parallelogram are equal. ∴ ∠BAD = ∠BED …(i) and AD = BE …(ii) But AD = BC [Given] …(iii)  ∴ From (ii) and (iii), we have BE = BC ⇒ ∠BCE = ∠BEC … (iv) [Angles opposite to equal sides of a triangle are equal] Now, ∠BED + ∠BEC = 180° [Linear pair] ⇒ ∠BAD + ∠BCE = 180° [Using (i) and (iv)] i.e., A pair of opposite angles of a quadrilateral ABCD is 180°. ∴ ABCD is cyclic. ⇒ The trapezium ABCD is cyclic. |
Question 9. 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 figure). Prove that ∠ ACP = ∠QCD. Solution – Since, angles in the same segment of a circle are equal. ∴ ∠ACP = ∠ABP …(i) Similarly, ∠QCD = ∠QBD …(ii) Since, ∠ABP = ∠QBD …(iii) [Vertically opposite angles] ∴ From (i), (ii) and (iii), we have ∠ACP = ∠QCD |
| Question 10. 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 ∆ABC, and two circles described with diameter as AB and AC respectively. They intersect at a point D, other than A. Let us join A and D.  ∵ AB is a diameter. ∴∠ADB is an angle formed in a semicircle. ⇒ ∠ADB = 90° ……(i) Similarly, ∠ADC = 90° ….(ii) Adding (i) and (ii), 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. |
| Question 11. ABC and ADC are two right angled triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD. Solution – We have ∆ABC and ∆ADC such that they are having AC as their common hypotenuse and ∠ADC = 90° = ∠ABC ∴ Both the triangles are in semi-circle. Case – I: If both the triangles are in the same semi-circle. Case – II : If both the triangles are not in the same semi-circle. |
| Question 12. 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 is 180°. ⇒ ∠A + ∠C = 180° …(i) But ∠A = ∠C …(ii) [Opposite angles of a parallelogram are equal] From (i) and (ii), we have ∠A = ∠C = 90°  Similarly, ∠B = ∠D = 90° ⇒ Each angle of the parallelogram ABCD is 90°. Thus, ABCD is a rectangle. |
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