RowQ
The Vault
RowQ
The Vault
CBSE Class 9 Maths · 11 questions · 26 marks
Every point of a circle is the same distance from its centre, and almost every theorem in this chapter is that one fact in disguise. You will learn how chords behave, why a diameter is special, and how angles standing on the same arc are forced to be equal. Once you can spot the arc an angle stands on, cyclic quadrilateral problems solve themselves.
A chord of length 24 cm is drawn in a circle of radius 13 cm. Its distance from the centre is:
Answer
The distance is 5 cm. The perpendicular from the centre bisects the chord, so half the chord is 24/2 = 12 cm. This forms a right triangle with the radius as hypotenuse: d² + 12² = 13². d² = 169 - 144 = 25, so d = 5 cm.
An arc of a circle subtends an angle of 130° at the centre O. The angle it subtends at a point on the major arc is:
Answer
The angle is 65°. The angle subtended by an arc at the centre is twice the angle it subtends at any point on the remaining part of the circle. So the angle at the major arc = (1/2)(130°) = 65°.
ABCD is a cyclic quadrilateral with ∠A = 95°. The measure of ∠C is:
Answer
∠C = 85°. ∠A and ∠C are opposite angles of a cyclic quadrilateral, and opposite angles of a cyclic quadrilateral are supplementary. So ∠C = 180° - 95° = 85°.
PQ is a diameter of a circle and R is any point on the circle other than P or Q. Then ∠PRQ equals:
Answer
∠PRQ = 90°. The diameter PQ subtends a straight angle of 180° at the centre. The angle at a point on the remaining circumference is half of that, so ∠PRQ = (1/2)(180°) = 90°. This is the standard result that the angle in a semicircle is a right angle.
Assertion (A): The angle in a semicircle is a right angle. Reason (R): The angle subtended by an arc at the centre of a circle is twice the angle subtended by the same arc at any point on the remaining part of the circle.
Answer
Both A and R are true and R is the correct explanation of A. Reason R is the central-angle theorem, which is true. Applying it to a diameter, the semicircular arc subtends 180° at the centre, so the angle at any point on the remaining circumference is (1/2)(180°) = 90°. That is exactly Assertion A, so R does explain A.
In a circle of radius 17 cm, a chord is at a distance of 8 cm from the centre. Find the length of the chord.
Answer
Drop a perpendicular from the centre O to the chord, meeting it at M. Then OM = 8 cm and M is the mid-point of the chord. In the right triangle formed, OM² + (half chord)² = radius². (half chord)² = 17² - 8² = 289 - 64 = 225, so half the chord = 15 cm. Length of the chord = 2 × 15 = 30 cm.
Two parallel chords of lengths 16 cm and 12 cm are drawn on the same side of the centre of a circle of radius 10 cm. Find the distance between the two chords.
Answer
For the 16 cm chord, half its length is 8 cm. Its distance from the centre is d₁, where d₁² = 10² - 8² = 100 - 64 = 36, so d₁ = 6 cm. For the 12 cm chord, half its length is 6 cm. Its distance is d₂, where d₂² = 10² - 6² = 100 - 36 = 64, so d₂ = 8 cm. Both chords lie on the same side of the centre, so the distance between them is the difference of these distances. Distance = 8 - 6 = 2 cm.
In a cyclic quadrilateral PQRS, ∠P = (2x + 15)° and ∠R = (3x - 30)°. Find x and both angles.
Answer
∠P and ∠R are opposite angles of a cyclic quadrilateral, so they are supplementary. (2x + 15) + (3x - 30) = 180 5x - 15 = 180, so 5x = 195 and x = 39. ∠P = 2(39) + 15 = 93° and ∠R = 3(39) - 30 = 87°. Check: 93° + 87° = 180°, as required.
(a) Prove that 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. (b) In a circle with centre O, points B and C lie on the circle and ∠BOC = 104°. A is a point on the major arc BC. Find ∠BAC and ∠OBC.
Answer
(a) Let arc BC subtend ∠BOC at the centre O and ∠BAC at a point A on the remaining circumference. Join AO and produce it to a point D inside the circle's far side. In triangle OAB, OA = OB (radii), so the angles opposite them are equal: ∠OAB = ∠OBA. ∠BOD is an exterior angle of triangle OAB, so ∠BOD = ∠OAB + ∠OBA = 2∠OAB. Similarly in triangle OAC, OA = OC, so ∠OAC = ∠OCA, and the exterior angle ∠COD = 2∠OAC. Adding, ∠BOD + ∠COD = 2(∠OAB + ∠OAC). That is, ∠BOC = 2∠BAC. Proved. (b) ∠BAC = (1/2)∠BOC = (1/2)(104°) = 52°. In triangle OBC, OB = OC (radii), so it is isosceles and ∠OBC = ∠OCB. ∠OBC + ∠OCB + 104° = 180°, so 2∠OBC = 76° and ∠OBC = 38°.
(a) Prove that the opposite angles of a cyclic quadrilateral are supplementary. (b) In cyclic quadrilateral ABCD, ∠A = 70° and ∠B = 85°. Find ∠C and ∠D, and hence find the exterior angle formed by producing side AB beyond B.
Answer
(a) Let ABCD be a cyclic quadrilateral with centre O. Join OB and OD. The arc BCD (the one not containing A) subtends ∠BOD at the centre and ∠A at the point A, so ∠BOD = 2∠A. The arc BAD (the one not containing C) subtends the reflex angle ∠BOD at the centre and ∠C at the point C, so reflex ∠BOD = 2∠C. The two central angles together make a complete turn: ∠BOD + reflex ∠BOD = 360°. So 2∠A + 2∠C = 360°, giving ∠A + ∠C = 180°. Since the angle sum of the quadrilateral is 360°, it follows that ∠B + ∠D = 360° - 180° = 180° as well. Proved. (b) ∠C = 180° - ∠A = 180° - 70° = 110°. ∠D = 180° - ∠B = 180° - 85° = 95°. Check: 70 + 85 + 110 + 95 = 360°. Producing AB beyond B gives an exterior angle that forms a linear pair with ∠B, so it equals 180° - 85° = 95°. Note this equals the interior opposite angle ∠D, illustrating the exterior-angle property of a cyclic quadrilateral.
Read the following and answer the questions that follow: A circular park has centre O and radius 25 m. A straight paved walkway AB of length 40 m is laid across the park as a chord. Later a second straight walkway CD, parallel to the first but on the opposite side of the centre, is laid with length 30 m. (a) Find the distance of walkway AB from the centre O. (b) Find the distance of walkway CD from the centre O. (c) Find the distance between the two walkways. (d) What is the greatest possible length of a straight walkway across this park, and where must it lie?
Answer
(a) The perpendicular from O to AB bisects it, so half of AB is 20 m. Distance² = 25² - 20² = 625 - 400 = 225, so the distance of AB from O is 15 m. (b) Half of CD is 15 m. Distance² = 25² - 15² = 625 - 225 = 400, so the distance of CD from O is 20 m. (c) The two chords lie on opposite sides of the centre, so the distance between them is the sum of their distances from O. Distance = 15 + 20 = 35 m. (d) The longest chord of a circle is its diameter, so the greatest possible walkway is 2 × 25 = 50 m long, and it must pass through the centre O. This matches the rule that the nearer a chord is to the centre, the longer it is — a chord at distance 0 is the longest of all.
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