Geometry - Study Mode

[#96] Two tangents AP and AQ are drawn to a circle with centre O from an external point A, where P and Q are points on the circle. If AP = 12 cm and ∠PAQ = 60°, then the length of chord PQ is:
Correct Answer

(A) 12 cm

Explanation

Solution: In ΔAPO 2 unit → 12 1 unit → 6 PT = 6 PQ = 2 × PT = 2 × 6 = 12 cm

[#97] Two tangents PA and PB are drawn from an external point P to a circle with centre O at the points A and B respectively on it, such that ∠APB = 120° and AP = 12.5 cm. The length of OP is:
Correct Answer

(B) 25 cm

Explanation

Solution: ∠APO = 60° ∠POA = 30° 1 unit → 12.5 2 unit → 12.5 × 2 = 25 PO = 25 cm

[#98] In the given figure, PT is a common tangent to three circles at points A, B and C respectively. The radius of the small, medium and large circles is 4 cm, 6 cm and 9 cm. O 1 , O 2 and O 3 are the centre of the three circles what is the value (in cm) of PC?
Correct Answer

(A) 18√6

Explanation

Solution: ΔPO 1 A is similar to ΔPO 2 B AO 1 : BO 2 = 2 : 3 O 1 O 2 = 10 = 3 - 2 = 1 unit 1 unit → 10 PO 2 = 3 units → 30 ΔPO 3 C is similar to ΔPO 2 B. BO 2 : CO 3 = 2 : 3 2 units → 30 1 unit → 15 PO 3 = 45 CO 3 = 9 $$eqalign{
& { ext{PC}} = sqrt {{{45}^2} - {9^2}} cr
& = sqrt {2025 - 81} cr
& = sqrt {1944} cr
& = 18sqrt 6 cr} $$

[#99] In ΔPQR, S is a point on the side QR such that ∠QPS = $$frac{1}{2}$$ ∠PSR, ∠QPR = 78° and ∠PRS = 44°. What is the measure of ∠PSQ?
Correct Answer

(C) 64°

Explanation

Solution: ∠QPS = $$frac{1}{2}$$∠PSR = x (Let) ∠PSR = 2x ∠PQS + ∠QPS = 2x ∠QPR = 78° and ∠PRS = 44° In ΔPQR ∠P + ∠Q + ∠R = 180° 78° + x + 44° = 180° x = 180° - 122° x = 58° ∠PSQ = 180° - 2x = 180° - 2 × 58° = 180° - 116° = 64°

[#100] A secant is drawn from a point P to a circle so that it meets the circle first at A, then goes through the centre, and leaves the circle at B. If the length of the tangent from P to the circle is 12 cm, and the radius of the circle is 5 cm, then the distance from P to A is:
Correct Answer

(A) 8 cm

Explanation

Solution: PQ = 12 cm r = 5 cm PQ 2 = PA × PB 12 2 = x(x + 10) 144 = x(x + 10) x 2 + 10x - 144 = 0 x 2 + (18x - 8x) -144 = 0 x(x + 18) - 8(x + 18) = 0 x = 8, x = -18 PA = 8 cm