SSC CGL 2024 Tier-I · 19 September 2024, Shift 1 · Quantitative Aptitude · Question 5 of 100

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 10 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

AB is a chord of length 32 cm of a circle of radius 20 cm, the tangents at A and B intersect at a point T. Find length TA (rounded off to two digits after decimal).

(a)21.33 cm
(b)36.50 cm
(c)26.67 cm
(d)19.93 cm

Answer

Why

Correct — C. Rule: a tangent meets the radius at the point of contact at 90°, and OT bisects chord AB at right angles at its midpoint M.

AM = 32 ⁄ 2 = 16 cm
OM = √(20² − 16²) = √144 = 12 cm

Triangle OAT is right-angled at A with AM as the altitude to hypotenuse OT:
OA² = OM × OT → 400 = 12 × OT → OT = 100⁄3 cm

TA = √(OT² − OA²) = √(10000⁄9 − 400) = √(6400⁄9)
= 80⁄3 = 26.67 cm → option (c)

Why the others are wrong

(a)
21.33 cm — 21.33 cm is 64⁄3, which is TM — the distance from T to the midpoint of the chord. It is a real length in the figure, just not the tangent.
(b)
36.50 cm — 36.50 cm is longer than OT itself (100⁄3 = 33.33 cm). TA is a leg of right triangle OAT whose hypotenuse is OT, so it cannot exceed 33.33 cm.
(d)
19.93 cm — 19.93 cm falls below the 20 cm radius. Since OT = 33.33 cm is more than 20√2, the leg TA comes out larger than OA, so any value under 20 is impossible.

Concept

Two standard circle facts do all the work. The radius is perpendicular to the tangent at the point of contact, and the line joining the centre to the external point bisects the chord of contact at right angles.

That makes OAT a right triangle with the right angle at A and AM its altitude, so the relation OA² = OM × OT applies.

You can also finish without OT: TM = 100⁄3 − 12 = 64⁄3, and TA = √(TM² + AM²) = √(4096⁄9 + 256) = 80⁄3.

The question asks for two decimal places, so 80⁄3 = 26.666… is reported as 26.67 cm. Keep the fraction until the last line — rounding OT to 33.33 early costs you the second decimal.

Key facts

A radius drawn to the point of contact is perpendicular to the tangent there.
The perpendicular from the centre to a chord bisects it, so a 32 cm chord in a 20 cm circle sits 12 cm from the centre.
In a right triangle the leg is the geometric mean of the hypotenuse and its adjacent segment, giving OA² = OM × OT.

Study next

Length of the chord of contact AB when the external distance OT is givenThe power of a point, PA × PB = PT², for a secant and a tangent from the same pointTwo circles touching internally or externally and their common tangent lengths

Common traps

Taking the half-chord as 32 cm instead of 16 cm.
Reporting TM = 21.33 cm, which is a length in the figure but not the tangent.
Assuming TA equals AB ⁄ 2 or some other convenient part of the picture.

SSC asks the tangent-from-an-external-point configuration with numbers chosen to give a clean 12-16-20 triangle, then demands a decimal so the fraction 80⁄3 has to be carried. The companion secant setup appears at Quant Q.15 in this shift.

Related PYQs

No directly related past PYQ was found.