SSC CGL 2024 Tier-I · 23 September 2024, Shift 3 · Quantitative Aptitude · Question 3 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.

If in a circle a chord of length 14 cm is at a distance of 24 cm from its centre, then the length of the radius of the circle is:

(a)28 cm
(b)25 cm
(c)24 cm
(d)31 cm

Answer

Why

Correct — B. The perpendicular dropped from the centre of a circle to a chord bisects that chord, so half the chord is 14 ⁄ 2 = 7 cm.

That perpendicular, the half-chord and the radius form a right triangle with the radius as hypotenuse:
r² = 7² + 24²
= 49 + 576 = 625
r = √625 = 25 cm → option (b)

7, 24, 25 is a Pythagorean triple, so the surd closes cleanly.

Why the others are wrong

(a)
28 cm — 28 cm is just 2 × 14, the chord doubled. It never uses the 24 cm distance at all, and any answer that ignores a given number is suspect.
(c)
24 cm — 24 cm is the centre-to-chord distance, not the radius. That distance is a leg of the right triangle and the radius is its hypotenuse, so the radius must be strictly longer.
(d)
31 cm — 31 cm is 7 + 24, adding the two legs instead of combining them. Pythagoras adds the squares of the legs, and √625 = 25, not 31.

Concept

Three lengths sit in one right triangle here: the radius r, the perpendicular distance d from the centre to the chord, and half the chord.

Because the perpendicular from the centre bisects the chord, the relation is always r² = d² + (chord ⁄ 2)². Give any two of the three and the third follows.

Recognising the Pythagorean triple is the time-saver: 7 and 24 as legs means 25 as hypotenuse, with no square root to evaluate by hand.

The numbers look strange at first: the chord (14 cm) is shorter than its distance from the centre (24 cm). That is perfectly legal — short chords sit far from the centre, and a chord through the centre is the longest one there is.

Key facts

A perpendicular from the centre of a circle to a chord bisects the chord.
r² = d² + (chord ⁄ 2)², where d is the centre-to-chord distance.
7, 24, 25 is a Pythagorean triple, in the same family as 3, 4, 5 and 5, 12, 13.

Study next

Equal chords are equidistant from the centre, and the converseLength of a tangent from an external point, which uses the same right triangleTwo parallel chords and the distance between them, on the same or opposite sides of the centre

Common traps

Putting the full chord 14 into the Pythagorean step instead of the half-chord 7.
Taking 24 cm as the radius because it is the larger of the two numbers given.

The same right triangle is set with the numbers changed at 18 Sep 2024, 09:00, Quant Q.25 — a 32 cm chord 12 cm from the centre, radius 20 cm. It reappears behind tangent lengths at 19 Sep 2024, 09:00, Quant Q.5, where a 32 cm chord in a radius-20 circle sets up the tangents.

Related PYQs

No directly related past PYQ was found.

Topic: Circle and its chords, tangents, angles subtended by chords of a circle, common tangents to two or more circles — all 67 SSC CGL questions on it, every solved shift.