SSC CGL 2024 Tier-I · 17 September 2024, Shift 2 · Quantitative Aptitude · Question 2 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 6 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 and CD are two parallel chords drawn in a circle with centre O. The distance between the two chords is 21 cm. If the lengths of AB and CD are 24 cm and 18 cm, respectively, then the radius of the circle is equal to:

(a)15 cm
(b)18 cm
(c)24 cm
(d)20 cm

Answer

Why

Correct — A. Drop a perpendicular from O to each chord; a perpendicular from the centre bisects the chord it meets.

Half of AB = 12, half of CD = 9
Distance to AB = √(r² − 144), distance to CD = √(r² − 81)

The gap of 21 cm is their sum, so the chords straddle O:
√(r² − 144) + √(r² − 81) = 21

Test r = 15:
√(225 − 144) = √81 = 9
√(225 − 81) = √144 = 12
9 + 12 = 21 → option (a)

Why the others are wrong

(b)
18 cm — 18 cm is simply the length of chord CD reused as a radius. At r = 18 the chords sit √180 ≈ 13.4 cm and √243 ≈ 15.6 cm from O, about 29 cm apart, not 21.
(c)
24 cm — 24 cm copies the length of chord AB. That radius puts the chords roughly 20.8 cm and 22.2 cm from the centre, a gap near 43 cm.
(d)
20 cm — 20 cm tempts because 12-16-20 is a Pythagorean triple, so AB would sit a neat 16 cm from O — but CD would then be √319 ≈ 17.9 cm away, giving about 33.9 cm, not 21.

Concept

The perpendicular from the centre to a chord bisects it, so every chord gives a right triangle: legs of half-chord and distance-from-centre, hypotenuse r.

That is the relation r² = a² + d² for a chord of length 2a lying d from the centre.

Two parallel chords admit two arrangements. On opposite sides of O the two distances add to the gap between the chords; on the same side they subtract.

The stem never says which side of O the chords lie on, so both arrangements have to be checked.

The same-side gap is √(r² − 81) − √(r² − 144), which is largest at r = 12 (about 7.9 cm) and shrinks as r grows. It can never reach 21 cm, so the chords must straddle the centre.

Key facts

The perpendicular from the centre of a circle to a chord bisects that chord.
For a chord of length 2a lying d from the centre, r² = a² + d².
Parallel chords on opposite sides of the centre have distances that add to the gap between them, while chords on the same side have distances that subtract.
At r = 15 cm the 24 cm chord lies 9 cm from O and the 18 cm chord lies 12 cm from O.

Study next

Chord length from its distance to the centre, and the reverseEqual chords of a circle and their equal distances from the centreCommon chord of two intersecting circles and the line joining the centres

Common traps

Assuming the chords lie on the same side of O and subtracting the two distances
Using the full chord length instead of half of it in the Pythagorean step
Reusing a printed chord length, 18 or 24, as the radius

SSC sets the parallel-chord pair both ways round, and the arrangement is what the question really tests.

A same-side version, where the distances subtract, runs at 09 Sep 2024, 09:00, Quant Q.4: AB = 10 cm, CD = 24 cm and PQ = 7 cm give a diameter of 26 cm.

Related PYQs

No directly related past PYQ was found.