SSC CGL 2024 Tier-I · 18 September 2024, Shift 3 · Quantitative Aptitude · Question 13 of 98

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.

Two equal circles pass through each other’s centre. If the radius of each of the circle is 13 cm, then what is the length of the common chord?

Answer

Why

Correct — A. Each circle passes through the other centre, so the distance between the two centres is one radius: O₁O₂ = 13 cm.

The line of centres bisects the common chord at right angles, so each centre stands 13 ⁄ 2 = 6.5 cm from the chord.

Half-chord = √(13² − 6.5²)
= √(169 − 42.25) = √126.75
= 13√3 ⁄ 2

Full chord = 2 × 13√3 ⁄ 2 = 13√3 cm → option (a).

Why the others are wrong

(b)
7√3 is the answer for a radius of 7. The common chord is r√3, so it scales with the radius, and the paper gives r = 13.
(c)
13√2 is the chord that subtends a right angle at the centre. Here the two centres and one intersection point form an equilateral triangle, so the chord subtends 120° and 2r sin 60° = r√3.
(d)
13√6 is about 31.8 cm, longer than the diameter of 26 cm. No chord can exceed the diameter, so this one falls before any calculation.

Concept

Two equal circles that each pass through the other centre make a fixed picture, not a fresh calculation.

The centres are r apart, and joining each centre to one intersection point gives a triangle with all three sides equal to r — equilateral. The two centres and the two intersection points make a rhombus of side r, and the common chord is one of its diagonals.

So the half-chord is r√3 ⁄ 2 and the whole common chord is r√3, whatever r happens to be. For r = 13 that is 13√3, about 22.5 cm — comfortably under the 26 cm diameter.

The property doing the work is the same one behind every chord sum: the perpendicular from the centre bisects the chord, so radius, perpendicular distance and half-chord form a right triangle.

Key facts

When two equal circles of radius r each pass through the other centre, the common chord is r√3.
The line joining the two centres is the perpendicular bisector of the common chord.
Chord length = 2r sin(θ ⁄ 2), where θ is the angle the chord subtends at the centre.
No chord can be longer than the diameter, 2r.

Study next

Chord length from the perpendicular distance to the centre, using half-chord² + distance² = r²Intersecting circles — line of centres, common chord and the rhombus they formEquilateral-triangle relations that appear inside circle figures

Common traps

Taking the distance between centres as 2r rather than r, which makes the circles touch and the chord vanish.
Computing the half-chord, 13√3 ⁄ 2, and marking it without doubling.
Choosing an option longer than the 26 cm diameter.

SSC builds chord questions on the right triangle of radius, perpendicular distance from the centre and half-chord — see 13 Sep 2024, 12:30, Quant Q.18 and 19 Sep 2024, 12:30, Quant Q.16, which both give the distance and ask for a length.

A chord is also asked at Quant Q.23 of this shift, where extending it both ways makes a secant.

Related PYQs

No directly related past PYQ was found.