SSC CGL 2025 Tier-I · 17 September 2025, Shift 2 · Quantitative Aptitude · Question 23 of 99

We transcribed this text from the pictures on a candidate's response sheet for this shift; the answer is the key SSC colour-marked on that sheet. Solution last reviewed 13 September 2026.

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SSC does not publish CGL question papers or a public answer key. From 2025 the candidate response sheet prints every question and option as a picture, so the text here is transcribed from those pictures for this shift and checked against them independently; where the layout carries meaning, the picture is shown as well. Each answer is the key SSC colour-marked on that sheet. SSC released a tentative key for challenge and later its final one, both only to candidates behind a login, and the sheet does not say which of the two it carries. The solutions are written by UnlockIAS.

In a circle, a chord AB is 12 cm long and is at a distance of 4 cm from the center. What is the radius of the circle?

(a)√52 cm
(b)√65 cm
(c)8 cm
(d)None of these

Answer

Why

Correct — A. The perpendicular from the centre to a chord bisects it.

Halve the chord: 12 ÷ 2 = 6 cm
Right triangle: r² = 6² + 4²
Add: 36 + 16 = 52
Take the root: r = √52 cm (= 2√13 ≈ 7.21 cm) → option (a)

Why the others are wrong

(b)
√65 cm — √65 cm needs a half-chord of 7 cm, since 65 − 4² = 49. That is a 14 cm chord; this one is 12 cm, so the half-chord is 6 and r² = 52.
(c)
8 cm — 8 cm gives r² = 64, so the half-chord would be √(64 − 16) = √48 ≈ 6.93 cm and the chord about 13.9 cm, not 12 cm.
(d)
None of these — √52 is on the list, so None of these cannot be right. 6² + 4² = 52 exactly; the option prints the root unsimplified rather than as 2√13.

Concept

The perpendicular from the centre bisects the chord. So the radius to one end, the distance d and half the chord make a right triangle, with the radius as hypotenuse.

Pythagoras gives r² = d² + (c⁄2)², where c is the chord. Any two of r, d and c fix the third.

√52 simplifies to 2√13 ≈ 7.21 cm. The option prints the unsimplified root, so compare r² values instead of hunting for 2√13.

Key facts

The perpendicular from the centre of a circle to a chord bisects the chord.
Radius² = (distance from centre)² + (half-chord)².
Equal chords of a circle are equidistant from the centre.

Study next

Chord length from radius and distance: c = 2√(r² − d²)Distance between two parallel chords, on the same or opposite sides of the centreChord length from the angle it subtends at the centre

Common traps

Using the full chord instead of half: 12² + 4² = 160 gives √160, which is not offered.
Simplifying √52 to 2√13, not seeing that form printed, and choosing None of these.

The same right triangle decides 18 Sep 2024, 09:00, Quant Q.25: a 32 cm chord at 12 cm gives r = √(16² + 12²) = 20 cm.

22 Sep 2025, 09:00, Quant Q.22 asks for it approximately: a 24 cm chord at 7 cm gives √193 ≈ 13.9, keyed 14 cm.

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 140 SSC CGL questions on it, every solved shift.