SSC CGL 2025 Tier-I · 20 September 2025, Shift 1 · Quantitative Aptitude · Question 22 of 100

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.

A circle that has a radius of 5 centimeters, and there is a chord measuring 8 centimeters in length. What is the distance from the center of the circle to the chord?

(a)3 centimeters
(b)4 centimeters
(c)5 centimeters
(d)6 centimeters

Answer

Why

Correct — A. The perpendicular from the centre bisects the chord, making a right triangle with the radius as hypotenuse.

Half-chord = 8 ÷ 2 = 4 cm
Distance² = radius² − half-chord²
= 5² − 4² = 25 − 16 = 9
Distance = √9 = 3 cm → option (a)

Why the others are wrong

(b)
4 centimeters — 4 cm is the half-chord, the other leg of the right triangle. The distance from the centre is the leg you solve for: √(5² − 4²) = 3 cm.
(c)
5 centimeters — 5 cm is the radius. A line 5 cm from the centre only touches the circle, so the chord would have length zero, not 8 cm.
(d)
6 centimeters — 6 cm is more than the radius of 5 cm. A line that far from the centre misses the circle, so it cannot carry an 8 cm chord.

Concept

The perpendicular from a circle's centre to a chord bisects the chord. That builds a right triangle from three lengths:

the radius r, as hypotenuse
half the chord, c⁄2
the distance d from the centre

So r² = (c⁄2)² + d², and any two of them give the third. Here the sides are 3, 4 and 5.

Use the half-chord, not the full chord: 5² − 8² = −39 is negative, the quickest sign that the halving was skipped.

Key facts

The perpendicular from the centre to a chord bisects the chord.
r² = (half-chord)² + d², where d is the chord's distance from the centre.
Equal chords are equidistant from the centre, and a longer chord lies closer to it.

Study next

Radius from a chord and its distance from the centreTwo parallel chords on the same side or opposite sides of the centreCommon chord of two intersecting circles

Common traps

Putting the full chord, 8, into Pythagoras instead of the half-chord, 4.
Picking 4 cm because it appears in the working: it is the half-chord, not the distance.

18 Sep 2024, 09:00, Quant Q.25 runs the same 3-4-5 shape, scaled by 4, in reverse: a 32 cm chord 12 cm from the centre gives radius √(16² + 12²) = 20 cm.

19 Sep 2024, 12:30, Quant Q.16 asks for the chord: radius 10 cm and distance 6 cm give half-chord 8 cm, so the chord is 16 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.