SSC CGL 2025 Tier-I · 23 September 2025, Shift 1 · Quantitative Aptitude · Question 25 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 triangle ABC has a circle inscribed within it with sides a, b, c. The area of the triangle is A. What is the radius of the inscribed circle?

(a)A⁄(a + b + c)
(b)2A⁄(a + b + c)
(c)(a + b + c)⁄(2A)
(d)A⁄(2(a + b + c))

Answer

Why

Correct — B.
Join the incentre I to A, B and C. That splits ABC into three triangles, each of height r on one side.

Area A = ½ar + ½br + ½cr
= ½r(a + b + c)
r = 2A⁄(a + b + c) → option (b).

With s = (a + b + c)⁄2, this is r = A⁄s.

Why the others are wrong

(a)
A⁄(a + b + c) — A⁄(a + b + c) divides by the whole perimeter instead of the semi-perimeter, so it is exactly half the true inradius.
(c)
(a + b + c)⁄(2A) — (a + b + c)⁄(2A) is the correct formula turned upside down. Its unit is 1⁄length, so it cannot be a radius.
(d)
A⁄(2(a + b + c)) — A⁄(2(a + b + c)) is one quarter of the true inradius: it halves where the formula doubles.

Concept

The incentre is the same distance, r, from all three sides, so the triangle splits into three triangles of height r. Adding their areas gives Area = r × s, where s = (a + b + c)⁄2.

Read the other way, the same relation gives the area when r and the sides are known.

The stem places 'with sides a, b, c' after the circle, but the sides belong to the triangle, and it uses A for both a vertex and the area. In the formula, A is the area.

Key facts

Inradius r = Area ⁄ s, where s = (a + b + c)⁄2.
For a right triangle with legs p, q and hypotenuse h, r = (p + q − h)⁄2.
For an equilateral triangle of side a, r = a⁄(2√3).
Circumradius R = abc⁄(4 × Area).

Study next

Heron's formula for the area from three sidesR : r = 2 : 1 in an equilateral triangleTangent lengths from the vertices to the incircle: s − a, s − b, s − c

Common traps

Dividing by the perimeter instead of the semi-perimeter, which halves the radius, as option (a) does.
Inverting the ratio. A radius must come out in units of length, and (a + b + c)⁄(2A) does not.

The formula is keyed at 19 Sep 2025, 09:00, Quant Q.24 (right triangle with legs 5 and 12 → r = 30⁄15 = 2) and run backwards at 26 Sep 2025, 12:30, Quant Q.23 (sides 5, 12, 13 with inradius 2 cm → area = 2 × 15 = 30 cm²).

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.