SSC CGL 2025 Tier-I · 19 September 2025, Shift 1 · Quantitative Aptitude · Question 24 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 11 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 is inscribed in a right triangle with legs of length 5 and 12. What is the radius of the circle?

(a)2
(b)3
(c)4
(d)5

Answer

Why

Correct — A.
Hypotenuse: √(5² + 12²) = √169 = 13
Area: ½ × 5 × 12 = 30
Semi-perimeter: s = (5 + 12 + 13) ⁄ 2 = 15

Inradius: r = Area ⁄ s = 30 ⁄ 15 = 2
Check with the right-triangle form r = (a + b − c) ⁄ 2: (5 + 12 − 13) ⁄ 2 = 2
r = 2 → option (a)

Why the others are wrong

(b)
3 — 3 fails the area check: r × s must equal the area, and 3 × 15 = 45, not 30.
(c)
4 — 4 forgets to halve. 5 + 12 − 13 = 4 is twice the radius, so r = 4 ⁄ 2 = 2. Check: 4 × 15 = 60, double the area of 30.
(d)
5 — 5 is the shorter leg, not the radius. A circle of radius 5 would need r × s = 5 × 15 = 75, against an area of 30.

Concept

The incircle touches all three sides. Joining its centre to the vertices splits the triangle into three triangles of height r, and adding their areas gives Area = r × s, where s is the semi-perimeter.

In a right triangle, the centre, the two contact points on the legs and the right-angle vertex form a square of side r. That gives the shortcut r = (a + b − c) ⁄ 2, with c the hypotenuse.

Do not confuse it with the circumradius. A right triangle's hypotenuse is a diameter of its circumcircle, so here R = 13 ⁄ 2 = 6.5.

Key facts

Inradius r = Area ⁄ s for any triangle, s being the semi-perimeter.
For a right triangle with legs a, b and hypotenuse c, r = (a + b − c) ⁄ 2.
The circumradius of a right triangle is half its hypotenuse.

Study next

Inradius and circumradius of an equilateral triangle: a ⁄ (2√3) and a ⁄ √3A sphere inscribed in a cone, through the incircle of its cross-sectionTangent lengths from each vertex to the incircle: s − a, s − b, s − c

Common traps

Forgetting to halve a + b − c, which gives 4
Using the circumradius, half the hypotenuse, for an inscribed circle

15 Sep 2025, 12:30, Quant Q.16 uses r = Area ⁄ s on a cone's cross-section: a triangle with base 14, height 24 and slant sides 25, so r = 168 ⁄ 32 = the keyed 5.25 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 84 SSC CGL questions on it, every solved shift.