SSC CGL 2025 Tier-I · 17 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 13 September 2026.

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A circle is inscribed within a right triangle. Considering that the lengths of the two legs measure 6 cm and 8 cm,what is the radius of the inscribed circle?

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

Answer

Why

Correct — A.
Hypotenuse = √(6² + 8²) = √100 = 10 cm
Area = ½ × 6 × 8 = 24 cm²
Semi-perimeter s = (6 + 8 + 10) ⁄ 2 = 12 cm

Inradius r = Area ÷ s = 24 ÷ 12 = 2 cm → option (a)

Right-triangle shortcut, same result: r = (6 + 8 − 10) ⁄ 2 = 2 cm.

Why the others are wrong

(b)
3 cm — 3 cm fails the area check: r × s would be 3 × 12 = 36 cm², but the triangle's area is ½ × 6 × 8 = 24 cm².
(c)
4 cm — 4 cm is 48 ÷ 12, which takes 6 × 8 = 48 as the area and drops the ½. The area is 24 cm², so r = 24 ÷ 12 = 2 cm.
(d)
5 cm — 5 cm is half the hypotenuse, the circumradius of a right triangle, not the radius of the circle drawn inside it.

Concept

The inscribed circle touches all three sides, and its radius r is the perpendicular distance from its centre to each side.

Join that centre to the three corners and the triangle splits into three triangles, each with height r and one side as its base. So Area = ½ × r × (a + b + c) = r × s, where s is the semi-perimeter.

In a right triangle with legs a, b and hypotenuse c, this simplifies to r = (a + b − c) ⁄ 2.

A second check: at the right angle, the incircle's centre and its two touching points form a square of side r. So r equals the tangent length from that corner, s − 10 = 12 − 10 = 2 cm.

Key facts

Inradius of any triangle: r = Area ÷ s, with s the semi-perimeter.
In a right triangle, r = (a + b − c) ⁄ 2, with c the hypotenuse.
In a right triangle, the circumradius is half the hypotenuse.
6, 8, 10 is the 3, 4, 5 triple doubled.

Study next

Circumradius of a right triangle, half the hypotenuseInradius of an equilateral triangle, side × √3 ⁄ 6Tangent lengths from a vertex to the incircle, s minus the opposite side

Common traps

Taking 6 × 8 = 48 as the area and forgetting the ½, which doubles r to 4 cm
Confusing the inscribed circle with the circumscribed one, whose radius here is 10 ÷ 2 = 5 cm

19 Sep 2025, 09:00, Quant Q.24 asks the same with legs 5 and 12: the hypotenuse is 13, so r = (5 + 12 − 13) ⁄ 2 = 2.

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.