SSC CGL 2025 Tier-I · 16 September 2025, Shift 2 · Quantitative Aptitude · Question 17 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.

Where this paper and its answers come from

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.

The area of a regular hexagon is made of how many equilateral triangles?

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

Answer

Why

Correct — D. Join the centre of a regular hexagon to its six vertices.

That cuts it into 6 triangles, one on each side.
Angle at the centre = 360° ÷ 6 = 60°
The two sides meeting at the centre are equal, so each base angle = (180° − 60°) ÷ 2 = 60°

All three angles are 60°, so every triangle is equilateral → option (d)

Why the others are wrong

(a)
3 — 3 is the number of rhombi you get by joining the centre to alternate vertices. Each rhombus holds two equilateral triangles, so the count is still 6.
(b)
4 — 4 is the number of triangles the diagonals from one vertex make (n − 2 for a hexagon). Each of those has a 120° or a 90° angle, so none is equilateral.
(c)
5 — 5 equal triangles around a centre would need centre angles of 360° ÷ 5 = 72°, making them isosceles, not equilateral. That is a regular pentagon's split.

Concept

A regular hexagon has six equal sides and six 120° angles. Joining its centre to the vertices splits it into six congruent triangles whose sides all equal the hexagon's side a.

That is where the area formula comes from: 6 × (√3⁄4)a² = (3√3⁄2)a². It also means the side of a regular hexagon equals the radius of its circumscribed circle.

Smaller equilateral triangles can also tile a hexagon: splitting each of the six into four gives 24. The question means the split through the centre, with triangles as wide as the hexagon's side.

Key facts

Each interior angle of a regular hexagon is 120°, and each angle at its centre is 60°.
Area of a regular hexagon of side a = 6 × (√3⁄4)a² = (3√3⁄2)a².
The side of a regular hexagon equals the radius of its circumscribed circle.

Study next

Area of a regular hexagon inscribed in a circle of given radiusVolume of a prism with a regular hexagonal baseNumber of diagonals of a polygon: n(n − 3)⁄2

Common traps

Using the n − 2 triangulation from one vertex and answering 4, though those triangles are not equilateral.
Mixing up the hexagon's interior angle (120°) with its centre angle (60°).

21 Sep 2025, 16:00, Quant Q.9 puts the six triangles to work: a hexagon inscribed in a circle of radius 14 cm has side 14 cm, so its area is 6 × (√3⁄4) × 14² = 294√3 cm², keyed as 509.21 cm² (with √3 ≈ 1.732).

18 Sep 2025, 09:00, Quant Q.15 uses the base area 6 × (√3⁄4) × 10² = 150√3 cm² for a hexagonal prism 32 cm tall in all, giving 4800√3 cm³.

Related PYQs

No directly related past PYQ was found.

Topic: Triangle, Quadrilaterals, Regular Polygons, Circle — all 97 SSC CGL questions on it, every solved shift.