SSC CGL Preparation: Concepts and Practice

Learning outcome

Select triangle-area methods, calculate perimeter separately, and check whether stated sides form a triangle.

Concepts and assumptions

Use ideal exact lengths, consistent units and non-degenerate triangles. Three positive side lengths form a triangle only when the two shortest sum to more than the longest. Equality produces a straight, zero-area arrangement.

A triangle’s perimeter is the sum of its three sides. Area instead depends on a base and its corresponding perpendicular height. Two congruent copies form a parallelogram of the same base and height, so:

Triangle area = base × perpendicular height ÷ 2.

Height is the perpendicular distance from the opposite vertex to the base line. Its foot may lie outside an obtuse triangle. A sloping side is not automatically a height. Base and height alone generally do not determine perimeter.

In a right triangle, the two legs meeting at the right angle provide a valid base-height pair. The hypotenuse satisfies hypotenuse² = leg₁² + leg₂².

With three valid sides a, b and c, Heron’s formula avoids needing a supplied height. Set s = (a + b + c)/2; area is the square root of s(s − a)(s − b)(s − c).

For an equilateral triangle of side a, an altitude bisects the base. Pythagoras gives height = a√3/2, hence area = a²√3/4. Retain square roots exactly here; no rounding is needed.

Worked examples

Example 1 — Base and height. A triangle has base 14 cm and perpendicular height 9 cm. Find its area. Is its perimeter determined?

Area = 14 × 9/2 = 63 cm². The other two sides are unknown, so perimeter cannot be determined from these data alone.

Example 2 — A right triangle. The perpendicular legs are 9 cm and 12 cm. Find area and perimeter.

Hypotenuse² = 9² + 12² = 81 + 144 = 225. Hypotenuse = 15 cm. Area = 9 × 12/2 = 54 cm². Perimeter = 9 + 12 + 15 = 36 cm.

Example 3 — Three sides. A triangle has sides 13 cm, 14 cm and 15 cm. Find its area and height to the 14 cm side.

Validity check: 13 + 14 > 15. Semiperimeter s = (13 + 14 + 15)/2 = 21 cm. Area = √(21 × 8 × 7 × 6) = √7056 = 84 cm². Since 84 = 14 × height/2, height = 168/14 = 12 cm.

Common mistakes

Using two non-perpendicular sides as base and height; taking half the perimeter as area; applying Heron before checking side validity; mistaking a height for a boundary side.

Practice questions

1. A triangle has area 72 cm² and base 18 cm. Find its perpendicular height.
2. A right triangle’s perpendicular legs are 5 m and 12 m. Find area and perimeter.
3. A triangle has sides 10 cm, 10 cm and 12 cm. Find area and perimeter.
4. An equilateral triangle has side 12 cm. Find its perimeter and exact area.

Worked solutions

1. From 72 = 18 × height/2, height = 144/18 = 8 cm.
2. Hypotenuse = √(25 + 144) = 13 m. Area = 5 × 12/2 = 30 m²; perimeter = 5 + 12 + 13 = 30 m.
3. The sides are valid because 10 + 10 > 12. Perimeter = 32 cm, so s = 16 cm. Area = √(16 × 6 × 6 × 4) = √2304 = 48 cm².
4. Perimeter = 3 × 12 = 36 cm. Height = √(12² − 6²) = √108 = 6√3 cm. Area = 12 × 6√3/2 = 36√3 cm², exactly.