In an urban project, triangular plots with side ratios 3:4:5 are allocated. If the perimeter is 60 m, find the area.

Correct — D.
Sides are 3k, 4k, 5k, so perimeter = 12k
12k = 60, so k = 5
Sides: 15 m, 20 m, 25 m

Right-angle test: 15² + 20² = 225 + 400 = 625 = 25²
So the legs are 15 m and 20 m

Area = ½ × 15 × 20 = 150 m² → option (d)

Why the others are wrong

• (a)
  180 m² — 180 m²: the area of a 3k-4k-5k triangle is ½ × 3k × 4k = 6k², and 6k² = 180 needs k = √30 ≈ 5.48 — a perimeter near 65.7 m, not 60 m.
• (b)
  120 m² — 120 m² needs 6k² = 120, so k = √20 ≈ 4.47 and a perimeter near 53.7 m. With a 60 m perimeter the legs are 15 and 20, and half their product is 150.
• (c)
  100 m² — 100 m² needs k² = 100⁄6 ≈ 16.7, so k ≈ 4.08 and a perimeter near 49 m. The given 60 m fixes k at 5 and the area at 6 × 25 = 150.

3 : 4 : 5 is a Pythagorean triple: 3² + 4² = 5², so any triangle with sides in this ratio has a right angle opposite the longest side.

The perimeter fixes the scale. Divide it by the sum of the terms, 3 + 4 + 5 = 12, to get k.

A right triangle's area is half the product of the two legs, so a 3k-4k-5k triangle has area ½ × 3k × 4k = 6k².

Key facts

• Sides 3k, 4k, 5k form a right triangle, with the right angle opposite 5k
• Area of a right triangle = ½ × leg × leg
• A 3 : 4 : 5 triangle with scale k has perimeter 12k and area 6k²
• Other triples worth knowing: 5 : 12 : 13, 8 : 15 : 17 and 7 : 24 : 25

Common traps

• Using the hypotenuse in ½ × a × b — the sides that meet at the right angle are 15 m and 20 m, not 25 m
• Multiplying the unscaled area 6 by k = 5 to get 30 — area grows with k², so it is 6 × 25

No directly related past PYQ was found.