Find area and perimeter of triangle with sides 5cm, 12cm, 13cm

Question

A triangle has sides of length 5 cm, 12 cm, and 13 cm. Find its perimeter and area.

(NCERT Class 7)

Solution — Step by Step

Perimeter = sum of all sides:

P = 5 + 12 + 13 = \mathbf{30 \text{ cm}}

Let’s check the Pythagorean condition: does $a^2 + b^2 = c^2$ ?

5^2 + 12^2 = 25 + 144 = 169 = 13^2

Yes! This is a right-angled triangle with the right angle between the sides of 5 cm and 12 cm, and hypotenuse 13 cm.

For a right triangle, the two legs serve as base and height:

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 12 = \mathbf{30 \text{ cm}^2}

Why This Works

The 5-12-13 triangle is a well-known Pythagorean triplet — a set of three positive integers that satisfy $a^2 + b^2 = c^2$ . Recognising it as a right triangle lets us use the simple $\frac{1}{2} \times b \times h$ formula instead of Heron’s formula.

Other common Pythagorean triplets: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25). Knowing these speeds up problem solving significantly.

Alternative Method — Heron’s Formula

If we didn’t notice it was a right triangle, we could use Heron’s formula:

Semi-perimeter: $s = \frac{30}{2} = 15$

\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{15 \times 10 \times 3 \times 2} = \sqrt{900} = 30 \text{ cm}^2

Same answer — which confirms our calculation.

Always check for a right angle first by testing $a^2 + b^2 = c^2$ (where $c$ is the longest side). If it works, the area calculation becomes much simpler than Heron’s formula. This saves time in exams.

Common Mistake

Students sometimes use $\frac{1}{2} \times 12 \times 13$ (base times hypotenuse) instead of $\frac{1}{2} \times 5 \times 12$ (base times height). The height must be perpendicular to the base. In a right triangle, the two legs are perpendicular to each other — the hypotenuse is NOT the height for either leg.

Question

Solution — Step by Step

Why This Works

Alternative Method — Heron’s Formula

Common Mistake

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