The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways. Class 9

The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways. Class 9

Question 1.

The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways. Class 9

Solution:

Sides of triangle = 7 cm, 24 cm, 25 cm

Method 1: Using Area of Right-angled Triangle Property

Consider: 7² + 24² = 49 + 576 = 625 = 25²

∴ Triangle is right-angled with hypotenuse 25 cm

Area of triangle = $\frac{1}{2}$ × Product of sides containing right-angled

= $\frac{1}{2}$ × 7 × 24

= 84

$\frac{1}{2}$ Area of triangle = 84 cm²

Method 2: Using Heron's Formula

∵ Perimeter (s) = $\frac{7+24+25}{2}$ = 28

$\begin{aligned} & \text { Area of the triangle }=\sqrt{s(s-a)(s-b)(s-c)} \\ & =\sqrt{28(28-7)(28-24)(28-25)} \\ & =\sqrt{28 \times 21 \times 4 \times 3} \\ & =\sqrt{4 \times 7 \times 7 \times 3 \times 4 \times 3} \\ & =4 \times 7 \times 3=84\end{aligned}$

∴ Area = 84 cm²

∴ Area of the triangle = 84 cm²