A wire, 112 cm long, is bent to form a right angled triangle. If the hypotenuse is 50 cm long, find the area of the triangle.

Sum of length of other two sides + Hypotenuse = 112

or, Sum of length of other two sides = 112 - Hypotenuse

∴ Sum of length of other two sides = 112 - 50 = 62 cm

Let the length of perpendicular = x cm

So, the length of base = (62 - x) cm

For right angled triangle,

(Perpendicular)² + (Base)² = (Hypotenuse)²

$\therefore x^2 + (62 - x)^2 = (50)^2 \\[1em] \Rightarrow x^2 + 3844 + x^2 - 124x = 2500 \\[1em] \Rightarrow 2x^2 - 124x + 3844 - 2500 = 0 \\[1em] \Rightarrow 2x^2 - 124x + 1344 = 0 \\[1em] \Rightarrow 2(x^2 - 62x + 672) = 0 \\[1em] \Rightarrow x^2 - 62x + 672 = 0 \\[1em] \Rightarrow x^2 - 48x - 14x + 672 = 0 \\[1em] \Rightarrow x(x - 48) - 14(x - 48) = 0 \\[1em] \Rightarrow (x - 48)(x - 14) = 0 \\[1em] \Rightarrow x - 48 = 0 \text{ or } x - 14 = 0 \\[1em] x = 48 \text{ or } x = 14$

∴ The length of other two sides are 48 cm and 14 cm.

$\text{Area} = \dfrac{1}{2} \times \text{Perpendicular} \times \text{Base} \\[1em] = \dfrac{1}{2} \times 48 \times 14 \\[1em] = 336$

Hence, the area of triangle is 336 cm^2.