Question

Find the area of an isosceles triangle having the base x cm and one side y cm.

Answer 1

Let ABC be an isosceles triangle such that AB = AC = y cm and BC = x cm
Construction: Draw $AD\perp BC$ which bisects BC at D
$\Rightarrow BD=\frac{x}{2}cm$

In $△ABD$,
$A{D}^2+B{D}^2=A{B}^2$
$AD=\sqrt{A{B}^2-B{D}^2}$
$=\sqrt{y^2-{\left(\frac{x}{2}\right)}^2}$
$=\sqrt{y^2-\frac{x^2}{4}}$
$\therefore \text{Area of triangle}=\frac{1}{2}\times \text{base}\times \text{altitude}$
$=\frac{1}{2}\times BC\times AD$
$=\frac{1}{2}\times x\times \sqrt{y^2-\frac{x^2}{4}}$
$=\frac{x}{2}\sqrt{y^2-\frac{x^2}{4}}$
Hence, the required area is $\frac{x}{2}\sqrt{y^2-\frac{x^2}{4}}$