Question

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

Answer 1

In isosceles $\Delta$ ABC,
AB= AC= y cm
BC= x cm

$\therefore$ Draw AD $\perp$ BC which bisects BC at D
$\therefore BD=\frac{x}{2}\therefore AD=\sqrt{A{B}^2-B{D}^2}=\sqrt{y^2-\frac{x^2}{4}}$
$\therefore$ Area = $\frac{Base\times altitude}{2}=\frac{1}{2}BC\times AD$
=$\frac{1}{2}\times x\times \sqrt{y^2-\frac{x^2}{4}=\frac{x}{2}}\sqrt{y^2-\frac{x^2}{4}}c{m}^2$