Question

The area of an isosceles triangle having base x cm and one side y cm is _________.

• A. $\frac{x}{2}\sqrt{y^2-\frac{x^2}{2}}c{m}^2$
• B. $\frac{x}{2}\sqrt{\frac{4y^2-x^2}{4}}c{m}^2$
• C. $\frac{x^2-y^2}{4}c{m}^2$
• D. $x\sqrt{4y^2-x^2}c{m}^2$

Answer 1

Answer: (B)

$\frac{x}{2}\sqrt{\frac{4y^2-x^2}{4}}c{m}^2$

If h is the height of the triangle, then

$h^2=y^2-{\frac{x}{4}}^2\Rightarrow h=\sqrt{\frac{4y^2-x^2}{4}}cm$

$\therefore Area=\frac{1}{2}\times base\times h=\frac{x}{2}\sqrt{\frac{4y^2-x^2}{4}}c{m}^2$