Question

The area of an isosceles triangle whose base is ${}^{\prime }{b}^{\prime }$ and equal sides are of length ${}^{\prime }{a}^{\prime }$ is:

• A. $\frac{a}{4}\sqrt{4b^2-a^2}$
• B. $\frac{b}{4}\sqrt{4a^2-b^2}$
• C. $\frac{b}{2}\sqrt{4b^2-a^2}$
• D. $\frac{a}{2}\sqrt{4a^2-b^2}$

Answer 1

Answer: (B)

$\frac{b}{4}\sqrt{4a^2-b^2}$

In an isosceles triangle, the altitude bisects the base.

$Altitude=\sqrt{(sidelengthofequalsides{)}^2+{\left(\frac{base}{2}\right)}^2}=\sqrt{a^2-\frac{b^2}{4}}$

Area of isosceles triangle :

$=\frac{1}{2}\times altitude\times base=\frac{1}{2}\times \sqrt{a^2-\frac{b^2}{4}}\times b=\frac{1}{4}\times b\sqrt{4a^2-b^2}$