Question

What is the area of an isosceles triangle with equal sides 'b' and base 'a'?

• A. $\frac{a}{2}\sqrt{b^2-a^2}$
• B. $\frac{1}{2}a\sqrt{4b^2-a^2}$
• C. $\frac{a}{4}\sqrt{b^2-a^2}$
• D. $\frac{a}{4}\sqrt{4b^2-a^2}$

Answer 1

The correct option is D $\frac{a}{4}\sqrt{4b^2-a^2}$

ABC is an isosceles triangle $AM\perp BC$, and AM bisect bas 'a' into two equal parts.

$\Rightarrow AM=\sqrt{b^2-{\left(a/2\right)}^2}=\sqrt{b^2-\frac{a^2}{4}}$

$\therefore$ Area of isosceles $△ABC=\frac{1}{2}\times a\times \sqrt{b^2-\frac{a^2}{4}}=\frac{1}{2}\times a\times \sqrt{\frac{{4b}^2-a^2}{4}}$

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