Question

If in an isosceles triangle, $a$ is the length of the base and $b$ the length of one of the equal sides, then its area is

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

Answer 1

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

In an isosceles triangle, the height- line divides the triangle into two equal areas and the base into two equal halves.
Given base $=a$
each equal side $=b.$

Using Pythagoras Theorem, we get
Therefore, height $=\sqrt{b^2-{\left(\frac{a}{2}\right)}^2}=\sqrt{\frac{{4b}^2-a^2}{4}}$
Therefore, area of the triangle $=\frac{1}{2}\times \text{base}\times \text{height}$
$=\frac{1}{2}\times a\times \sqrt{\frac{{4b}^2-a^2}{4}}$
$=\frac{a}{4}\sqrt{{4b}^2-a^2}$

Hence, option A.