The area of the isosceles triangle is


The area of the isosceles triangle is (5/4) √11 cm2 if the perimeter is 11 cm and the base is 5 cm.



Let equal sides of an isosceles triangle be $b$.

$\therefore$ Perimeter of a triangie, $\quad 2 s=b+b+5 \quad[\because 2 s=a+b+c]$

$\therefore$$11=2 b+5$

$\Rightarrow \quad 2 b=11-5 \Rightarrow 2 b=6$

$\Rightarrow \quad b=\frac{6}{2}=3 \mathrm{~cm}$

We know that, area of an isosceles triangle

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

Here, sides of triangle are $a=5 \mathrm{~cm}$ and $b=3 \mathrm{~cm}$

$\therefore \quad$ Area of an isosceles triangle $=\frac{5 \sqrt{4(3)^{2}-(5)^{2}}}{4}$

$=\frac{5 \sqrt{4 \times 9-25}}{4}$

$=5 \frac{\sqrt{36-25}}{4}$

$=\frac{5 \sqrt{11}}{4} \mathrm{~cm}^{2}$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now