Question:
The base of an isosceles triangle measures 80 cm and its area is 360 cm2. Find the perimeter of the triangle.
Solution:
Let $\triangle P Q R$ be an isosceles triangle and $P X \perp Q R$.
Now,
Area of triangle $=360 \mathrm{~cm}^{2}$
$\Rightarrow \frac{1}{2} \times Q R \times P X=360$
$\Rightarrow h=\frac{720}{80}=9 \mathrm{~cm}$
Now,
$Q X=\frac{1}{2} \times 80=40 \mathrm{~cm}$ and $P X=9 \mathrm{~cm}$
Also,
$P Q=\sqrt{Q X^{2}+P X^{2}}$
$a=\sqrt{40^{2}+9^{2}}=\sqrt{1600+81}=\sqrt{1681}=41 \mathrm{~cm}$
∴ Perimeter = 80 + 41 + 41 = 162 cm
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