Question

The area of an isosceles triangle is $48c{m}^2$. If the length of the altitude to the non-equal side is 8 cm, then the length of the equal sides of the triangle is

• A. 12 cm
• B. 6 cm
• C. 10 cm
• D. 8 cm

Answer 1

The correct option is C 10 cm



Let ΔABC be an isosceles triangle with AB = AC = a and BC = b.
Also, let AD be the altitude to the non-equal side BC.
$\therefore$ AD = 8 cm
Now, area of ΔABC is 48 cm${}^2$.
$\Rightarrow \frac{1}{2}\times BC\times AD=48$
$\Rightarrow \frac{1}{2}\times b\times 8=48$
$\Rightarrow b=12cm$
Since area of an isosceles triangle with sides a, a, and b is given by $\frac{b}{4}\sqrt{4a^2-b^2}$
$\therefore 48=\frac{12}{4}\sqrt{4a^2-(12{)}^2}$
$\Rightarrow 16=\sqrt{4a^2-144}$
$\Rightarrow 4a^2=(16{)}^2+144=400$
$\Rightarrow a=10cm$
Thus, the length of the equal sides of the triangle is 10 cm.
Hence, the correct answer is option (c).