Question

The area of an isosceles triangle is $50\sqrt{5}c{m}^2$ and the length of its equal sides is 15 cm. The base of this triangle can be

• A. 30 cm
• B. 20 cm
• C. 25 cm
• D. 13 cm

Answer 1

The correct option is B 20 cm



Let $\Delta ABC$ be an isosceles triangle with equal sides be denoted as a and un-equal sides be denoted as b.
$\therefore a=15cm$
Now, area of isosceles ΔABC is given by = $\frac{b}{4}\sqrt{4a^2-b^2}$
$\therefore 50\sqrt{5}=\frac{b}{4}\sqrt{4(15{)}^2-b^2}$
$\Rightarrow (200\sqrt{5}{)}^2=b^2(900-b^2)$
$\Rightarrow b^4–900b^2+200000=0$
$\Rightarrow b^4–500b^2–400b^2+200000=0$
$\Rightarrow (b^2–500)(b^2–400)=0$
$\Rightarrow b^2–500=0$ or $b^2–400=0$
$\Rightarrow b=\pm 10\sqrt{5}cm$ or $b=\pm 20cm$
$\Rightarrow b=10\sqrt{5}cm$ or $b=20cm$
(Neglecting negative values)
Hence, the correct answer is option (b).