Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m, then the altitude of the triangle is
(a)$10 \sqrt{2} \mathrm{m}$
(b)100 m
(c)$100 \sqrt{2} \mathrm{m}$
(d)200 m
Answer
Answer (as printed): D
Explanation
Let the altitude of the triangle be $h_{1}$ and base of each be $b$. Then, $\frac{1}{2} \times b_{1} \times h_{1}=b \times h_{2}$, where $h_{2}=100 \mathrm{m}$ $\Leftrightarrow h_{1}=2 h_{2}=(2 \times 100) \mathrm{m}=200 \mathrm{m}$.