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 :
Options:
A . 10$$\sqrt 2 $$ m
B . 100 m
C . 100$$\sqrt 2 $$ m
D . 200 m
Answer: Option D Let the altitude of the triangle be h1 and base of each be b.Then, $$\frac{1}{2}$$ × b1 × h1 = b × h2, where h2 = 100 m ⇔ h1 = 2 h2 ⇔ h1 = (2 × 100) m ⇔ h1 = 200 m
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