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1. 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 $\sqrt{2}$ m

B. 100 m

C. 100 $\sqrt{2}$ m

D. 200 m

Solution:

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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