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(D) Let the base of both the triangle and the parallelogram be $b$.Let the altitude of the triangle be $h_1$ and the altitude of the parallelogram be

Vedclass · Accepted Answer

$200 \, m$

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(D) Let the base of both the triangle and the parallelogram be $b$.Let the altitude of the triangle be $h_1$ and the altitude of the parallelogram be $h_2$.The area of the triangle is given by $A_t = \frac{1}{2} 	imes b 	imes h_1$.The area of the parallelogram is given by $A_p = b 	imes h_2$.According to the problem,the areas are equal,so $A_t = A_p$.Therefore,$\frac{1}{2} 	imes b 	imes h_1 = b 	imes h_2$.Dividing both sides by $b$ (assuming $b 
eq 0$),we get $\frac{1}{2} h_1 = h_2$,which implies $h_1 = 2 h_2$.Given that the altitude of the parallelogram $h_2 = 100 \, m$.Substituting this value,we get $h_1 = 2 	imes 100 \, m = 200 \, m$.

$A$ 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

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