Question

A triangle and a parallelogram are constructed on the base such that their areas are equal if the altitude of a parallelogram is 100 m then the altitude of a triangle is

Answer 1

Let the base of the triangle and parallelogram be denoted by 'b' and their respective altitudes be denoted by 'h1' and 'h2'.

Given that the area of the triangle is equal to the area of the parallelogram, we have:

Area of triangle = (1/2)bh1 Area of parallelogram = b*h2

Since the areas are equal, we have:

(1/2)bh1 = b*h2

Simplifying the above equation, we get:

h1 = 2*h2

Substituting the given value of h2 as 100 m, we get:

h1 = 2*100 = 200 m

Therefore, the altitude of the triangle is 200 meters