Question 102

The area of a trapeziumis $$\frac{1}{2}(a^2 - b^2)$$ sq. units. where a and b are lengths of the parallel sides. Then, the distance between its parallel sides is

Solution

As per the question,

The area of the trapezium is $$\frac{1}{2}(a^2 - b^2)$$ sq. units.

We know that, the area of the trapezium$$=\dfrac{(a+b)(d)}{2}$$

Hence,

$$\Rightarrow \dfrac{(a+b)(d)}{2}=\frac{1}{2}(a^2 - b^2)$$

$$\Rightarrow d=\dfrac{2(a^2-b^2)}{2(a+b)}$$

$$\Rightarrow d=\dfrac{(a+b)(a-b)}{(a+b)}=(a-b)$$

Hence the distance between the parallel sides $$d=(a-b)$$

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