Question

The lengths of the shorter and longer parallel sides of a trapezium are $x$ cm and $y$ cm respectively. If the area of the trapezium is $(x^2-y^2)$, then the height of the trapezium is:

• A. $x$
• B. $(x+y)$
• C. $y$
• D. $2(x-y)$

Answer 1

The correct option is D $2(x-y)$

Area of trap$=\frac{1}{2}(x+y)\times h$

$\Rightarrow (x^2-y^2)=\frac{1}{2}(x+y)\times h$

$\Rightarrow h=\frac{2(x-y)(x+y)}{(x+y)}=2(x-y)$