Question : The lengths of the parallel sides of a trapezium are $x$ cm and $y$ cm and the area of the trapezium is $\frac{1}{2}\left(x^2-y^2\right) cm^2$. What is the distance between the parallel sides (in cm )?
Option 1: $x$ - $y$
Option 2: $x$ + $y$
Option 3: $y$
Option 4: $x$
Correct Answer: $x$ - $y$
Solution :
If the lengths of the parallel sides of a trapezium are $x$ cm and $y$ cm and its height is h cm then, the area of the trapezium = $\frac{1}{2}(x+y) × {h}$.
${x}^2-{y}^2 = ({x}+{y})({x} - {y})$
The area of the trapezium = $\frac{1}{2}({x}^2- {y}^2)$ = $\frac{1}{2}({x}+{y})({x} - {y})$
⇒ $\frac{1}{2}({x}+{y}){h}$ = $\frac{1}{2}({x}+{y})({x} - {y})$
⇒ $h = x – y$
$\therefore$ The distance between the parallel sides (in cm) is $x – y$.
Hence, the correct answer is $x$ - $y$.
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