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If the parallel sides of a trapezium are doubled, distance between them remaining same, then find the ratio of the area of the new trapezium to the area of the given trapezium.
Given: The parallel sides of a trapezium are doubled, distance between them remaining same.
To do: To find the ratio of the area of the new trapezium to the area of the given trapezium.
Solution:
Let $a$ and $b$ be the sides of the trapezium. let $h$ be the height of the of the trapezium.
Therefore, area of the trapezium, $A_1=\frac{1}{2}\times( a+b)\times h$
According to question, $2a$ and $2b$ are the sides of trapezium.
Area of new trapezium, $A_2=\frac{1}{2}\times( 2a+2b)\times h$
Ratio of the area of the new trapezium to the area of the given trapezium $\frac{A_2}{A_1}=\frac{\frac{1}{2}\times( a+b)\times h}{\frac{1}{2}\times( 2a+2b)\times h}$
$\Rightarrow \frac{A_2}{A_1}=\frac{\frac{1}{2}\times( a+b)\times h}{\frac{1}{2}\times2( a+b)\times h}$
$\Rightarrow \frac{A_2}{A_1}=\frac{1}{2}$
$\Rightarrow A_1:A_2=1:2$
thus, ratio of the area of the new trapezium to the area of the given trapezium$=1:2$
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