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$=1:2$