Question

Prove that the area of a triangle formed by joining the mid-point of one of the non-parallel sides of a trapezium to the ends of the opposite sides is half that of the area of trapezium.

Answer 1

Given : $ABCD$ is a trapezium. Hence, $AB\parallel DC$.

$E$ is mid-point of $BC$.

Proof: We know that, the median divides the triangle into equal parts.

$\therefore$ $ar\left(△ABE\right)=\frac{1}{2}\times area\left(△ABC\right)$ ----- ( 1 )

Similarly $ar\left(△DEC\right)=\frac{1}{2}\times ar\left(△BDC\right)$ --- ( 2 )

$△BDC$ and $△ADC$ situated between parallel lines $AB$ and $DC$, with common base $DC$

The triangles formed between same pair of parallel lines and with the same base are equal in area.

$\therefore$ $ar\left(△BDC\right)=ar\left(△ADC\right)$

$\therefore$ $ar\left(△DEC\right)=\frac{1}{2}\times ar\left(△ADC\right)$ ---- ( 3 )

Adding equation ( 1 ) and ( 3 ) we get,

$ar\left(△ABE\right)+ar\left(△DEC\right)=\frac{1}{2}\times \left[ar\right(△ABC)+ar(△ADC\left)\right]$

$\Rightarrow$ $ar\left(△ABE\right)+ar\left(△DEC\right)=\frac{1}{2}\times ar\left(trapeziumABCD\right)$