Question

Prove that the area of a trapezium is half the sum of the parallel sides multiplied by the disrance between them

Answer 1

$\textbf{ \huge{ Hello Mate! }}$

Given : The quadrilateral is trapezium.

To Prove : ar ( trap. ABCD ) = $\frac{1(a+b)(h)}{2}$

To Construct : Join AC and BD

Proof : ar( triangle ABD ) = 1/2 × base × height

ar( triangle ABD ) = 1/2 × AB × AL ___(1)

Similarily,

ar( triangle BDC ) = 1/2 × CD × AL ___(2)

Adding both equation we get

ar( tri ABD + BDC ) = ( 1/2 × AB + AL ) + ( 1/2 × AB × AL )

Since ar( □ ABCD ) = ar( tri ABD + BCD )

ar( □ ABCD ) = 1/2 × AL ( AB + CD )

$\textsf{\red{ Hence the area of a trapezium is half the sum of the parallel sides multiplied by the disrance between them }}$

$\textbf{ Have great future ahead! }$