Question

If a triangle and a parallelogram are on the same base and between the same parallels, then prove that area of the triangles is equal to half the area of the parallelogram ?

Answer 1

Solution:

For parallelogram $ABCD$

Since $DM\perp AB$

Height is $DM$, Base is $AB$

Area$ABCD=Base\times height$

Area $ABCD=AB\times DM.........\left(1\right)$

For $△PAB$

Since $PN\perp AB$

Height is $PN$, Base is $AB$

Area$△PAB=\frac{1}{2}Base\times height$

Area$△PAB=\frac{1}{2}AB\times PN$

Since opposite sides of Parallelogram are parallel

so, $AB\parallel CD$

distance between parallel lines are equal

$PN=DM$

Area$△PAB=\frac{1}{2}AB\times DM.......\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

Area$△PAB=\frac{1}{2}AreaABCD$

Hence proved.