Question

Prove that area of a $\mathrm{triangle}=\frac{1}{2}\times \mathrm{base}\times \mathrm{altitude}$.

Answer 1

In ∆ABC, BC is the base and AL is the corresponding height.
Construction: Through A and C, draw AD|| BC and CD || BA, intersecting each other at D.

AD ​|| BC and CD ​|| BA
So, BCDA is a parallelogram. Its diagonal AC divides it into two triangles of equal areas.
∴ ar(​∆ABC ) = $\frac{1}{2}$ × ar(​​​||gm BCDA)
= $\frac{1}{2}$ × BC × AL [ ∵ ar (​​​​||gm BCDA) = BC​× AL]
∴ Area of a triangle = ​$\frac{1}{2}$× base × height
Hence, proved.