Question

Show that the area of a triangle is half the product of its base (or any side) and the corresponding attitude (height).

Answer 1

Let ABC be a triangle.
Draw AD || BC such that CD = BA.
Now ABCD is a parallelogram, one of whose diagonals is AC.
$\text{We know}△ABC\cong △ACD$
$\text{So}ar\left(△ABC\right)=ar\left(△ACD\right)$
$\because$ Congruent triangles have equal area)
$\text{Therefore,}ar△ABC=\frac{1}{2}ar\left(ABCD\right)$
$\text{Draw AL}\perp BC$
$\text{We know}ar\left(ABCD\right)=BC\times AL$
$\text{We have}ar\left(△ABC\right)=\frac{1}{2}ar\left(ABCD\right)$
$=\frac{1}{2}\times BC\times AL$
$\text{So}ar△ABC=\frac{1}{2}\times \text{base}BC\times \text{Corresponding attitude}AE$