# What is the area of the triangle ABC?

## Points $M$ and $N$ are sides' midpoints. $B N$ ⊥ $C M$; $B N = 8$cm and $C M = 12$cm I know, that the answer is $64 c {m}^{2}$, but I need a solution step by step.

Oct 31, 2017

area of $\Delta A B C = 64 {\text{ cm}}^{2}$

#### Explanation:

Let $| A B C |$ denote area of $\Delta A B C$
As $A M : A N = A B : A C , \implies M N \mathmr{and} B C$ are parallel,
$\implies \Delta A B C \mathmr{and} \Delta A M N$ are similar,
as $A M : A B = 1 : 2$,
$| A M N | : | A B C | = 1 : {2}^{2} = 1 : 4$
=> |AMN|:|MNCB|=1:3
given $B N$ is perpendicular to $C M$,
$\implies | M N C B | = \frac{B N \cdot C M}{2} = \frac{8 \cdot 12}{2} = 48$
$\implies | A M N | = \frac{| M N C B |}{3} = \frac{48}{3} = 16$
$\implies | A B C | = | A M N | + | M N C B | = 16 + 48 = 64 {\text{ cm}}^{2}$