Question

Prove that the ratio of the area of two similar triangles is equal to the ratio of squares on the corresponding sides.

Answer 1

$Given:△ABC\sim △PQR$

To prove :$\frac{ar\left(ABC\right)}{ar\left(PQR\right)}={\left(\frac{AB}{PQ}\right)}^2={\left(\frac{BC}{QR}\right)}^2={\left(\frac{AC}{PR}\right)}^2$

Proof:-

$arABC=\frac{1}{2}\times base\times height=\frac{1}{2}\times BC\times AM---\left(i\right)$

$ar\left(PQR\right)=\frac{1}{2}\times base\times hight=\frac{1}{2}\times QP\times PN----\left(ii\right)$

$\frac{ar\left(ABC\right)}{ar\left(PQR\right)}=\frac{BC\times AM}{QR\times PN}----\left(A\right)$

In $△ABM$ and $△PQN,<B=<Q,<M=<N$

$△ABM\sim △PQN$

$\therefore \frac{AB}{PQ}=\frac{AM}{PN}---\left(B\right)$

From $\left(A\right)$

$\frac{ar\left(ABC\right)}{ar\left(PQR\right)}=\frac{BC\times AM}{QR\times PN}=\frac{BC}{QR}\times \frac{AB}{PQ}---\left(C\right)$

Now, given $△ABC\sim △PQR\Rightarrow \frac{AB}{PQ}=\frac{BC}{QP}=\frac{AC}{PR}$

Putting in $\left(C\right)\frac{ar\left(ABC\right)}{ar\left(PQR\right)}={\left(\frac{AB}{PQ}\right)}^2$

Similarly

$\Rightarrow \frac{ar\left(ABC\right)}{ar\left(PQR\right)}={\left(\frac{AB}{PQ}\right)}^2{\left(\frac{BC}{QR}\right)}^2={\left(\frac{AC}{PR}\right)}^2$

Hence proved