Question

Given $∆ABC~∆PQR,\mathrm{if}\frac{AB}{PQ}=\frac{1}{3},\mathrm{then}\mathrm{find}\frac{\mathrm{Area}\left(∆ABC\right)}{\mathrm{Area}\left(∆PQR\right)}$.

Answer 1

Given: $△\mathrm{ABC}~△\mathrm{PQR}$
Also, $\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{1}{3}$
We know if two triangles are similar then the ratio of the areas of two similar triangles is equal to the square of the ratio of their
corresponding sides.
$\frac{\mathrm{ar}△\mathrm{ABC}}{\mathrm{ar}△\mathrm{PQR}}={\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)}^2={\left(\frac{1}{3}\right)}^2=\frac{1}{9}$