Question

It is given that ΔABC ~ΔPQR, with $\frac{BC}{QR}=\frac{1}{3},\mathrm{then}\frac{ar\left(\mathrm{\Delta }PRQ\right)}{ar\left(∆BCA\right)}$= __________.

Answer 1

ΔABC ~ ΔPQR (Given)

$\therefore \frac{\mathrm{ar}\left(\mathrm{\Delta PQR}\right)}{\mathrm{ar}\left(∆\mathrm{ABC}\right)}={\left(\frac{\mathrm{QR}}{\mathrm{BC}}\right)}^2$ (The ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides)

$\Rightarrow \frac{\mathrm{ar}\left(\mathrm{\Delta PQR}\right)}{\mathrm{ar}\left(∆\mathrm{ABC}\right)}={\left(\frac{3}{1}\right)}^2=\frac{9}{1}$ $\left(\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{1}{3}\right)$

Or $\frac{\mathrm{ar}\left(\mathrm{\Delta PRQ}\right)}{\mathrm{ar}\left(∆\mathrm{BCA}\right)}=9$

It is given that ΔABC ~ ΔPQR, with $\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{1}{3},\mathrm{then}\frac{\mathrm{ar}\left(\mathrm{\Delta PRQ}\right)}{\mathrm{ar}\left(∆\mathrm{BCA}\right)}$ = _____9_____.