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Question

# The sides of triangle ABC measure 5, 6 and 7 units, while the perimeter of $△\mathrm{PQR}$ is 360 units . If $△\mathrm{ABC}$ is similar to $△\mathrm{PQR}$, then find the sides of $△\mathrm{PQR}$.

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Solution

## We have to find the sides of $△\mathrm{PQR}$ It is given that $△\mathrm{ABC}~△\mathrm{PQR}\phantom{\rule{0ex}{0ex}}$. $\mathrm{Therefore},\mathrm{by}\mathrm{the}\mathrm{property}\mathrm{of}\mathrm{similarity},\phantom{\rule{0ex}{0ex}}\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}}\phantom{\rule{0ex}{0ex}}\mathrm{This}\mathrm{is}\mathrm{also}\mathrm{equal}\mathrm{to}\frac{\mathrm{AB}+\mathrm{BC}+\mathrm{AC}}{\mathrm{PQ}+\mathrm{QR}+\mathrm{PR}}.\phantom{\rule{0ex}{0ex}}\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AB}+\mathrm{BC}+\mathrm{AC}}{\mathrm{PQ}+\mathrm{QR}+\mathrm{PR}}\phantom{\rule{0ex}{0ex}}\therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AB}+\mathrm{BC}+\mathrm{AC}}{\mathrm{PQ}+\mathrm{QR}+\mathrm{PR}}\phantom{\rule{0ex}{0ex}}⇒\frac{5}{\mathrm{PQ}}=\frac{5+6+7}{360}\phantom{\rule{0ex}{0ex}}⇒\mathrm{PQ}=\frac{360×5}{18}\phantom{\rule{0ex}{0ex}}⇒\mathrm{PQ}=100\mathrm{units}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore \frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AB}+\mathrm{BC}+\mathrm{AC}}{\mathrm{PQ}+\mathrm{QR}+\mathrm{PR}}\phantom{\rule{0ex}{0ex}}⇒\frac{6}{\mathrm{QR}}=\frac{18}{360}\phantom{\rule{0ex}{0ex}}⇒\mathrm{QR}=\frac{360×6}{18}\phantom{\rule{0ex}{0ex}}⇒\mathrm{QR}=120\mathrm{units}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore \frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AB}+\mathrm{BC}+\mathrm{AC}}{\mathrm{PQ}+\mathrm{QR}+\mathrm{PR}}\phantom{\rule{0ex}{0ex}}⇒\frac{7}{\mathrm{PR}}=\frac{18}{360}$ $⇒PR=\frac{360×7}{18}\phantom{\rule{0ex}{0ex}}⇒PR=140\mathrm{units}$ Thus, the sides are 100, 120 and 140 units.

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