∆ABC ∼ ∆PQR such that ar(∆ABC) = 4 ar(∆PQR). If BC = 12 cm, then QR =


$\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ such that $\operatorname{ar}(\triangle \mathrm{ABC})=4 \operatorname{ar}(\triangle \mathrm{PQR})$. If $\mathrm{BC}=12 \mathrm{~cm}$, then $\mathrm{QR}=$

(a) 9 cm
(b) 10 cm
(c) 6 cm
(d) 8 cm


Given: In Δ ABC and ΔPQR

$\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$

$\operatorname{Ar}(\triangle \mathrm{ABC})=4 \mathrm{Ar}(\triangle \mathrm{PQR})$

$\mathrm{BC}=12 \mathrm{~cm}$

To find: Measure of QR

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\frac{\operatorname{Ar}(\triangle \mathrm{ABC})}{\operatorname{Ar}(\Delta \mathrm{PQR})}=\frac{\mathrm{BC}^{2}}{\mathrm{QR}^{2}}$

$\frac{4 \operatorname{Ar}(\Delta \mathrm{PQR})}{\operatorname{Ar}(\triangle \mathrm{PQR})}=\frac{12^{2}}{\mathrm{QR}^{2}}(\operatorname{Given} \mathrm{Ar}(\triangle \mathrm{ABC})=4 \mathrm{Ar}(\triangle \mathrm{PQR}))$

$\frac{4}{1}=\frac{12^{2}}{Q R^{2}}$


$\mathrm{QR}=6 \mathrm{~cm}$

Hence the correct answer is (c)

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