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  • If \Delta ABC\sim \Delta QRP,\frac{ar(ABC)}{ar(PQR)}=\frac{9}{4},AB=18\; cm and BC=15\; cm, then PR is equal to (A) 10\; cm (B) 12\; cm (C) \frac{20}{3}\; cm (D) 8\; cm

\text{If }\Delta ABC\sim \Delta QRP,\frac{ar(ABC)}{ar(PQR)}=\frac{9}{4},AB=18\; cm  If BC = 15 cm then PR is equal to?

\text{(A)} 10\; cm

\text{(B)} 12\; cm

\text{(C)} \frac{20}{3}\; cm

\text{(D)} 8\; cm

Answers (1)

Answer : [A]

\Delta ABC \sim \Delta QRP \; \text {and}\; AB = 18 \; cm, BC = 15 \; cm

\therefore \frac{AC}{PQ}=\frac{AB}{QR}=\frac{BC}{RP}

\frac{ar(ABC)}{ar\left ( PQR \right )}=\frac{BC^{2}}{RP^{2}}\; \; \; \; \; \; \; \; ....(1)

[Q If triangles are similar, their areas are proportional to the squares of the corresponding sides.]

\frac{ar(ABC)}{ar\left ( PQR \right )}=\frac{9}{4}

[Using equation (1)]

\Rightarrow \frac{BC^{2}}{RP^{2}}=\frac{9}{4}                         

\frac{(15)^{2}}{RP^{2}}=\frac{9}{4}

RP^{2}=\frac{225\times 4}{9}=100

RP=\sqrt{100}

RP=10\; cm

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infoexpert23

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