Question

IfABC~QRP,ar(ABC)/ar(PQR)=9/4,AB=18cm and BC =15cm,then PR=??

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$PR = 10\:cm$

Step-by-step explanation:

Given ∆ABC ~ ∆QRP,

AB=18 cm , BC = 15 cm,

$\frac{Area\:of \triangle \:ABC}{Area \: of \triangle PQR }= \frac{9}{4}--(1)$

/* we know that,

The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides */

$\frac{Area\:of \: \triangle \:ABC}{Area \: of \ triangle\:PQR }= \frac{9}{4}$

$\implies \frac{BC^{2}}{RP^{2}}=\frac{9}{4}$

$\implies \frac{15^{2}}{RP^{2}}=\frac{9}{4}$

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

$\implies 25 × 4 = RP^{2}$

$\implies RP = \sqrt{100}$

$\implies RP = 10$

Therefore,

$PR = 10\:cm$

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