Prove the following

Question:

If $\triangle A B C \sim \triangle Q R P, \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\Delta P Q R)}=\frac{9}{4}, A B=18 \mathrm{~cm}$

and $B C=15 \mathrm{~cm}$, then $P R$ is equal to

(a) $10 \mathrm{~cm}$

(b) $12 \mathrm{~cm}$

(c) $\frac{20}{3} \mathrm{~cm}$

(d) $8 \mathrm{~cm}$

Solution:

(a) Given, Δ ABC ~Δ QRP, AB = 18cm and BC = 15cm

We know that, the ratio of area of two similar triangles is equal to the ratio of square of their corresponding sides.

$\therefore$ $\frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta Q R P)}=\frac{(B C)^{2}}{(R P)^{2}}$

But given, $\frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta P Q R)}=\frac{9}{4}$ [given]

$\Rightarrow$ $\frac{(15)^{2}}{(R P)^{2}}=\frac{9}{4}$ $[\because B C=15 \mathrm{~cm}$, given]

$\Rightarrow$ $(R P)^{2}=\frac{225 \times 4}{9}=100$ 

$\therefore$ $R P=10 \mathrm{~cm}$

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