question_answer

If \[\Delta ABC\tilde{\ }\Delta QRP,\] \[\frac{ar(ABC)}{ar(PQR)}=\frac{9}{4},\] \[AB=18\,cm\] and \[BC=15\,cm,\] then PR is equal to: (NCERT EXEMPLAR)

A) \[10\,cm\]

B) \[12\,cm\]

C) \[\frac{20}{3}\,cm\]

D) \[8\,cm\]

Correct Answer: A

Solution :

[a] Given, \[\Delta ABC\tilde{\ }\Delta QRP,\] \[AB=18cm\] and \[BC=15cm\]

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

\[\therefore \,\,\,\,\frac{ar(\Delta ABC)}{ar(\Delta QRP)}=\frac{{{(BC)}^{2}}}{{{(RP)}^{2}}}\]

But     \[\frac{ar(\Delta ABC)}{ar(\Delta PQR)}=\frac{9}{4}\]             (Given)

\[\Rightarrow \,\,\,\frac{{{(15)}^{2}}}{{{(RP)}^{2}}}=\frac{9}{4}\,\,\,\,\Rightarrow \,\,\,{{(RP)}^{2}}=\frac{225\times 4}{9}=100\]

\[\therefore \,\,\,\,\,\,\,\,\,RP=1\,0cm\]