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) |
\[\Rightarrow \,\,\,\frac{{{(15)}^{2}}}{{{(RP)}^{2}}}=\frac{9}{4}\,\,\,\,\Rightarrow \,\,\,{{(RP)}^{2}}=\frac{225\times 4}{9}=100\] |
\[\therefore \,\,\,\,\,\,\,\,\,RP=1\,0cm\] |
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