A) \[1:9\]
B) \[1:3\]
C) \[1:2\]
D) \[1:4\]
Correct Answer: A
Solution :
[a] In \[\Delta PST\] and \[\Delta PQR,\] |
\[\angle PST=\angle Q\] (Corresponding angle) |
\[\angle P=\angle P\] (Common) |
\[\Delta PST\tilde{\ }\Delta PQR\] (AA similarity) |
So, \[\frac{ar(\Delta PST)}{ar(\Delta PQR)}=\frac{P{{T}^{2}}}{P{{R}^{2}}}=\frac{P{{T}^{2}}}{{{(PT+TR)}^{2}}}\] |
(Given, \[PT=2\,cm,\,\,TR=4cm\]) |
\[=\frac{{{(2)}^{2}}}{{{(2+4)}^{2}}}=\frac{4}{36}=\frac{1}{9}\] |
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