A) \[\frac{25}{4}\]
B) \[\frac{25}{4}(\sqrt{2}+1)\]
C) \[\frac{25}{2\sqrt{2}}\]
D) \[\frac{25}{2}\]
Correct Answer: C
Solution :
\[B(5,\,0)\] is a point on the semi-circle with origin as the center. Hence, we can say that the radius of the semi- circle is 5. \[\therefore \] \[OC=OP=5\] Now, we project a line from P on OA such that is perpendicular to OA. This is shown as follows: Here, \[OB\bot OA\]and \[m\angle POA={{45}^{o}}\] Hence, \[\Delta \,POM\]forms a right isosceles triangle \[\Rightarrow \] \[PM=OM.\] Apply Pythagoras theorem \[P{{M}^{2}}+O{{M}^{2}}=O{{P}^{2}}\] \[\therefore \] \[P{{M}^{2}}+P{{M}^{2}}={{(5)}^{2}}\] or \[2P{{M}^{2}}={{(5)}^{2}}\] or \[P{{M}^{2}}=\frac{25}{2}\] or \[PM=\frac{5}{\sqrt{2}}\] Now, on observing we see that OC forms the base of \[\Delta OPC\] and PM forms the altitude. We know, area of triangle \[=\frac{1}{2}\times b\times h\] \[\therefore \] Area of \[\Delta OPC\]\[=\frac{1}{2}\times PM\times OC\] \[=\frac{1}{2}\times \frac{5}{\sqrt{2}}\times 5=\frac{25}{2\sqrt{2}}\]You need to login to perform this action.
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