question_answer

In the given figure, ABC is a semi-circle with its center as origin. What is the area of AOPC?

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 \[\Delta OPC\]\[=\frac{1}{2}\times PM\times OC\]                 \[=\frac{1}{2}\times \frac{5}{\sqrt{2}}\times 5=\frac{25}{2\sqrt{2}}\]