## Geometry: Common Core (15th Edition)

$6.25 \pi \ m^2$
Let $A$ be the area of a sector of a circle . The area $(A)$ of a sector of a circle whose radius is $r$ is given by: $A=\pi r^2 \times \dfrac{Measure \ of \ the \ arc}{360^{\circ}}$ or, $A_{Top}=\pi r^2 \times \dfrac{m \widehat{TP}}{360^{\circ}}..(1)$ Radius, $r=5 \ m$ and $m \widehat{TP}=90^{\circ}$ Plug the data in the equation (1) to obtain: $A_{TOP}=\pi (5)^2 \times \dfrac{90^{\circ}}{360^{\circ}}$ Therefore, we get , $Area=6.25 \pi \ m^2$