In the Rectangle

$x=\sqrt { 25^{ 2 }-7^{ 2 } } =24\\ \theta =\tan ^{ -1 }{ \left( \frac { 24 }{ 7 } \right) } \approx 73.74$

Area of shaded Regon = Area of triangle - Area of sector

the Area $=\frac { 1 }{ 2 } \times 24\times 7-49\pi \frac { 73.74 }{ 360 } \approx 52.468$

Draw $OX$ to form right triangle $OXY$ . By pythagorean theorem, we get

$XY=\sqrt{25^2-7^2}=24$

Compute for $\angle YOX$ , we get

$\angle YOX = \sin^{-1} \dfrac{24}{25} \approx 73.74$

From the diagram, we can see that the area of the shaded region is equal to the area of the right triangle minus the area of the circular sector. We have

$area~of~shaded~region = area~of~triangle~-~area~of~circular~sector = \dfrac{1}{2}bh - \dfrac{\angle YOX}{360} \pi r^2 = \dfrac{1}{2}(7))(24)-\dfrac{73.74}{360} (\pi)(7^2) \approx \large{\boxed{52.468}}$