Trigonometry! #19

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We know that the angle subtended by a chord on the major arc is half of the angle subtended by it at the centre. $\therefore \angle AOB = 45^{\circ}$ .

Now all we have to do is find the area of the segment.

Area of $\Delta AOB = \frac{1}{2}ab\sin \theta = \left(\frac{1}{2}\right)(10)(10)(\sin(45^{\circ})) = 25\sqrt{2}$

Area of sector $O-AB = \frac{\theta}{360^{\circ}} \times \pi r^{2} = \frac{45^{\circ}}{360^{\circ}} \times \pi \times 100 = \frac{25\pi}{2}$

Hence the answer, $\frac{25\pi}{2} - 25\sqrt{2} = \frac{50\pi}{4} - \frac{50}{\sqrt{2}} = \boxed{50\left(\frac{\pi}{4} - \frac{1}{\sqrt{2}} \right)}$