1. **Problem statement:** We have a semicircle PQRS with center O and radius $r$. The semicircle is divided into three arcs of equal length by points P, Q, R, and S. We need to find the area of the shaded region under chord PR within the semicircle, expressed in terms of $r$. 2. **Understanding the problem:** The semicircle has total arc length $\pi r$ (half the circumference of a full circle). Since PQ, QR, and RS arcs are equal, each arc length is $\frac{\pi r}{3}$. 3. **Find the central angles:** The arc length $s$ relates to the radius and central angle $\theta$ by $s = r\theta$ (with $\theta$ in radians). So each arc corresponds to an angle: $$\theta = \frac{s}{r} = \frac{\pi r/3}{r} = \frac{\pi}{3}$$ Thus, the points P, Q, R, S divide the semicircle into three arcs each subtending $\frac{\pi}{3}$ radians. 4. **Coordinates of points:** Place the semicircle centered at origin O with diameter along the x-axis from P to S. - P at angle $\pi$ (180°): $(-r,0)$ - Q at angle $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$ - R at angle $\pi - 2\times \frac{\pi}{3} = \frac{\pi}{3}$ - S at angle 0: $(r,0)$ Coordinates: $$Q = (r\cos(\frac{2\pi}{3}), r\sin(\frac{2\pi}{3})) = \left(-\frac{r}{2}, \frac{r\sqrt{3}}{2}\right)$$ $$R = (r\cos(\frac{\pi}{3}), r\sin(\frac{\pi}{3})) = \left(\frac{r}{2}, \frac{r\sqrt{3}}{2}\right)$$ 5. **Find area of shaded region:** The shaded region is the area under chord PR inside the semicircle. This is the area of the sector PR minus the area of triangle PRS. - The central angle for arc PR is $2\times \frac{\pi}{3} = \frac{2\pi}{3}$. - Area of sector PR: $$A_{sector} = \frac{1}{2} r^2 \theta = \frac{1}{2} r^2 \times \frac{2\pi}{3} = \frac{\pi r^2}{3}$$ - Area of triangle PRS: Points: $$P = (-r,0), R = \left(\frac{r}{2}, \frac{r\sqrt{3}}{2}\right), S = (r,0)$$ Using the formula for area of triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$: $$A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$ Substitute: $$= \frac{1}{2} |-r\left(\frac{r\sqrt{3}}{2} - 0\right) + \frac{r}{2}(0 - 0) + r(0 - \frac{r\sqrt{3}}{2})|$$ $$= \frac{1}{2} |-r \times \frac{r\sqrt{3}}{2} + 0 - r \times \frac{r\sqrt{3}}{2}| = \frac{1}{2} |-\frac{r^2 \sqrt{3}}{2} - \frac{r^2 \sqrt{3}}{2}| = \frac{1}{2} |-r^2 \sqrt{3}| = \frac{r^2 \sqrt{3}}{2}$$ 6. **Calculate shaded area:** $$A_{shaded} = A_{sector} - A_{triangle} = \frac{\pi r^2}{3} - \frac{r^2 \sqrt{3}}{2}$$ 7. **Final answer:** $$\boxed{\frac{\pi r^2}{3} - \frac{r^2 \sqrt{3}}{2}}$$ Using $\pi = 3.142$, the area in terms of $r$ is: $$\frac{3.142 r^2}{3} - \frac{r^2 \times 1.732}{2} = 1.047 r^2 - 0.866 r^2 = 0.181 r^2$$ So the shaded area is approximately $0.181 r^2$ cm².