**Problem 723**

Locate the centroid of the shaded area in Fig. P-723.

**Solution 723**

$A_1 = 90(60) = 5400 \, \text{ mm}^2$

$x_1 = \frac{1}{2}(90) = 45 \, \text{ mm}$

$y_1 = \frac{1}{2}(60) = 30 \, \text{ mm}$

For quarter circle

$A_2 = \frac{1}{4}\pi r^2 = \frac{1}{4}\pi (30^2) = 706.86 \, \text{ mm}^2$

$x_2 = \dfrac{4r}{3\pi} = \dfrac{4(30)r}{3\pi} = 12.73 \, \text{ mm}$

$y_2 = 60 - \dfrac{4r}{3\pi} = 60 - \dfrac{4(30)}{3\pi} = 47.27 \, \text{ mm}$

For the triangle

$A_3 = \frac{1}{2}(45)(60) = 1350 \, \text{ mm}^2$

$x_3 = 45 + \frac{2}{3}(45) = 75 \, \text{ mm}$

$y_3 = \frac{1}{3}(60) = 20 \, \text{ mm}$

For the shaded region

$A = A_1 - A_2 - A_3 = 5400 - 706.86 - 1350$

$A = 3343.14 \, \text{ mm}^2$

$A\bar{x} = \Sigma ax$

$3343.14\bar{x} = 5400(45) - 706.86(12.73) - 1350(75)$

$\bar{x} = 39.71 \, \text{ mm}$ *answer*

$A\bar{y} = \Sigma ay$

$3343.14\bar{y} = 5400(30) - 706.86(47.27) - 1350(20)$

$\bar{y} = 30.39 \, \text{ mm}$ *answer*