In the given figure,there is a large semicirc | vedclass

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(D) $1$. The diameter of the large semicircle is $AB = 56 \ cm$,so its radius $R = 28 \ cm$.$2$. The diameter of each of the four small semicircles is

Vedclass · Accepted Answer

$394.24$

Vedclass · Answer

(D) $1$. The diameter of the large semicircle is $AB = 56 \ cm$,so its radius $R = 28 \ cm$.$2$. The diameter of each of the four small semicircles is $d = 56 / 4 = 14 \ cm$,so their radius $r = 7 \ cm$.$3$. Let the radius of the shaded circle be $x$. The center of the large semicircle is at the midpoint of $AB$. Let this be the origin $(0,0)$. The large semicircle is defined by $x^2 + y^2 = 28^2$ for $y \ge 0$.$4$. The centers of the two middle small semicircles are at $(-7, 0)$ and $(7, 0)$.$5$. The shaded circle is tangent to the large semicircle,so the distance from the origin to its center $(0, y_c)$ is $28 - x$. Thus,the center is at $(0, 28 - x)$.$6$. The shaded circle is also tangent to the small semicircles. The distance between the center of the shaded circle $(0, 28 - x)$ and the center of a small semicircle $(7, 0)$ is $r + x = 7 + x$.$7$. Using the distance formula: $(7 - 0)^2 + (0 - (28 - x))^2 = (7 + x)^2$.$8$. $49 + (28 - x)^2 = (7 + x)^2$.$9$. $49 + 784 - 56x + x^2 = 49 + 14x + x^2$.$10$. $784 = 70x$,which gives $x = 784 / 70 = 11.2 \ cm$.$11$. The area of the shaded circle is $\pi x^2 = (22/7) 	imes 11.2 	imes 11.2 = 22 	imes 1.6 	imes 11.2 = 394.24 \ cm^2$.

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