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(B) Let the side of the cube be $a$ and the radius of the hemisphere be $r$. Since the hemisphere is placed on the cube,the diameter of the hemisphere

Vedclass · Accepted Answer

$3462.67$

Vedclass · Answer

(B) Let the side of the cube be $a$ and the radius of the hemisphere be $r$. Since the hemisphere is placed on the cube,the diameter of the hemisphere equals the side of the cube,so $2r = a$,or $r = a/2$.The total height of the figure is $a + r = 221 \ cm$.Substituting $r = a/2$,we get $a + a/2 = 221$,which implies $3a/2 = 221$,so $a = 442/3 \ cm$ and $r = 221/3 \ cm$.The curved surface area of the hemisphere is $2\pi r^2$ and the total surface area of the cube is $6a^2$.The ratio is given as $\frac{2\pi r^2}{6a^2} = \frac{11}{42}$.Substituting $r = a/2$,we get $\frac{2\pi (a/2)^2}{6a^2} = \frac{2\pi (a^2/4)}{6a^2} = \frac{\pi}{12} = \frac{11}{42}$.Using $\pi \approx 22/7$,we have $\frac{22/7}{12} = \frac{22}{84} = \frac{11}{42}$,which matches the given ratio.The total volume $V$ is the sum of the volume of the cube and the hemisphere:$V = a^3 + \frac{2}{3}\pi r^3 = a^3 + \frac{2}{3}\pi (a/2)^3 = a^3 + \frac{\pi a^3}{12} = a^3(1 + \frac{\pi}{12})$.Substituting $a = 442/3 \approx 147.33$:$V = (147.33)^3 	imes (1 + \frac{22/7}{12}) = 3198545.6 	imes (1 + 0.2619) \approx 4035955 \ cm^3$.Wait,re-evaluating the height: If $a=14$,$r=7$,height $= 21$. Here height is $221$. If $a=147.33$,the volume is very large. Checking options,it seems the height was meant to be $21 \ cm$. If $a+r=21$,$1.5a=21 \implies a=14, r=7$.$V = 14^3 + \frac{2}{3} 	imes \frac{22}{7} 	imes 7^3 = 2744 + 718.67 = 3462.67 \ cm^3$.

$A$ hemisphere is placed on top of a cube. Its front view is shown in the given figure. The total height of the figure is $221 \ cm$. The ratio of the curved surface area of the hemisphere to the total surface area of the cube is $11:42$. What is the total volume (in $cm^3$) of the figure?

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