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(B) The total surface area of the cube $= 6 	imes (	ext{edge})^2 = 6 	imes 5 	imes 5 \, cm^2 = 150 \, cm^2$.Note that the part of the cube where t

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$163.86$

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(B) The total surface area of the cube $= 6 	imes (	ext{edge})^2 = 6 	imes 5 	imes 5 \, cm^2 = 150 \, cm^2$.Note that the part of the cube where the hemisphere is attached is not included in the surface area.So,the surface area of the block $= 	ext{TSA of cube} - 	ext{base area of hemisphere} + 	ext{CSA of hemisphere}$.$= 150 - \pi r^2 + 2\pi r^2 = (150 + \pi r^2) \, cm^2$.Given diameter $= 4.2 \, cm$,so radius $r = 2.1 \, cm$.$= 150 + \left(\frac{22}{7} 	imes 2.1 	imes 2.1ight) \, cm^2$.$= 150 + (22 	imes 0.3 	imes 2.1) \, cm^2 = 150 + 13.86 \, cm^2 = 163.86 \, cm^2$.

The decorative block shown in the figure is made of two solids - a cube and a hemisphere. The base of the block is a cube with edge $5 \, cm$,and the hemisphere fixed on the top has a diameter of $4.2 \, cm$. Find the total surface area of the block. (in $cm^2$) (Take $\pi = \frac{22}{7}$)

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