Question

The radius of a hemispherical balloon increases from 6cm to 12cm
as air is being pumped into it. the ratios of the surface areas of the balloon in the two cases is​

Answer 1

Answer:

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Answer 2

Solution

$case \: 1 \\ surface \: area\: of \: the \: balloon \: with \: radius\: 6 \: cm \: = 3\pi \: r {}^{2} \\ case \: 2 \\ surface \: area \: of \: the \: balloon \: with \: radius \: 12 \: cm \: = 3\pi {R}^{2} \\ ratio \: of \: the \: surface \: areas \: in \: the \: above \: 2 \: cases \: = \: \frac{3\pi6 \times 6}{3\pi12 \times 12} \\ = \frac{1}{4} = 1:4$