the decorative block shown in figure …
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Sia ? 6 years, 4 months ago
Let S be the total surface area of the decorative block. Then,
S = Total surface area of the cube - Base area of hemisphere + Curved surface area of hemisphere
{tex}\Rightarrow S = \left( 6 \times 5 \times 5 - \pi r ^ { 2 } + 2 \pi r ^ { 2 } \right) \mathrm { cm } ^ { 2 }{/tex}
{tex}\Rightarrow S = \left( 150 + \pi r ^ { 2 } \right) \mathrm { cm } ^ { 2 }{/tex}
{tex}\Rightarrow S = \left\{ 150 + \frac { 22 } { 7 } \times ( 2.1 ) ^ { 2 } \right\} \mathrm { cm } ^ { 2 }{/tex}
{tex}\Rightarrow S = \{ 150 + 22 \times 0.3 \times 2.1 \} \mathrm { cm } ^ { 2 }{/tex}
{tex}\Rightarrow{/tex} S = (150 + 13.86) cm2 = 163.86 cm2
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