a chord of a circle of …
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Sia ? 6 years, 6 months ago
Area of minor sector = {tex}\frac { \theta } { 360 } \times \pi r ^ { 2 }{/tex}
{tex}= \frac { 90 } { 360 } \times 3.14 \times 10 \times 10 = 78.5 \mathrm { cm } ^ { 2 }{/tex}
Area of {tex}\triangle O A B = \frac { O A \times O B } { 2 }{/tex}
{tex}= \frac { 10 \times 10 } { 2 } = 50 \mathrm { cm } ^ { 2 }{/tex}
{tex}\therefore {/tex} Area of the minor segment
= Area of minor sector - Area of {tex}\triangle O A B{/tex}
{tex}= 78.5 \mathrm { cm } ^ { 2 } - 50 \mathrm { cm } ^ { 2 } = 28.5 \mathrm { cm } ^ { 2 }{/tex}
{tex}= 3.14 \times 10 \times 10 - 78.5{/tex}
{tex}= 314 - 78.5 = 235.5 \mathrm { cm } ^ { 2 }{/tex}
Alternative method
Area of major sector = {tex}\frac { 360 - \theta } { 360 } \times \pi r ^ { 2 }{/tex}
{tex}= \frac { 360 - 90 } { 360 } \times 3.14 \times ( 10 ) ^ { 2 } = 235.5 \mathrm { cm } ^ { 2 }{/tex}
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