Question

A chord of circle of radius 10 cm subtends a right angle at the centre. Find the corresponding area of minor segment

Answer 1

$\huge\bf\red{\underline{\underline{Given}}}\::$

• $\sf\gray{\:OA\:=\:OB}$
• $\sf\gray{\:Radius\:=\:10cm}$
• $\sf\gray{\:\theta \:=\:90^{\circ}}$

$\huge\bf\red{\underline{\underline{To\:Find}}}\::$

• $\sf\gray{Corresponding \:area \:of \:minor\: segment}$

$\huge\bf\red{\underline{\underline{Solution}}}\::$

$\clubsuit\:\:\underline{\boxed{\sf{\blue{\:Refer\:the\: attachment\:for\:diagram\:}}}}\:\:\clubsuit$

$\star\:\sf\underline\orange{Area\:of\: segment\:APB\:=\:Area \: of \ sector \ OAPB \ - \ Area \ of \ \triangle AOB}$

$\to\:\sf\purple{Area \: of \ sector \ OAPB \ = \ \dfrac{\theta}{360^{\circ}} \:\times\:\pi r^{2}}$

$\to\:\sf\green{Area \: of \ sector \ OAPB \ = \ \dfrac{90}{360} \:\times\:3.14\:\times\:(10)^{2}}$

$\to\:\sf\purple{Area \: of \ sector \ OAPB \ = \ \dfrac{1}{4}\:\times\:3.14\:\times\:100}$

$\to\:\sf\green{Area \: of \ sector \ OAPB \ = \ \dfrac{1}{4}\:\times\:314}$

$\to\:\sf\purple{Area \: of \ sector \ OAPB \ = \ 78.5cm^{2}}$

$\sf\blue{In\:\triangle AOB}$

• $\sf\gray{\angle O \:=\:90^{\circ}}$

• $\sf\gray{Base\:=\:OA}$

• $\sf\gray{Height\:=\:OB}$

$\star\:\sf\underline\red{Now, \:Finding\:the\:area\:of\:\triangle AOB}$

$\mapsto\:\sf\orange{Area\:of\:\triangle AOB \:=\:\dfrac{1}{2}\:\times\:Base\:\times\: Height}$

$\mapsto\:\sf\pink{Area\:of\:\triangle AOB \:=\:\dfrac{1}{2}\:\times\:OA\:\times\: OB}$

$\mapsto\:\sf\orange{Area\:of\:\triangle AOB \:=\:\dfrac{1}{2}\:\times\:10\:\times\: 10}$

$\mapsto\:\sf\pink{Area\:of\:\triangle AOB \:=\:50cm^{2}}$

$\star\:\sf\underline\red{Area\:of\: segment\:APB\:=\:Area\:of\: sector\:-\:</p><p>Area\:of\:\triangle AOB}$

$\hookrightarrow\:\sf\blue{Area\:of\: segment\:APB\:=\:78.5\:-\:50}$

$\hookrightarrow\:{\underline{\boxed{\tt{\purple{\:Area\:of\: segment\:APB\:=\:28.5cm^{2}\:}}}}}$

Answer 2

$\pink{\large{\underline{\underline{ \rm{Given: }}}}}$

◕ Radius of the circle = 10 cm

◕ Angle subtended by chord at center i.e., θ = 90°

$\pink{\large{\underline{\underline{ \rm{To \: Find: }}}}}$

◉ Corresponding area of the minor segment.

$\pink{\large{\underline{\underline{ \rm{Solution: }}}}}$

Area of the minor segment = Area of the sector OAB - Area of ∆ ABC formed with the radius and chord.

Area of the minor segment =

$\green{ \underline{ \boxed{ \tt{r</strong><strong>[</strong><strong>\</strong><strong>d</strong><strong>frac{\pi \theta}{360 \degree} - \</strong><strong>d</strong><strong>frac{1}{2} \sin\theta</strong><strong>]</strong><strong>}}}}$

$\tt = { \dfrac{\pi {r}^{2} \theta }{360 \degree} - \dfrac{1}{2} {r}^{2} \sin\theta}$

Substitute the values in the formula, we have:

$\tt{ = 3.14 \times \dfrac{10 \times 10 \times 90 \degree}{360 \degree} - \dfrac{1}{2} \times 10 \times 10 \times \sin90 \degree}$

Sin 90° = 1;

$\tt{ = 3.14 \times \dfrac{10 \times 10}{4} - \dfrac{1}{2} \times 10 \times 10 \times 1}$

$\tt{ = 3.14 \times \dfrac{100}{4} - \dfrac{100}{2} }$

Further solving it, we have:

$\tt{ = 3.14 \times 25 - 50}$

$\tt{ = 78.5 - 50}$

$\tt{ = 28.5 \: {cm}^{2} }$

Area of the minor segment = $\green{ \underline{ \boxed{ \tt{28.5 \: {cm}^{2} }}}}$

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If we need to find Area of major segment.

Area of the major segment = Area of the circle - Area of the minor segment.

$\tt{ = \pi {r}^{2} - 28.5}$

On substituting the values of π and r², we have

$\tt = 3.14 \times 10 \times 10 - 28.5$

$\tt{ = 3.14 \times 100 - 28.5}$

Removing Decimal:-

$\tt = \dfrac{314}{100} \times 100 - 28.5$

$\tt{ = 314 - 28.5}$

$\tt = 285.5 \: {cm}^{2}$

∴ Area of the major segment = $\green{ \underline{ \boxed{ \tt{285.5 \: {cm}^{2} }}}}$

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