Question

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Please answer my question.

Class - 10th
Chapter - Area related to Circles.

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Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the Concept of Perimeter of Semicircle has been used. We need to find the Perimeter of shaded region. We see we are given two semicircles which have two semicircles inside them. If we add the perimeter of CQD from Perimeter of BSD, then we get perimeter of shaded region in semi - circle BSD. Similarly we can get perimeter of shaded region in semi - circle ARC by adding perimeter of APB from Perimeter of ARC. Now adding both perimeter of shaded region will give perimeter of shaded region in total which is our answer.

Let do it  !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{Perimeter\;of\;Semi\:-\:Circle\;=\;\bf{\dfrac{2\pi r}{2}}}}}$

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★ Solution :-

From figure,

» Diameter of semi - circle APB = 7 cm

» Diameter of semi - circle CQD = 7 cm

» Diameter of semi - circle ARC = 14 cm

» Diameter of semi - circle BSD = 14 cm

• Radius of semi - circle APB = ½ × 7 cm

• Radius of semi - circle CQD = ½ × 7 cm

• Radius of semi - circle ARC = ½ × 14 cm = 7 cm

• Radius of semi - circle BSD = ½ × 14 cm = 7 cm

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~ For the Perimeter of APB and Perimeter of ARC ::

We know that,

$\\\;\sf{:\rightarrow\;\;Perimeter\;of\;Semi\:-\:Circle\;=\;\bf{\dfrac{2\pi r}{2}}}$

• Perimeter of ARC :-

$\\\;\sf{:\Longrightarrow\;\;Perimeter\;of\;Semi\:-\:Circle_{(ARC)}\;=\;\bf{\dfrac{2\:\times\:\dfrac{22}{7}\:\times\:(7)}{2}}}$

$\\\;\sf{:\Longrightarrow\;\;Perimeter\;of\;Semi\:-\:Circle_{(ARC)}\;=\;\bf{\dfrac{2\:\times\:22}{2}}}$

$\\\;\bf{:\Longrightarrow\;\;Perimeter\;of\;Semi\:-\:Circle_{(ARC)}\;=\;\bf{\orange{22\;\;cm}}}$

• Perimeter of APB :-

$\\\;\sf{:\Longrightarrow\;\;Perimeter\;of\;Semi\:-\:Circle_{(APB)}\;=\;\bf{\dfrac{2\:\times\:\dfrac{22}{7}\:\times\:\dfrac{7}{2}}{2}}}$

$\\\;\sf{:\Longrightarrow\;\;Perimeter\;of\;Semi\:-\:Circle_{(APB)}\;=\;\bf{\dfrac{22}{2}}}$

$\\\;\bf{:\Longrightarrow\;\;Perimeter\;of\;Semi\:-\:Circle_{(APB)}\;=\;\bf{\blue{11\;\;cm}}}$

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~ For the Perimeter of BSD and Perimeter of CQD ::

We know that,

$\\\;\sf{:\rightarrow\;\;Perimeter\;of\;Semi\:-\:Circle\;=\;\bf{\dfrac{2\pi r}{2}}}$

• Perimeter BSD :-

$\\\;\sf{:\Longrightarrow\;\;Perimeter\;of\;Semi\:-\:Circle_{(BSD)}\;=\;\bf{\dfrac{2\:\times\:\dfrac{22}{7}\:\times\:(7)}{2}}}$

$\\\;\sf{:\Longrightarrow\;\;Perimeter\;of\;Semi\:-\:Circle_{(BSD)}\;=\;\bf{\dfrac{2\:\times\:22}{2}}}$

$\\\;\bf{:\Longrightarrow\;\;Perimeter\;of\;Semi\:-\:Circle_{(BSD)}\;=\;\bf{\orange{22\;\;cm}}}$

• Perimeter of CQD :-

$\\\;\sf{:\Longrightarrow\;\;Perimeter\;of\;Semi\:-\:Circle_{(CQD)}\;=\;\bf{\dfrac{2\:\times\:\dfrac{22}{7}\:\times\:\dfrac{7}{2}}{2}}}$

$\\\;\sf{:\Longrightarrow\;\;Perimeter\;of\;Semi\:-\:Circle_{(CQD)}\;=\;\bf{\dfrac{22}{2}}}$

$\\\;\bf{:\Longrightarrow\;\;Perimeter\;of\;Semi\:-\:Circle_{(CQD)}\;=\;\bf{\blue{11\;\;cm}}}$

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~ For Perimeter of Shaded Region ::

This is given as,

$\;\sf{\green{\mapsto\;\;Peri.\;of\;Shaded\;Region\;=\;\bf{Peri.(APB)\;+\;Peri.(ARC)\;+\;Peri.(BSD)\;+\;Peri.(CQD)}}}$

Here Peri. denotes Perimeter

By applying values, we get

$\\\;\sf{\mapsto\;\;Perimeter\;of\;Shaded\;Region\;=\;\bf{11\;+\;22\;+\;22+\;11}}$

$\\\;\sf{\mapsto\;\;Perimeter\;of\;Shaded\;Region\;=\;\bf{\red{66\;\;cm}}}$

$\\\;\underline{\boxed{\tt{Perimeter\;of\;Shaded\;Region\;=\;\bf{\purple{66\;\;cm}}}}}$

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★ More to know about Circles and Semi - Circles ::

$\\\;\sf{\leadsto\;\;Area\;of\;Semi\:-\:Circle\;=\;\dfrac{\pi r^{2}}{2}}$

$\\\;\sf{\leadsto\;\;Area\;of\;Semi\:-\:Circle\;=\;\pi r^{2}}$

$\\\;\sf{\leadsto\;\;Perimeter\;of\;Circle\;=\;2\pi r}$

• Half of Semicircle is known as Quadrant.

• A circle is a line joining all the points on the plane which are at equal distance from a centre and that line has no starting or ending point.

• Line drawn from center of circle to any point on circle is known as radius of circle.

• Line which divided the circle into two equal parts is known as Diameter of the Circle.

Answer 2

$\Large{\underline{\bf{\color{cyan}GiVeN,}}}$

$\bf\pink{In\:the\:attachment\:figure,}$

• Diameter of semi-circles APB & CQD are equal, i.e. 7 cm.

$\longmapsto\:\:\bf\blue{Radius\:of\:semi-circles\:APB\:\&\:CQD\:is,}$

⇒ $\:\:\bf{\dfrac{7}{2}}$

⇒ $\:\:\bf{3.5\:cm}$

• Diameter of semi-circles ARC & BSD are equal, i.e. 14 cm.

$\longmapsto\:\:\bf\purple{Radius\:of\:semi-circles\:ARC\:\&\:BSD\:is,}$

⇒ $\:\:\bf{\dfrac{14}{2}}$

⇒ $\:\:\bf{7\:cm}$

$\Large{\underline{\bf{\color{indigo}To\:FiNd,}}}$

• The perimeter of the shaded region.

$\Large{\underline{\bf{\color{lime}CaLcUlAtIoN,}}}$

$\bf\red{We\:know\:that,}$

☆ Perimeter of circle is also known as Circumference of circle.

$\red\bigstar\:\:{\underline{\green{\boxed{\bf{\color{peru}Perimeter_{circle}\:=\:2\:\pi\:r\:}}}}}$

$\bf\purple{So,}$

✅ Perimeter of semi-circle is,

⇒ Perimeter of semi-circle = ½ × (Perimeter of circle)

$\pink\bigstar\:\:{\underline{\blue{\boxed{\bf{\color{olive}Perimeter_{semi-circle}\:=\:\pi\:r\:}}}}}$

$\bf\orange{Thus,}$

☆ Perimeter of the shaded region is,

$\longrightarrow\:\:\bf\green{Perimeter\:of\:semi-circle\:(ARC\:+\:CQD\: +\: BSD \:+\: APB)\:}$

$:\implies\:\:\bf{\pi\times{7}\:+\:\pi\times{3.5}\: +\:\pi\times{7}\:+\:\pi\times{3.5}\:}$

$:\implies\:\:\bf{\pi\:\Big(7\:+\:3.5\: +\:7\:+\:3.5\Big)\:}$

$:\implies\:\:\bf{\dfrac{22}{7}\times{21}\:}$

$:\implies\:\:\bf{22\times{3}\:}$

$:\implies\:\:\bf\blue{66\:cm}$

$\Large\bold\therefore$ The perimeter of the shaded region is 66 cm.