Question

Find the area and the perimeter of the shaded region in figurer
The diamensions are in centimetres
​

Answer 1

$\underline{\large{\sf\ \ \ \ \ SOLUTION:-\ \ \ \ }}$

• Radius of Big Semicircle(R) = 14cm

• Radius of Small semicircle(r) = 14/2= 7cm

We have to find the perimeter and the area of shaded region of the shaded region

• First find the Area of Shaded region :-

$\star\sf\ \ Area\ of\ shaded\ region = \big(Area\ of\ Big_{Semicircle}\big)-\big(Area\ of\ Small_{semicircle}\big)\\ \\ \\ \star\sf\ \ Area\ of\ semicircle= \dfrac{\pi r^2}{2}\\ \\ \\ :\implies\sf\ Area\ of\ shaded\ region = \dfrac{\pi R^2}{2}-\dfrac{\pi r^2}{2}\\ \\ \\ :\implies\sf\ Area\ = \dfrac{\pi }{2}\Big(R^2-r^2\Big)\\ \\ \\:\implies\sf\ Area\ = \dfrac{\cancel{22}}{7\times \cancel{2}}\times \Big\{(R+r)(R-r)\Big\}\\ \\ \\ :\implies\sf \ Area\ = \dfrac{11}{7}\times \Big\{(14+7)(14-7)\Big\}\\ \\ \\ :\implies\sf\ Area\ = \dfrac{11}{\not{7}}\times 21\times \not{7}\\ \\ \\ :\implies\sf\ Area\ = 11\times 21\\ \\ \\ :\implies\sf\ Area\ = 231cm^2\\ \\ \\ \underline{\purple{\sf \ Area\ of\ shaded\ Region= 231cm^2}}$

Now the perimeter of shaded region :-

Perimeter= Circumference of big semicircle + Circumference of small semicircle + Radius of big semi Circle.

$\bigstar\sf\ \ \ Perimeter\ of\ semicircle = \dfrac{2\pi r}{2}\\ \\ \\ \dashrightarrow\sf\ Perimeter\ of\ semicircle= \pi r\\ \\ \\ :\implies\sf\ Perimeter= (\pi R+ \pi r)+ R\\ \\ \\ :\implies\sf\ \ Perimeter= \pi(R+r)+R\\ \\ \\ :\implies\sf\ \ Perimeter= \dfrac{22}{7}\times \Big(14+7\Big)+ 14\\ \\ \\ :\implies\sf\ \ Perimeter= \Bigg\{\dfrac{22}{\cancel{7}}\times \cancel{21} \Bigg\}+14\\ \\ \\ :\implies\sf\ \ Perimeter= \big(22\times 3\big)+14\\ \\ \\ :\implies\sf\ \ Perimeter= 66+14\\ \\ \\ :\implies\sf\ Perimeter= 80cm\\ \\ \\ \underline{\bigstar{\textsf{\textbf{Perimeter\ of\ shaded\ region= 80cm}}}}$

Answer 2

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${\large{\bold{\rm{\underbrace{Requied \; Solution}}}}}$

★ Find the area and the perimeter of the shaded region in figure (figure is already given in the attachment) (the dimensions are in centimetres).

${\large{\bold{\rm{\underline{Full \; Solution}}}}}$

~ As according to the figure we are already able to see that the figure is of a semi-circle and in that semi-circle one more semi-circle is drawn. Means there is a big semi-circle and a small semi-circle. Small semi-circle = 14 cm(Radius) the left area according to the figure = 14 cm too(Diameter). Means,

★ The radius R of big semi-circle = 14 cm

★ Radius r of small semi-circle = 14/2 cm

• Radius r of small semi-circle = 7 cm

[Radius = Diameter/2]

~ As it's given that we have to find out the area and the perimeter of the shaded region according to the figure. So firstly let us find the area of the shaded region.

➨ Area of the shaded region = Area of big semi-circle and the area of small semi-circle.

${\small{\boxed{\bold{\bf{Area \: of \: semi-circle \: = \dfrac{\pi r^{2}}{2}}}}}}$

➨ Henceforth, the area of the shaded region be the given -

${\small{\boxed{\bold{\bf{Area \: of \: shaded \: region \: = \dfrac{\pi R^{2}}{2} - \dfrac{\pi r^{2}}{2}}}}}}$

➨ Combining the formula together we get the following,

${\small{\boxed{\bold{\bf{Area \: of \: shaded \: region \: = \dfrac{\pi}{2}(R^{2}-r^{2})}}}}}$

➨ Now let's put the values according to this above formula.

${\sf{:\implies \dfrac{22}{7 \times 2} \times (R+r)(R-r)}}$

${\sf{:\implies \dfrac{11}{7} \times (R+r)(R-r)}}$

${\sf{:\implies \dfrac{11}{7} \times (14+7)(14-7)}}$

${\sf{:\implies \dfrac{11}{7} \times (21)(7)}}$

${\sf{:\implies \dfrac{11}{7} \times 147}}$

${\sf{:\implies 11 \times 21}}$

${\sf{:\implies 231 \: cm^{2}}}$

${\underline{\bold{\frak{231 \: cm^{2} \: is \: area \: of \: shaded \: region}}}}$

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~ As it's given that we have to find out the area and the perimeter of the shaded region according to the figure. So now let us find the perimeter of the shaded region.

➨ Perimeter of the shaded region = Circumference of big semi-circle + Circumference of small semi-circle + Radius of big semi-circle

${\small{\boxed{\bold{\bf{Perimeter \: of \: semi-circle \: = \dfrac{2 \pi r}{2}}}}}}$

${\small{\boxed{\bold{\bf{Perimeter \: of \: semi-circle \: = \pi r}}}}}$

${\small{\boxed{\bold{\bf{Perimeter \: of \: semi-circle \: = (\pi R + \pi r) + R}}}}}$

➨ Combining the formula together we get the following,

${\small{\boxed{\bold{\bf{Perimeter \: of \: semi-circle \: = \pi(R+r)+R}}}}}$

➨ Now let's put the values according to this above formula.

${\sf{:\implies 3.14 \times (14+7) + 14}}$

${\sf{:\implies 3.14 \times 21 + 14}}$

${\sf{:\implies 65.94 + 14}}$

${\sf{:\implies 79.94 \: cm}}$

${\underline{\bold{\frak{79.94 \: cm \: is \: perimeter \: of \: shaded \: region}}}}$

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${\large{\bold{\rm{\underbrace{Additional \; knowledge}}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle = \: \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Circumference \: of \: circle \: = \: 2 \pi r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$

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