Question

In the given figure , AB = 21 cm is the diameter of a semicircle with C as the centre, ADC and CEB are semicircles with diameters AC and CB respectively. Find the area of the shaded region​

Answer 1

Question :

In the given figure , AB = 21 cm is the diameter of a semicircle with C as the centre, ADC and CEB are semicircles with diameters AC and CB respectively. Find the area of the shaded region

Solution :

Semi Circle = sc

Area of the semi circle = $\sf \dfrac{\pi r^2}{2}$

Area of the sc AFB :

Diameter = 21 cm

Radius of AFB = $\sf \dfrac{21}{2}\ cm$

$\longrightarrow \sf Area= \dfrac{\pi (\frac{21}{2})^2}{2}$

$\longrightarrow \sf Area= \dfrac{441 \pi}{8}\ cm^2$

Note : Diameter of both sm CED and sc ADC are equal , so areas are also equal .

Let's do ,

Area of the sc CEB :

Diameter = $\sf \dfrac{21}{2}\ cm$

Radius of CEB = $\sf \dfrac{21}{4}\ cm$

$\longrightarrow \sf Area= \dfrac{\pi (\frac{21}{4})^2}{2}$

$\longrightarrow \sf Area= \dfrac{441 \pi}{32}\ cm^2$

Area of the sc ACD :

Diameter = $\sf \dfrac{21}{2}\ cm$

Radius of ACD = $\sf \dfrac{21}{4}\ cm$

$\longrightarrow \sf Area= \dfrac{\pi (\frac{21}{4})^2}{2}$

$\longrightarrow \sf Area= \dfrac{441 \pi}{32}\ cm^2$

Area of the shaded region :

$\longrightarrow$ Area of sm ( AFB - CED + ADC )

$\sf \longrightarrow \dfrac{441 \pi}{8} - \dfrac{441 \pi}{32} + \dfrac{441 \pi}{32}$

$\sf \longrightarrow \dfrac{441 \pi}{8}$

$\sf \longrightarrow \pink{173.09\ cm^2 }\ \; \bigstar$

So , Area of the shaded region = 173.09 cm²

Answer 2

Answer:

Question :

In the given figure , AB = 21 cm is the diameter of a semicircle with C as the centre, ADC and CEB are semicircles with diameters AC and CB respectively. Find the area of the shaded region

Solution :

Semi Circle = sc

Area of the semi circle = \sf \dfrac{\pi r^2}{2}2πr2

Area of the sc AFB :

Diameter = 21 cm

Radius of AFB = \sf \dfrac{21}{2}\ cm221 cm

\longrightarrow \sf Area= \dfrac{\pi (\frac{21}{2})^2}{2}⟶Area=2π(221)2

\longrightarrow \sf Area= \dfrac{441 \pi}{8}\ cm^2⟶Area=8441π cm2

Note : Diameter of both sm CED and sc ADC are equal , so areas are also equal .

Let's do ,

Area of the sc CEB :

Diameter = \sf \dfrac{21}{2}\ cm221 cm

Radius of CEB = \sf \dfrac{21}{4}\ cm421 cm

\longrightarrow \sf Area= \dfrac{\pi (\frac{21}{4})^2}{2}⟶Area=2π(421)2

\longrightarrow \sf Area= \dfrac{441 \pi}{32}\ cm^2⟶Area=32441π cm2

Area of the sc ACD :

Diameter = \sf \dfrac{21}{2}\ cm221 cm

Radius of ACD = \sf \dfrac{21}{4}\ cm421 cm

\longrightarrow \sf Area= \dfrac{\pi (\frac{21}{4})^2}{2}⟶Area=2π(421)2

\longrightarrow \sf Area= \dfrac{441 \pi}{32}\ cm^2⟶Area=32441π cm2

Area of the shaded region :

\longrightarrow⟶ Area of sm ( AFB - CED + ADC )

\sf \longrightarrow \dfrac{441 \pi}{8} - \dfrac{441 \pi}{32} + \dfrac{441 \pi}{32}⟶8441π−32441π+32441π

\sf \longrightarrow \dfrac{441 \pi}{8}⟶8441π

\sf \longrightarrow \pink{173.09\ cm^2 }\ \; \bigstar⟶173.09 cm2 ★

So , Area of the shaded region = 173.09 cm²