Question

Find the area of semicircle of radius 4cm.​

Answer 1

$\star \; {\underline{\boxed{\orange{\pmb{\pmb{\pmb{\textbf{\textsf{ Given \;}}}}}}}}}$

$\begin{gathered} \\\end{gathered}$

• Radius = 4cm

$\begin{gathered} \\\end{gathered}\begin{gathered}\begin{gathered} \\ \\\end{gathered}\star \;{\underline{\boxed{\purple{\pmb{\pmb{\pmb{\textbf{\textsf{ To\;Find \; :- }}}}}}}}}\end{gathered}$

$\begin{gathered} \\\end{gathered}$

• Area of semicircle = ?

$\begin{gathered}\begin{gathered} \\ \qquad{\rule{200pt}{2pt}} \end{gathered}\end{gathered}$

$\begin{gathered} \\\end{gathered}\star \;{\underline{\boxed{\red{\pmb{\pmb{\pmb{\textbf{\textsf{ SoluTion \; :- }}}}}}}}}$

$\begin{gathered} \\ \\\end{gathered}$

❍ Formula Used :

• ${\underline{\boxed{\pmb{\pmb{\sf{ \: \sf \:Area_{(Semicircle)}=\dfrac{1}{2}×\dfrac{22}{7}×r^2 }}}}}}$

$\begin{gathered}\begin{gathered} \\ \end{gathered}\end{gathered}\begin{gathered}\begin{gathered} \\ \\\end{gathered}\dag \;{\underline{\underline{\sf{ \; Calculating \; the \; Area}}}}\end{gathered}$

$\begin{gathered} \\ \qquad \; \dashrightarrow \sf{ Area_{(Semicircle)}=\dfrac{1}{2}×\dfrac{22}{7}×r^2}\end{gathered}$

$\begin{gathered} \\ \qquad \; \dashrightarrow \sf{ Area_{(Semicircle)}=\dfrac{1}{2}×\dfrac{22}{7}×4 {}^{2} }\end{gathered}$

$\begin{gathered} \\ \qquad \; \dashrightarrow \sf{ Area_{(Semicircle)}=\dfrac{1}{ \cancel{2}}×\dfrac{ \cancel{22}}{7}×4 \times 4}\end{gathered}$

$\begin{gathered} \\ \qquad \; \dashrightarrow \sf{ Area_{(Semicircle)}=\dfrac{11}{7}×4 \times 4}\end{gathered}$

$\begin{gathered} \\ \qquad \; \dashrightarrow \sf{ Area_{(Semicircle)}=\dfrac{176}{7} \times 4}\end{gathered}$

$\begin{gathered} \\ \qquad \; \dashrightarrow \sf{ Area_{(Semicircle)}=25.14cm \: \: \: { \color{lightgreen}(Approx.)}}\end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; {\underline{\boxed{\color{darkblue}{\pmb{\pmb{\frak { Area_{(Semicircle)}=25.14cm}}}}}}} \; \bigstar \\ \\ \end{gathered}$

$\begin{gathered}\begin{gathered} \\ \qquad{\rule{200pt}{2pt}} \end{gathered}\end{gathered}$

∴ Area of the semicircle is 25.14cm.

$\begin{gathered}\begin{gathered} \\ \qquad{\rule{200pt}{2pt}} \end{gathered}\end{gathered}$

Answer 2

Answer:

\star \; {\underline{\boxed{\orange{\pmb{\pmb{\pmb{\textbf{\textsf{ Given \;}}}}}}}}}⋆

Given

Given

Given

Given

Given

Given

Given

Given

\begin{gathered}\begin{gathered} \\\end{gathered}\end{gathered}

Radius = 4cm

\begin{gathered}\begin{gathered} \\\end{gathered}\begin{gathered}\begin{gathered} \\ \\\end{gathered}\star \;{\underline{\boxed{\purple{\pmb{\pmb{\pmb{\textbf{\textsf{ To\;Find \; :- }}}}}}}}}\end{gathered}\end{gathered}

⋆

ToFind :-

ToFind :-

ToFind :-

ToFind :-

ToFind :-

ToFind :-

ToFind :-

ToFind :-

\begin{gathered}\begin{gathered} \\\end{gathered}\end{gathered}

Area of semicircle = ?

\begin{gathered}\begin{gathered}\begin{gathered} \\ \qquad{\rule{200pt}{2pt}} \end{gathered}\end{gathered}\end{gathered}

\begin{gathered}\begin{gathered} \\\end{gathered}\star \;{\underline{\boxed{\red{\pmb{\pmb{\pmb{\textbf{\textsf{ SoluTion \; :- }}}}}}}}}\end{gathered}

⋆

SoluTion :-

SoluTion :-

SoluTion :-

SoluTion :-

SoluTion :-

SoluTion :-

SoluTion :-

SoluTion :-

\begin{gathered}\begin{gathered} \\ \\\end{gathered}\end{gathered}

❍ Formula Used :

{\underline{\boxed{\pmb{\pmb{\sf{ \: \sf \:Area_{(Semicircle)}=\dfrac{1}{2}×\dfrac{22}{7}×r^2 }}}}}}

Area

(Semicircle)

=

2

1

×

7

22

×r

2

Area

(Semicircle)

=

2

1

×

7

22

×r

2

Area

(Semicircle)

=

2

1

×

7

22

×r

2

Area

(Semicircle)

=

2

1

×

7

22

×r

2

\begin{gathered}\begin{gathered}\begin{gathered} \\ \end{gathered}\end{gathered}\begin{gathered}\begin{gathered} \\ \\\end{gathered}\dag \;{\underline{\underline{\sf{ \; Calculating \; the \; Area}}}}\end{gathered}\end{gathered}

†

CalculatingtheArea

\begin{gathered}\begin{gathered} \\ \qquad \; \dashrightarrow \sf{ Area_{(Semicircle)}=\dfrac{1}{2}×\dfrac{22}{7}×r^2}\end{gathered}\end{gathered}

⇢Area

(Semicircle)

=

2

1

×

7

22

×r

2

\begin{gathered}\begin{gathered} \\ \qquad \; \dashrightarrow \sf{ Area_{(Semicircle)}=\dfrac{1}{2}×\dfrac{22}{7}×4 {}^{2} }\end{gathered}\end{gathered}

⇢Area

(Semicircle)

=

2

1

×

7

22

×4

2

\begin{gathered}\begin{gathered} \\ \qquad \; \dashrightarrow \sf{ Area_{(Semicircle)}=\dfrac{1}{ \cancel{2}}×\dfrac{ \cancel{22}}{7}×4 \times 4}\end{gathered}\end{gathered}

⇢Area

(Semicircle)

=

2

1

×

7

22

×4×4

\begin{gathered}\begin{gathered} \\ \qquad \; \dashrightarrow \sf{ Area_{(Semicircle)}=\dfrac{11}{7}×4 \times 4}\end{gathered}\end{gathered}

⇢Area

(Semicircle)

=

7

11

×4×4

\begin{gathered}\begin{gathered} \\ \qquad \; \dashrightarrow \sf{ Area_{(Semicircle)}=\dfrac{176}{7} \times 4}\end{gathered}\end{gathered}

⇢Area

(Semicircle)

=

7

176

×4

\begin{gathered}\begin{gathered} \\ \qquad \; \dashrightarrow \sf{ Area_{(Semicircle)}=25.14cm \: \: \: { \color{lightgreen}(Approx.)}}\end{gathered}\end{gathered}

⇢Area

(Semicircle)

=25.14cm(Approx.)

\begin{gathered} \\ \\ \end{gathered}

\begin{gathered} \begin{gathered} \qquad \; \dashrightarrow \; \; {\underline{\boxed{\color{darkblue}{\pmb{\pmb{\frak { Area_{(Semicircle)}=25.14cm}}}}}}} \; \bigstar \\ \\ \end{gathered}\end{gathered}

⇢

Area

(Semicircle)

=25.14cm

Area

(Semicircle)

=25.14cm

Area

(Semicircle)

=25.14cm

Area

(Semicircle)

=25.14cm

★

\begin{gathered}\begin{gathered}\begin{gathered} \\ \qquad{\rule{200pt}{2pt}} \end{gathered}\end{gathered}\end{gathered}

∴ Area of the semicircle is 25.14cm.

\begin{gathered}\begin{gathered}\begin{gathered} \\ \qquad{\rule{200pt}{2pt}} \end{gathered}\end{gathered}\end{gathered}