Question

Find the area of a circle whose diameter is 28 mm.


PLEASE ANSWER THIS QUESTION ​

Answer 1

Step-by-step explanation:

Radius (r) =d/2=28/2=14mm

NOW,

Here pie=3.14

Area of circle (A) =3.14*14*14

=615.44mm

Answer 2

Given : The Diameter of Circle is 28 mm .

Need To Find : Area of Circle .

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❒ Formula for finding Radius of Circle is given by :

• $\underline {\boxed{\sf{\bigstar Radius_{(Circle)} = \dfrac{Diameter}{2} \:units}}}$

Where,

• d is the Radius of Circle in mm .

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

$\qquad \qquad:\implies \sf{ Radius_{(Circle)} = \dfrac{\cancel {28}}{\cancel {2}}}$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  Radius = 14\: mm}}}}\:\bf{\bigstar}$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{\mathrm {  Radius \:of\:Circle \:is\:14\: mm}}}$

⠀⠀⠀⠀⠀Formula for Finding Area of circle is given by :

• $\underline {\boxed{\sf{\bigstar Area_{(Circle)} = \pi r^{2} \:sq.units }}}$

Where,

• r is the Radius of Circle in mm and $\pi = \dfrac{22}{7}$ .

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

$\qquad \qquad:\implies \sf{ Area_{(Circle)} = \pi \times 14 ^{2} }$

$\qquad \qquad:\implies \sf{ Area_{(Circle)} = \dfrac{22}{\cancel {7}} \times \cancel {14} \times 14 }$

$\qquad \qquad:\implies \sf { Area_{(Circle)} = 22 \times 2 \times 14 }$

$\qquad \qquad:\implies \sf{Area_{(Circle)} = 22 \times 28 }$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  Area_{(Circle)} = 616\:m^{2}}}}}\:\bf{\bigstar}$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {  Area \:of\:Circle \:is\:\bf{616\: mm^{2}}}}}$

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$\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}$

$\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}$

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