Question

find the radius and diameter of the circle given the area -1386m2​

Answer 1

$\large {\red {\underline {\overline {\mid {\sf {Aim:-{\mid }}}}}}}$

To find the radius and diameter of a circle whose area is 1386 m².

$\large {\red {\underline {\overline {\mid {\sf {Pre - requisite \: knowledge :-{\mid }}}}}}}$

• $\small{\orange {\boxed {\bf {Area \: of \: a \: circle = πr² }}}}$
• $\small{\orange {\boxed {\bf {Diameter = 2 × Radius }}}}$

$\large {\red {\underline {\overline {\mid {\sf {Calculation:-{\mid }}}}}}}$

$Area \: of \: circle = πr² \\ ⇒1386 m² = \frac {22}{7} × r² \\ ⇒ 1386 × \frac {7}{22} = r² \\⇒441 = r² \\⇒\sqrt{441} = r \\⇒r = 21 m$

$\small{\orange {\boxed {\bf {Radius = 21m \: and \: Diameter = 42m}}}}$

Answer 2

Given:

• area of a circle is 1386m²

To Find:

• the radius and diameter of the circle.

Solution:

◐ Now, we have given the area of the circle and it's said to find the Radius and the diameter of the circle.

${ \blue{ \underline{ \frak{As \: we \: know \: that \dag}}}}$

$\longrightarrow\blue{ \underline{ \boxed{ \pink{ \pmb{ \mathfrak{ area \: of \: a \: circle = \pi \: {r}^{2} }}} }\bigstar}}$

Where,

• ➪ π stands for ${\bf{\frac{22}{7}}}$
• ➪ r stands for ${\bf{ Radius}}$

Let's substitute the values now :

${ : \implies} \sf \: area _{(circle)} \: = \pi{r}^{2} \: \: \: \: \\ \\ \\ { : \implies} \sf 1386 {m}^{2} = \frac{22}{7} \times {r}^{2} \\ \\ \\ { : \implies} \sf {r}^{2} = \frac{1386 \times 7}{22} \: \: \: \: \: \: \\ \\ \\ { : \implies} \sf \: {r}^{2} = 441 {m}^{2} \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \sf r = \sqrt{441 {m}^{2} } \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \sf { \purple{ \underline{ \boxed{{ \frak{r = 21m}}}} \bigstar}} \: \: \: \: \: \: \: \: \: \:$

$\therefore{ \rm { \underline{the \: radius \: of \: the \: circle = 21m \bigstar}}}$

Now, let's find the diameter

• As we know that,

Diameter = 2[ radius]

Diameter = 2 [21m]

Diameter = 42m

Daigram of circle:

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.121,-2.121)(-0,-2.3)\put(0,0){\line(1,0){2.3}}\put(0.5,0.3){\bf\large 21cm}\end{picture}$

More to know:

• area of a square = side²

• area of a rectangle = length × breadth

• area of a triangle ½ base × hieght

• area of a parallelogram = base×hieght

hope this helped