find the circumference and area of the circle in which diameter is 22mm
Answers
the area of circle is 121π
circumference of circle is 22π
Step-by-step explanation:
diameter of circle= 22
radius of circle=diameter/2
radius(r)=11
circumference of circle=2*π*r
=2*π*11
=22π
area of circle=π*r*r
=π*11*11
=121π
Step-by-step explanation:
Given :
Ratio of lenhth and breadth of a rectangular field = 9 : 5
Area of the rectangular field = 14580 sq.m
To Find :
The cost of surrounding the field with a fence at the rate of 3.25 per m
Solution :
Area of a rectangle is given by ,
\begin{gathered} \\ \star \: {\boxed{\purple{\sf{Area_{(rectangle)} = length \times breadth}}}} \\ \\ \end{gathered}
⋆
Area
(rectangle)
=length×breadth
We are given that ratio of length and breadth of a rectangular field as 9 : 5. Let the ratio constant be "x". Then ,
Length = 9x
Breadth = 5x
We are given the area of rectangular field as 14580 sq.m.
Now ,
\begin{gathered} \\ : \implies \sf \: 14580 \: {m}^{2} = 9x \times 5x \\ \\ \end{gathered}
:⟹14580m
2
=9x×5x
\begin{gathered} \\ : \implies \sf \: 14580 \: {m}^{2} = 45 {x}^{2} \\ \\ \end{gathered}
:⟹14580m
2
=45x
2
\begin{gathered} \\ : \implies \sf \: {x}^{2} = \frac{14580 \: {m}^{2} }{45} \\ \\ \end{gathered}
:⟹x
2
=
45
14580m
2
\begin{gathered} \\ : \implies \sf \: {x}^{2} = 324 \: {m}^{2} \\ \\ \end{gathered}
:⟹x
2
=324m
2
\begin{gathered} \\ : \implies \sf \: x = \sqrt{324 \: {m}^{2} } \\ \\ \end{gathered}
:⟹x=
324m
2
\begin{gathered} \\ : \implies{\underline{\boxed {\red{\mathfrak{x = 18 \: m}}}}} \\ \\ \end{gathered}
:⟹
x=18m
Now ,
Length = 9x = 9(18) = 162 m
Breadth = 5x = 5(18) = 90 m
Perimeter of a rectangle is given by ,
\begin{gathered} \\ \star \: {\boxed{\purple{\sf{perimeter_{(rectangle)} = 2(length + breadth)}}}} \\ \\ \end{gathered}
⋆
perimeter
(rectangle)
=2(length+breadth)
Substituting the values we have ,
\begin{gathered} \\ : \implies \sf \: Perimeter_{(rectangular\:field)} = 2(162 + 90) \\ \\ \end{gathered}
:⟹Perimeter
(rectangularfield)
=2(162+90)
\begin{gathered} \\ : \implies \sf \:Perimeter_{(rectangular\:field)} = 2(252) \\ \\ \end{gathered}
:⟹Perimeter
(rectangularfield)
=2(252)
\begin{gathered} \\ : \implies{\underline{\boxed{\blue {\mathfrak{ Perimeter_{(rectangular \: field)} = 504 \: m}}}}} \: \bigstar\\ \\ \end{gathered}
:⟹
Perimeter
(rectangularfield)
=504m
★
We are given that cost of surrounding the field with fence per m as Rs. 3.25 .
Then , Cost of fencing the rectangular field per 504 m is
\begin{gathered} \\ \longrightarrow \sf \: 504 \times 3.25 = Rs.1638 \\ \\ \end{gathered}
⟶504×3.25=Rs.1638
Hence ,
The cost of surrounding the field with fence at the rate of Rs. 3.25 per m is Rs.1638