A road 14m wide surrounded a circular park whose circumference is 704m. Find the surface area of the road. Also, find the cost of paying the road at Rs 100 per \[{m^2}\]
Answer
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Hint: A road 14m wide surrounded by a circular both must be circular as well.
Circumference of a circle is the perimeter i.e. length of outer boundary of the circle.
Hence formula of circumference of a circle is
\[ = 2\pi r\] where \[l = \] radius of circle and \[\pi = \dfrac{{22}}{7}\]
As we know that road is surrounded by a circular park i.e., they are forming concentric circles.
Area of concentric circle \[ = \pi ({R^2} - {r^2})\], where
R= outer radius of the circle
R= Inner radius of the circle and \[R = width + r\]
Complete step by step solution:
Given,
Circumference of a circular park \[ = 704m\]
Width of a road\[ = 14m\]
Circumference of a circle \[ = 2\pi \]
\[
704 = 2 \times \dfrac{{22}}{7} \times r \\
704 \times 7 = 44r \\
r = \dfrac{{704 \times 7}}{{44}} \\
r = 16 \times 7 \\
r = 112m \\
\]
Inner radius of the park i.e. \[r = 112m\]
Outer radius of the park i.e. \[R = width + r\]
\[
\Rightarrow R = 14 + 112 = 126m \\
\Rightarrow R = 126m \\
\]
Area of the road \[
= \pi ({R^2} - {r^2}) \\
\Rightarrow \pi (R + r)(R - r) \\
\] By using identify \[[{a^2} - {b^2} = (a - b)(a + b)]\]
\[
\Rightarrow \dfrac{{22}}{7}(126 + 112)(126 - 112) \\
\Rightarrow \dfrac{{22}}{7} \times 238 \times 14 \\
\Rightarrow 22 \times 238 \times 2 \\
\Rightarrow 238 \times 44 \\
\Rightarrow 10472\,{m^2} \\
\]
Hence, Area of the road \[ = 10472\,{m^2}\]
Cost of paving \[1{m^2}\]road \[ = Rs\,100\]
Cost of paving \[10472\,{m^2}\]road \[
= Rs\,100 \times 10472 \\
= Rs\,1047200 \\
\]
Note: The circles with a common center point are known as concentric circles. In other words. It is defined as two or more circles that have the same center point. The region between two concentric circles of different radii is known as an annulus.
In the concept of concentric circles, one circle is inscribed in the other. Hence there are two types of radius i.e. Outer radius denoted by R and inner radius i.e. r.
Circumference of a circle is the perimeter i.e. length of outer boundary of the circle.
Hence formula of circumference of a circle is
\[ = 2\pi r\] where \[l = \] radius of circle and \[\pi = \dfrac{{22}}{7}\]
As we know that road is surrounded by a circular park i.e., they are forming concentric circles.
Area of concentric circle \[ = \pi ({R^2} - {r^2})\], where
R= outer radius of the circle
R= Inner radius of the circle and \[R = width + r\]
Complete step by step solution:
Given,
Circumference of a circular park \[ = 704m\]
Width of a road\[ = 14m\]
Circumference of a circle \[ = 2\pi \]
\[
704 = 2 \times \dfrac{{22}}{7} \times r \\
704 \times 7 = 44r \\
r = \dfrac{{704 \times 7}}{{44}} \\
r = 16 \times 7 \\
r = 112m \\
\]
Inner radius of the park i.e. \[r = 112m\]
Outer radius of the park i.e. \[R = width + r\]
\[
\Rightarrow R = 14 + 112 = 126m \\
\Rightarrow R = 126m \\
\]
Area of the road \[
= \pi ({R^2} - {r^2}) \\
\Rightarrow \pi (R + r)(R - r) \\
\] By using identify \[[{a^2} - {b^2} = (a - b)(a + b)]\]
\[
\Rightarrow \dfrac{{22}}{7}(126 + 112)(126 - 112) \\
\Rightarrow \dfrac{{22}}{7} \times 238 \times 14 \\
\Rightarrow 22 \times 238 \times 2 \\
\Rightarrow 238 \times 44 \\
\Rightarrow 10472\,{m^2} \\
\]
Hence, Area of the road \[ = 10472\,{m^2}\]
Cost of paving \[1{m^2}\]road \[ = Rs\,100\]
Cost of paving \[10472\,{m^2}\]road \[
= Rs\,100 \times 10472 \\
= Rs\,1047200 \\
\]
Note: The circles with a common center point are known as concentric circles. In other words. It is defined as two or more circles that have the same center point. The region between two concentric circles of different radii is known as an annulus.
In the concept of concentric circles, one circle is inscribed in the other. Hence there are two types of radius i.e. Outer radius denoted by R and inner radius i.e. r.
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