Answer
Verified
479.7k+ views
Hint: Area of circle is $\pi {{r}^{2}}$ where ‘r’ is the radius of the circle and $\pi =\dfrac{22}{7}$. Perimeter is defined as the continuous line forming the boundary of a closed geometric figure.
Complete step-by-step answer:
Here, we have a path which runs around a semi-circular plot as shown in diagram:-
Width of path = 3.5m
Perimeter of plot (semi-circle BC) =72m
As we know that perimeter of any semi-circle can be given as $\pi r+2r$,
where r= Radius of inner semi-circle
R= Radius of outer semi-circle
$\pi =\dfrac{22}{7}$
Hence, Perimeter of plot as shown in diagram =72
$\begin{align}
& \pi r+2r=72 \\
& \dfrac{22r}{7}+\dfrac{2r}{1}=72 \\
& \dfrac{22r+14r}{7}=72 \\
\end{align}$
or $\dfrac{36r}{7}=72$
So, we get
$r=\dfrac{72\times 7}{36}=14m$.
Now, we can determine the radius of semicircle AD or outer part of the path be OC+CD i.e. r+3.5.
Hence, R=r+3.5m
Where r=14m
So, R=14+3.5=17.5m
As we have to determine area of path which can be given by relation as,
Area of path (ABCD) =Area of outer semi-circle (AD)-Area of inner semi-circle (BC)
We have area of semicircle=$\dfrac{\pi {{r}^{2}}}{2}$
where r is the radius of a semi-circle.
Hence, area of path can be given as
Area of path = $\dfrac{\pi {{R}^{2}}}{2}-\dfrac{\pi {{r}^{2}}}{2}$
where R=17.5m and r=14m and $\pi =\dfrac{22}{7}$.
Area of path = $\dfrac{22}{7}\times \dfrac{1}{2}\times \left( {{R}^{2}}-{{r}^{2}} \right)=\dfrac{11}{7}\left( {{17.5}^{2}}-{{14}^{2}} \right)$
We have algebraic identity $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$
So, area of path can be rewritten as
Area of path = $\dfrac{11}{7}\left( 17.5-14 \right)\left( 17.5+14 \right)=\dfrac{11}{7}\times 3.5\times 31.5=11\times 0.5\times 31.5$
Area of path$=173.25{{m}^{2}}$
Note: One can go wrong while writing the perimeter of the semi-circle of the plot. He/she may apply the formula ‘$\pi r$’ in place of $\pi r+2r$ which will give the wrong answer. Hence, take care while writing the perimeter of the semi-circle.
Complete step-by-step answer:
Here, we have a path which runs around a semi-circular plot as shown in diagram:-
Width of path = 3.5m
Perimeter of plot (semi-circle BC) =72m
As we know that perimeter of any semi-circle can be given as $\pi r+2r$,
where r= Radius of inner semi-circle
R= Radius of outer semi-circle
$\pi =\dfrac{22}{7}$
Hence, Perimeter of plot as shown in diagram =72
$\begin{align}
& \pi r+2r=72 \\
& \dfrac{22r}{7}+\dfrac{2r}{1}=72 \\
& \dfrac{22r+14r}{7}=72 \\
\end{align}$
or $\dfrac{36r}{7}=72$
So, we get
$r=\dfrac{72\times 7}{36}=14m$.
Now, we can determine the radius of semicircle AD or outer part of the path be OC+CD i.e. r+3.5.
Hence, R=r+3.5m
Where r=14m
So, R=14+3.5=17.5m
As we have to determine area of path which can be given by relation as,
Area of path (ABCD) =Area of outer semi-circle (AD)-Area of inner semi-circle (BC)
We have area of semicircle=$\dfrac{\pi {{r}^{2}}}{2}$
where r is the radius of a semi-circle.
Hence, area of path can be given as
Area of path = $\dfrac{\pi {{R}^{2}}}{2}-\dfrac{\pi {{r}^{2}}}{2}$
where R=17.5m and r=14m and $\pi =\dfrac{22}{7}$.
Area of path = $\dfrac{22}{7}\times \dfrac{1}{2}\times \left( {{R}^{2}}-{{r}^{2}} \right)=\dfrac{11}{7}\left( {{17.5}^{2}}-{{14}^{2}} \right)$
We have algebraic identity $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$
So, area of path can be rewritten as
Area of path = $\dfrac{11}{7}\left( 17.5-14 \right)\left( 17.5+14 \right)=\dfrac{11}{7}\times 3.5\times 31.5=11\times 0.5\times 31.5$
Area of path$=173.25{{m}^{2}}$
Note: One can go wrong while writing the perimeter of the semi-circle of the plot. He/she may apply the formula ‘$\pi r$’ in place of $\pi r+2r$ which will give the wrong answer. Hence, take care while writing the perimeter of the semi-circle.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Change the following sentences into negative and interrogative class 10 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE