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Hint: Assume ‘r’ as the radius of the given shot put circle. Apply the formula to find the area of the circle given by: - \[A=\pi {{r}^{2}}\], A = Area of the circle and r = radius of the circle. Substitute the value of ‘r’ given in the question and find the value of A. Substitute the value of \[\pi =\dfrac{22}{7}\] as this will make our calculation easy.
Complete step by step answer:
The above figure can be assumed as the field of shot put circle. Here, we have assumed that the radius of the given field is ‘r’.
We know that area of a circle is given as: -
\[A=\pi {{r}^{2}}\]
Where, A = Area of the circle
r = radius of the circle
\[\pi \] = constant
Therefore, area of the shot put circle is given as: -
\[A=\pi {{r}^{2}}\] - (i)
We have been provided with the radius of the circle equal to 2.135m. Therefore, substituting r = 2.135m in equation (i), we get,
\[A=\pi \times {{\left( 2.135 \right)}^{2}}\]
Now, substituting, \[\pi =\dfrac{22}{7}\] in the above relation, we have,
\[\begin{align}
& \Rightarrow A=\dfrac{22}{7}\times 2.135\times 2.135 \\
& \Rightarrow A=14.32585{{m}^{2}} \\
\end{align}\]
Note: One may note that we have taken the value of \[\pi =\dfrac{22}{7}\] because the radius 2.135m can be easily cancelled by 7. This makes our calculation easy. We can also substitute \[\pi =3.14\], this will not make a difference in the answer but only numbers after decimal may change. One important thing to remember is that we must write the unit of area in the end which is ‘\[{{m}^{2}}\]’ here. Failing to do so may get marks deducted in subjective questions.
Complete step by step answer:
The above figure can be assumed as the field of shot put circle. Here, we have assumed that the radius of the given field is ‘r’.
We know that area of a circle is given as: -
\[A=\pi {{r}^{2}}\]
Where, A = Area of the circle
r = radius of the circle
\[\pi \] = constant
Therefore, area of the shot put circle is given as: -
\[A=\pi {{r}^{2}}\] - (i)
We have been provided with the radius of the circle equal to 2.135m. Therefore, substituting r = 2.135m in equation (i), we get,
\[A=\pi \times {{\left( 2.135 \right)}^{2}}\]
Now, substituting, \[\pi =\dfrac{22}{7}\] in the above relation, we have,
\[\begin{align}
& \Rightarrow A=\dfrac{22}{7}\times 2.135\times 2.135 \\
& \Rightarrow A=14.32585{{m}^{2}} \\
\end{align}\]
Note: One may note that we have taken the value of \[\pi =\dfrac{22}{7}\] because the radius 2.135m can be easily cancelled by 7. This makes our calculation easy. We can also substitute \[\pi =3.14\], this will not make a difference in the answer but only numbers after decimal may change. One important thing to remember is that we must write the unit of area in the end which is ‘\[{{m}^{2}}\]’ here. Failing to do so may get marks deducted in subjective questions.
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