
The circumference of lower and upper bases of the frustum cone is 3.5 and 1.5 ft. The slant height is 5 ft. What is the curved surface area?
A. $12.5f{{t}^{2}}$
B. $72.5f{{t}^{2}}$
C. $11.5f{{t}^{2}}$
D. $10.5f{{t}^{2}}$
Answer
495k+ views
Hint: We know that the curved surface area of a frustum cone is given by the formula $\pi l\left( R+r \right)$, where l is the slant height, R is the radii of a bigger circle and r is the radii of a smaller circle. So by substituting the values we will get the desired answer.
Complete step by step answer:
We have been given that the circumference of lower and upper bases of the frustum cone is 3.5 and 1.5 ft. The slant height is 5 ft.
We have to find the curved surface area of the frustum cone.
Now, we know that the curved surface area of a frustum cone is given by the formula $\pi l\left( R+r \right)$, where l is the slant height, R is the radii of a bigger circle and r is the radii of a smaller circle.
Now, we know that the circumferences of lower and upper bases of a frustum are given by the formula $2\pi R$ and $2\pi r$ respectively.
Now, we have
$\begin{align}
& \Rightarrow 2\pi R=3.5 \\
& \Rightarrow R=\dfrac{3.5}{2\pi } \\
\end{align}$
Also we have
$\begin{align}
& \Rightarrow 2\pi r=1.5 \\
& \Rightarrow r=\dfrac{1.5}{2\pi } \\
\end{align}$
Also given in the question the slant height $l=5ft.$
Now, substituting the values in the formula we will get
$\begin{align}
& \Rightarrow CSA=\pi l\left( R+r \right) \\
& \Rightarrow CSA=\pi \times 5\left( \dfrac{3.5}{2\pi }+\dfrac{1.5}{2\pi } \right) \\
\end{align}$
Now, simplifying the above obtained equation we will get
$\begin{align}
& \Rightarrow CSA=5\pi \left( \dfrac{3.5+1.5}{2\pi } \right) \\
& \Rightarrow CSA=5\pi \left( \dfrac{5}{2\pi } \right) \\
& \Rightarrow CSA=\left( \dfrac{25}{2} \right) \\
& \Rightarrow CSA=12.5f{{t}^{2}} \\
\end{align}$
Hence above is the required curved surface area of frustum cones.
So, the correct answer is “Option A”.
Note: The point to be noted is that curved surface area does not include the area of top and bottom. Also write the units with the given measures and also write the units with the final answer. Alternatively we can directly use the formula $CSA=\dfrac{1}{2}l\left( 2\pi R+2\pi r \right)$ to find the curved surface area when the circumference is given.
Complete step by step answer:
We have been given that the circumference of lower and upper bases of the frustum cone is 3.5 and 1.5 ft. The slant height is 5 ft.
We have to find the curved surface area of the frustum cone.
Now, we know that the curved surface area of a frustum cone is given by the formula $\pi l\left( R+r \right)$, where l is the slant height, R is the radii of a bigger circle and r is the radii of a smaller circle.
Now, we know that the circumferences of lower and upper bases of a frustum are given by the formula $2\pi R$ and $2\pi r$ respectively.
Now, we have
$\begin{align}
& \Rightarrow 2\pi R=3.5 \\
& \Rightarrow R=\dfrac{3.5}{2\pi } \\
\end{align}$
Also we have
$\begin{align}
& \Rightarrow 2\pi r=1.5 \\
& \Rightarrow r=\dfrac{1.5}{2\pi } \\
\end{align}$
Also given in the question the slant height $l=5ft.$
Now, substituting the values in the formula we will get
$\begin{align}
& \Rightarrow CSA=\pi l\left( R+r \right) \\
& \Rightarrow CSA=\pi \times 5\left( \dfrac{3.5}{2\pi }+\dfrac{1.5}{2\pi } \right) \\
\end{align}$
Now, simplifying the above obtained equation we will get
$\begin{align}
& \Rightarrow CSA=5\pi \left( \dfrac{3.5+1.5}{2\pi } \right) \\
& \Rightarrow CSA=5\pi \left( \dfrac{5}{2\pi } \right) \\
& \Rightarrow CSA=\left( \dfrac{25}{2} \right) \\
& \Rightarrow CSA=12.5f{{t}^{2}} \\
\end{align}$
Hence above is the required curved surface area of frustum cones.
So, the correct answer is “Option A”.
Note: The point to be noted is that curved surface area does not include the area of top and bottom. Also write the units with the given measures and also write the units with the final answer. Alternatively we can directly use the formula $CSA=\dfrac{1}{2}l\left( 2\pi R+2\pi r \right)$ to find the curved surface area when the circumference is given.
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