
It is required to make a hollow cone 24cm high and whose base radius is 7cm. Find the area of metal sheet required including the base. Also, find the capacity of the cone.
Answer
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Hint: We are given the radius and height of the cone. Using the given measurements and Pythagoras theorem, we can calculate the slant height. Once we have the slant height, we will calculate total surface area using the values that are given in the question and that we have found. Next, to calculate the capacity of the cone, we will calculate the volume simply by putting the values.
Formula used: 1) Total surface area of cone- $\pi r(l + r)$
2) Volume of cone - $\dfrac{1}{3}\pi {r^2}h$
Complete step-by-step answer:
We know that the radius $\left( r \right)$ of the base of the cone is $7cm$, height $(h)$ of the cone is $24cm$.
We can calculate the slant height using Pythagoras Theorem,
$ \Rightarrow {(AO)^2} + {(OB)^2} = {(AB)^2}$
Substituting the values given in question,
$ \Rightarrow {24^2} + {7^2} = {(AB)^2}$
Solving we get,
$ \Rightarrow 576 + 49 = {(AB)^2}$
$ \Rightarrow 625 = {(25)^2} = {(AB)^2}$
$ \Rightarrow AB = 25cm$
To find the total metal sheet required to make the base, we will have to find the total surface area of the cone.
We know total surface area of cone = $\pi r(l + r)$
Putting $l = 25cm$ and $r = 7cm$,
$ \Rightarrow 7\pi (25 + 7)$
Putting $\pi = \dfrac{{22}}{7}$,
$ \Rightarrow \dfrac{{22}}{7} \times 7 \times 32$
On solving we will have,
Total surface area = $22 \times 32 = 704c{m^2}$
Next, we have been asked to calculate the capacity of the cone. For this, we will calculate the volume of the cone.
We know, volume of cone = $\dfrac{1}{3}\pi {r^2}h$
Putting $\pi = \dfrac{{22}}{7}$, $r = 7cm$ and $h = 24cm$,
$ \Rightarrow \dfrac{1}{3} \times \dfrac{{22}}{7} \times {7^2} \times 24$
On solving we will have,
Volume = $1232c{m^3}$
Hence, we require $704c{m^2}$ of metal sheet and the capacity of the tank is $1232c{m^3}$.
Note: It must be noted that in this question we will find total surface area and not curved surface area. This is because it is mentioned in the question that a metal sheet is required for the base also. Total surface area includes base whereas curved surface area excludes base.
Formula used: 1) Total surface area of cone- $\pi r(l + r)$
2) Volume of cone - $\dfrac{1}{3}\pi {r^2}h$
Complete step-by-step answer:
We know that the radius $\left( r \right)$ of the base of the cone is $7cm$, height $(h)$ of the cone is $24cm$.
We can calculate the slant height using Pythagoras Theorem,
$ \Rightarrow {(AO)^2} + {(OB)^2} = {(AB)^2}$
Substituting the values given in question,
$ \Rightarrow {24^2} + {7^2} = {(AB)^2}$
Solving we get,
$ \Rightarrow 576 + 49 = {(AB)^2}$
$ \Rightarrow 625 = {(25)^2} = {(AB)^2}$
$ \Rightarrow AB = 25cm$
To find the total metal sheet required to make the base, we will have to find the total surface area of the cone.
We know total surface area of cone = $\pi r(l + r)$
Putting $l = 25cm$ and $r = 7cm$,
$ \Rightarrow 7\pi (25 + 7)$
Putting $\pi = \dfrac{{22}}{7}$,
$ \Rightarrow \dfrac{{22}}{7} \times 7 \times 32$
On solving we will have,
Total surface area = $22 \times 32 = 704c{m^2}$
Next, we have been asked to calculate the capacity of the cone. For this, we will calculate the volume of the cone.
We know, volume of cone = $\dfrac{1}{3}\pi {r^2}h$
Putting $\pi = \dfrac{{22}}{7}$, $r = 7cm$ and $h = 24cm$,
$ \Rightarrow \dfrac{1}{3} \times \dfrac{{22}}{7} \times {7^2} \times 24$
On solving we will have,
Volume = $1232c{m^3}$
Hence, we require $704c{m^2}$ of metal sheet and the capacity of the tank is $1232c{m^3}$.
Note: It must be noted that in this question we will find total surface area and not curved surface area. This is because it is mentioned in the question that a metal sheet is required for the base also. Total surface area includes base whereas curved surface area excludes base.
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