Calculate the total surface area of a cone of height $12\,cm$ and base radius $5\,cm$. Take $\pi =\dfrac{22}{7}$ ?
Answer
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Hint:Here we have to find the total surface area of the cone. Firstly we will find the slant height of the cone using the Pythagoras theorem. Then we will write the formula for total surface area and substitute all the values in it. Finally we will put the value of pie in the formula and simplify it to get our desired answer.
Complete step by step solution:
We have been given the following information about the cone:
Height $h=12\,cm$
Radius $r=5\,cm$…$\left( 1 \right)$
So the diagram of the cone will be as follows:
Now we will find the slant height of the cone denoted as $s$
As we can see in the diagram that $\Delta AOC$ is a right angle triangle so we will use the Pythagoras theorem such that:
$A{{C}^{2}}=A{{O}^{2}}+O{{C}^{2}}$
Where $AO=$ Height and $OC=$ Radius
$\Rightarrow {{s}^{2}}={{\left( 12 \right)}^{2}}+{{\left( 5 \right)}^{2}}$
Simplifying further we get,
$\Rightarrow {{s}^{2}}=144+25$
$\Rightarrow {{s}^{2}}=169$
Take square root both sides,
$\Rightarrow s=\sqrt{169}$
$\Rightarrow s=13\,cm$…$\left( 2 \right)$
So we can draw the cone as follows:
The total surface area of a cone is the sum of the area of the base and the lateral area given as follows:
Surface Area $=\pi {{r}^{2}}+\pi rs$
Substitute the value from equation (1) and (2) above,
Surface Area $=\pi \times {{5}^{2}}+\pi \times 5\times 13$
Surface Area $=25\times \pi +65\times \pi $
Put $\pi =\dfrac{22}{7}$ above,
Surface Area $=25\times \dfrac{22}{7}+65\times \dfrac{22}{7}$
Surface Area $=\dfrac{550}{7}+\dfrac{1430}{7}$
Surface Area $=\dfrac{550+1430}{7}$
Surface Area $=\dfrac{1980}{7}\,c{{m}^{2}}$
Surface Area $=282\dfrac{6}{7}\,c{{m}^{2}}$
Hence the total surface area of a cone of height $12\,cm$ and base radius $5\,cm$ is $282\dfrac{6}{7}\,c{{m}^{2}}$ .
Note:
Cone is a three-dimensional shape that has a circular base and a pointed top which is like a triangle but in $3-d$ . It has only one face which is circular base and no edges. Curved surface area and total surface area are calculated by different formula one shouldn’t get confused between them. Formula for curved surface area is $\pi rl$ where $r$ is the radius and $l$ is the height of the cone.
Complete step by step solution:
We have been given the following information about the cone:
Height $h=12\,cm$
Radius $r=5\,cm$…$\left( 1 \right)$
So the diagram of the cone will be as follows:
Now we will find the slant height of the cone denoted as $s$
As we can see in the diagram that $\Delta AOC$ is a right angle triangle so we will use the Pythagoras theorem such that:
$A{{C}^{2}}=A{{O}^{2}}+O{{C}^{2}}$
Where $AO=$ Height and $OC=$ Radius
$\Rightarrow {{s}^{2}}={{\left( 12 \right)}^{2}}+{{\left( 5 \right)}^{2}}$
Simplifying further we get,
$\Rightarrow {{s}^{2}}=144+25$
$\Rightarrow {{s}^{2}}=169$
Take square root both sides,
$\Rightarrow s=\sqrt{169}$
$\Rightarrow s=13\,cm$…$\left( 2 \right)$
So we can draw the cone as follows:
The total surface area of a cone is the sum of the area of the base and the lateral area given as follows:
Surface Area $=\pi {{r}^{2}}+\pi rs$
Substitute the value from equation (1) and (2) above,
Surface Area $=\pi \times {{5}^{2}}+\pi \times 5\times 13$
Surface Area $=25\times \pi +65\times \pi $
Put $\pi =\dfrac{22}{7}$ above,
Surface Area $=25\times \dfrac{22}{7}+65\times \dfrac{22}{7}$
Surface Area $=\dfrac{550}{7}+\dfrac{1430}{7}$
Surface Area $=\dfrac{550+1430}{7}$
Surface Area $=\dfrac{1980}{7}\,c{{m}^{2}}$
Surface Area $=282\dfrac{6}{7}\,c{{m}^{2}}$
Hence the total surface area of a cone of height $12\,cm$ and base radius $5\,cm$ is $282\dfrac{6}{7}\,c{{m}^{2}}$ .
Note:
Cone is a three-dimensional shape that has a circular base and a pointed top which is like a triangle but in $3-d$ . It has only one face which is circular base and no edges. Curved surface area and total surface area are calculated by different formula one shouldn’t get confused between them. Formula for curved surface area is $\pi rl$ where $r$ is the radius and $l$ is the height of the cone.
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