Question

How much canvas is required to make a tent of base area 440 square meters and height 12 meters?

Answer 1

Hint – In this particular question the canvas required is the total surface area of the tent which is the sum of the lateral surface area and the curved surface area so use this concept to reach the solution of the question.

Complete step-by-step solution -

The pictorial representation of the tent is shown above in the figure, as the shape of the tent is not given so we assumed the general shape of the tent which is in the form of a right circular cone.
Let r be the radius of the circular base of the tent, h be the height of the tent and L be the slant height of the tent.
It is given that the height of the tent = 12 meters.
Therefore, h = 12 m
It is also given that the area of the base = 440 square meter.
As we know that the area of the circle = $\pi {r^2}$ square units, where r is the radius of the circle.
Therefore, $\pi {r^2}$ = 440
$ \Rightarrow \dfrac{{22}}{7}{r^2} = 440$, $\left[ {\because \pi = \dfrac{{22}}{7}} \right]$
$ \Rightarrow {r^2} = \dfrac{{440 \times 7}}{{22}} = 140$
Now take square root on both sides we have,
$ \Rightarrow r = \sqrt {140} $ Meter
Now apply Pythagoras theorem in triangle AOB we have,
$ \Rightarrow {\left( {{\text{hypotenuse}}} \right)^2} = {\left( {{\text{perpendicular}}} \right)^2} + {\left( {{\text{base}}} \right)^2}$
$ \Rightarrow {\left( {{\text{AB}}} \right)^2} = {\left( {{\text{AO}}} \right)^2} + {\left( {{\text{OB}}} \right)^2}$
$ \Rightarrow {\left( L \right)^2} = {\left( {{\text{12}}} \right)^2} + {\left( {\sqrt {140} } \right)^2} = 144 + 140 = 284$
Now take square root on both sides we have,
$ \Rightarrow \left( L \right) = \sqrt {284} $ m
Now as we know that the total surface area of the right circular cone is the sum of the lateral surface area and the curved surface area.
So the lateral surface area of the cone = $\pi rL$
And the curved surface area of the cone = $\pi {r^2}$
So the total surface area (T.S.A) of the cone = $\pi rL$ + $\pi {r^2}$
Now substitute the values we have,
Therefore, T.S.A = $\dfrac{{22}}{7}\sqrt {140} \sqrt {284} + \dfrac{{22}}{7}\left( {140} \right)$
Now simplify this we have,
Therefore, T.S.A = $\dfrac{{22}}{7}\sqrt {2 \times 2 \times 7 \times 5} \sqrt {2 \times 2 \times 71} + 22\left( {20} \right)$
\[ \Rightarrow \dfrac{{22}}{7}\left( {2 \times 2} \right)\sqrt {35} \sqrt {71} + 440\]
\[ \Rightarrow \dfrac{{22}}{7}\left( 4 \right)\left( {5.916} \right)\left( {8.426} \right) + 440\]
\[ \Rightarrow \dfrac{{22}}{7}\left( 4 \right)\left( {5.916} \right)\left( {8.426} \right) + 440 = 1066.11\] Square meter.
So this is the required canvas to make the tent in square meters.

Note – Whenever we face such types of questions the key concept we have to remember is that the canvas required to make the tent is the total surface area of the tent and always remember the formula of the total surface area of the right circular cone which is stated above then first calculate the radius of the base then the slant height of the cone as above then substitute all these values in the formula as above and simplify we will get the required canvas to make the tent.