![SearchIcon](https://vmkt.vedantu.com/vmkt/PROD/png/bdcdbbd8-08a7-4688-98e6-4aa54e5e0800-1733305962725-4102606384256179.png)
Due to heavy floods in a state, thousands were rendered homeless. 50 schools collectively offered to the state government to provide a place and the canvas for 1500 tents to be fixed by the governments and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 m and height 3.5 m, with conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents costs Rs. 120 per sq. m , find the amount shared by each school to set up the tents. Enter the answer in the nearest integer. Use $\pi = \dfrac{{22}}{7}$
Answer
479.7k+ views
Hint:We will first find the area of the curved surface of the cylinder and cone and add them up to get the surface area of one tent and then multiply it by 1500 to get the surface area of 1500 tents. Now, multiply it by 120 to get the cost and divide that cost by 50 to get the share of each school.
Complete step-by-step answer:
Let us first write the formulas of curved surface area of the cylinder and the cone which we will require in the solution.
The curved surface area of a cylinder is given by the formula ${S_1} = 2\pi rh$ ……….(1), where r is the radius of the base of the cylinder and h is the height of the cylinder.
The curved surface area of a cone is given by the formula ${S_2} = \pi rl$ ………..(2), where r is the radius of the base of the cone and l is the slant height of the cone.
The slant height of a cone is given by $l = \sqrt {{r^2} + {h^2}} $ ………..(3), where r is the radius of the base of cone and h is the height of cone here.
Let us first find the slant height of the cone given to us.
Given that radius = 2.8 m and height = 2.1 m
Substituting in equation (3) we get,
$l = \sqrt {{{(2.8)}^2} + {{(2.1)}^2}} $
$ \Rightarrow l = \sqrt {7.84 + 4.41} m = \sqrt {12.25} m = 3.5m$
$ \Rightarrow l = 3.5m$
In question they said that the tent consists of two parts in which the upper part is in conical shape and lower part is in cylindrical shape, So sum of these two surface areas gives total surface area of tent.
Now, surface area of tent can be found by adding (1) and (2):
$ \Rightarrow S = {S_1} + {S_2}$
$ \Rightarrow S = 2\pi rh + \pi l = \pi r(2h + l)$ ………..(4)
Substituting value of radius,slant height and height of the cone in (4), we will get:-
$ \Rightarrow S = \dfrac{{22}}{7} \times 2.8 \times (7 + 3.5)$
Simplifying it to get:-
$ \Rightarrow S = \dfrac{{22}}{7} \times 2.8 \times 10.5 = 92.4{m^2}$ ………..(5)
Now, (5) gives us the surface area of one tent.
So, the surface area of 1500 tents will be $(92.4 \times 1500){m^2} = 138600{m^2}$.
Since, we are given that 1 square meter costs = Rs. 120
Hence, the cost of 138600 square meters will be = $Rs.(138600 \times 120) = Rs.16632000$.
Now, this whole is paid by 50 schools.
So, 1 school pays = $Rs.\dfrac{{1663200}}{{50}} = Rs.332640$.
Hence, one school pays Rs. 332640.
Note:The students may make it a bit easier by calculating that if 50 schools collaborate to make 1500 tents that means one school pays for $\dfrac{{1500}}{{50}} = 30$ tents. Now, after finding the cost of one tent by multiplying 92.4 by 120. We will multiply the result by 30 to get the share of one school.The students must not forget to find the value of slant height and use the normal height instead.And also should remember the formulas of surface area of cone and cylindrical shapes for solving these types of problems.
Complete step-by-step answer:
Let us first write the formulas of curved surface area of the cylinder and the cone which we will require in the solution.
The curved surface area of a cylinder is given by the formula ${S_1} = 2\pi rh$ ……….(1), where r is the radius of the base of the cylinder and h is the height of the cylinder.
The curved surface area of a cone is given by the formula ${S_2} = \pi rl$ ………..(2), where r is the radius of the base of the cone and l is the slant height of the cone.
The slant height of a cone is given by $l = \sqrt {{r^2} + {h^2}} $ ………..(3), where r is the radius of the base of cone and h is the height of cone here.
Let us first find the slant height of the cone given to us.
Given that radius = 2.8 m and height = 2.1 m
Substituting in equation (3) we get,
$l = \sqrt {{{(2.8)}^2} + {{(2.1)}^2}} $
$ \Rightarrow l = \sqrt {7.84 + 4.41} m = \sqrt {12.25} m = 3.5m$
$ \Rightarrow l = 3.5m$
In question they said that the tent consists of two parts in which the upper part is in conical shape and lower part is in cylindrical shape, So sum of these two surface areas gives total surface area of tent.
Now, surface area of tent can be found by adding (1) and (2):
$ \Rightarrow S = {S_1} + {S_2}$
$ \Rightarrow S = 2\pi rh + \pi l = \pi r(2h + l)$ ………..(4)
Substituting value of radius,slant height and height of the cone in (4), we will get:-
$ \Rightarrow S = \dfrac{{22}}{7} \times 2.8 \times (7 + 3.5)$
Simplifying it to get:-
$ \Rightarrow S = \dfrac{{22}}{7} \times 2.8 \times 10.5 = 92.4{m^2}$ ………..(5)
Now, (5) gives us the surface area of one tent.
So, the surface area of 1500 tents will be $(92.4 \times 1500){m^2} = 138600{m^2}$.
Since, we are given that 1 square meter costs = Rs. 120
Hence, the cost of 138600 square meters will be = $Rs.(138600 \times 120) = Rs.16632000$.
Now, this whole is paid by 50 schools.
So, 1 school pays = $Rs.\dfrac{{1663200}}{{50}} = Rs.332640$.
Hence, one school pays Rs. 332640.
Note:The students may make it a bit easier by calculating that if 50 schools collaborate to make 1500 tents that means one school pays for $\dfrac{{1500}}{{50}} = 30$ tents. Now, after finding the cost of one tent by multiplying 92.4 by 120. We will multiply the result by 30 to get the share of one school.The students must not forget to find the value of slant height and use the normal height instead.And also should remember the formulas of surface area of cone and cylindrical shapes for solving these types of problems.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Find the value of x if the mode of the following data class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
10 examples of friction in our daily life
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Prokaryotic Cells and Eukaryotic Cells
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
State and prove Bernoullis theorem class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What organs are located on the left side of your body class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Define least count of vernier callipers How do you class 11 physics CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)