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# A manufacturer involved ten children in colouring a playing top (lattu) which is shaped like a cone surmounted by a hemisphere. The entire top is of $5cm$ in height and the diameter of the top is $3.5cm$. Find the area they had to paint if $50$ playing tops were given to them. (Take $\pi = \dfrac{{22}}{7}$ )

Last updated date: 07th Aug 2024
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Hint: In this problem, we are given a playing top which is a combination of two shapes: a cone and a hemisphere. The cone is surmounted by the hemisphere. The height of the top and the diameter of the hemisphere is given to us in the problem itself. We have to find the area of the playing top. So, we will have to find the area of both the shapes. Sometimes we need to calculate the dimensions of the shapes from the given data. So we will calculate the height of the cone and that of the hemisphere separately. On adding their areas, we will get the area of the top. We are also given that $50$ playing tops are to coloured by the children. So, we can find the area to be painted by the children by multiplying the area of a playing top by $50$ .

So, the diameter of the top $= 5cm$
Since the hemisphere is surmounted over the cone. So, the radii of both the shapes will be equal. Hence, Radius of cone $=$ Radius of hemisphere
Also, Radius of hemisphere $=$ Radius of cone $=$ $\dfrac{{Diameter\,of\,top}}{2}$
So, we get, Radius of cone $=$ Radius of hemisphere $=$ $\dfrac{{3.5}}{2}cm = 1.75cm$
Now, the height of the entire top $= 5cm$
Also, on seeing the figure, the height of top is the sum of radius of the hemisphere and height of the cone.
So, we get, height of top $=$ height of cone $+$ radius of hemisphere
$\Rightarrow$ $5cm$ $=$ height of cone $+$ $1.75cm$
$\Rightarrow$ height of cone $= 3.25cm$
Now, we can find the area of the shapes separately.
We know that the surface area of a hemisphere is $2\pi {r^2}$ .
So, the surface area of hemisphere $= 2\pi {\left( {1.75} \right)^2}c{m^2}$
Also, we know that the curved surface area of a cone is $\pi rl$ , where l is the slant height of the cone.
We can calculate the slant height of the cone by substituting the values of height and radius of the cone into the formula ${l^2} = {h^2} + {r^2}$ .
So, we get, ${l^2} = {\left( {3.25} \right)^2} + {\left( {1.75} \right)^2}$
$\Rightarrow l = \sqrt {{{\left( {3.25} \right)}^2} + {{\left( {1.75} \right)}^2}} cm$
So, surface area of cone
$= \pi \left( {1.75} \right)\left( {\sqrt {{{\left( {3.25} \right)}^2} + {{\left( {1.75} \right)}^2}} } \right)c{m^2}$
Now, total surface area of playing top $=$ surface area of cone $+$ surface area of hemisphere
Hence, the total surface area of playing top
$= \pi \left( {1.75} \right)\left( {\sqrt {{{\left( {3.25} \right)}^2} + {{\left( {1.75} \right)}^2}} } \right)c{m^2} + 2\pi {\left( {1.75} \right)^2}c{m^2}$
Taking $1.75\pi$ common from both the terms, we get,
$= 1.75\pi \left[ {\sqrt {{{\left( {3.25} \right)}^2} + {{\left( {1.75} \right)}^2}} + 2\left( {1.75} \right)} \right]c{m^2}$
$= 1.75\pi \left[ {\sqrt {{{\left( {3.25} \right)}^2} + {{\left( {1.75} \right)}^2}} + 3.5} \right]c{m^2}$
Substituting the value of $\pi$ as $\dfrac{{22}}{7}$ and evaluating the squares, we get,
$= 1.75 \times \dfrac{{22}}{7} \times \left[ {\sqrt {10.5625 + 3.0625} + 3.5} \right]c{m^2}$
Simplifying the expression further, we get,
$= 0.25 \times 22 \times \left[ {3.7 + 3.5} \right]c{m^2}$
Computing the product of the terms,
$= 0.25 \times 22 \times 7.2\,c{m^2}$
$= 39.6\,c{m^2}$ (approx.)
So, the area of a single playing top is $39.6\,c{m^2}$ .
Now, the area of $50$ such playing tops $= 50 \times 39.6\,c{m^2}$
$= 1980\,c{m^2}$
So, the area to be painted by the children is approximately $1980\,c{m^2}$ .
So, the correct answer is “ $1980\,c{m^2}$ . ”.

Note: Please note that we have not taken the flat surfaces of the cone and hemisphere into consideration while calculating the area of the playing top as they are not to be painted by the children. We must be aware of the difference in meaning and formulae of the terms lateral surface area and total surface area. Calculations should be taken care of so as to be sure of the final answer.