Question

Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere (see Fig). The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. (Take $\pi =\frac{22}{7}$) [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 3 Marks

TSA of the top = CSA of hemisphere + CSA of cone

Now, the curved surface area of the hemisphere =

$=\frac{1}{2}\left(4\pi r^2\right)=2\pi r^2$

$=(2\times \frac{22}{7}\times \frac{3.5}{2}\times \frac{3.5}{2})c{m}^2$

Also the height of the cone = height of the top – height (radius) of the hemispherical part

$=(5-\frac{3.5}{2})cm=3.25cm$

So, the slant height of the cone
$(l=\sqrt{r^2+h^2}=\sqrt{{\left(\frac{3.5}{2}\right)}^2+(3.5{)}^2}cm=3.7cm$ ( Approx.)

Therefore , CSA of cone
$=\pi rl=(\frac{22}{7}\times \frac{3.5}{2}\times 3.7)c{m}^2$

This gives the total surface area of the top as
=CSA of hemisphere + CSA of cone

$=(2\times \frac{22}{7}\times \frac{3.5}{2}\times \frac{3.5}{2})c{m}^2$ + $(\frac{22}{7}\times \frac{3.5}{2}\times 3.7)c{m}^2$

$=\frac{22}{7}\times \frac{3.5}{2}(3.5+3.7)c{m}^2=\frac{11}{2}\times (3.5+3.7)c{m}^2$

$=39.6c{m}^2\left(approx\right)$

You may note that 'total surface area of the top' is not the sum of the total surface areas of the cone and hemisphere but it is sum of CSA of the cone and hemisphere.