Questions & Answers

Question

Answers

$

a)\,11.88\pi c{m^2} \\

b)\,1.88\pi c{m^2} \\

c)\,31.88\pi c{m^2} \\

d)\,none\,\,of\,these \\

$

Answer
Verified

It is given that a play top is in the form of a hemisphere surmounted on a cone.

The diameter of the hemisphere is $3.6cm$.

Since the hemisphere and the cone are evenly joined, the diameter of the bases will be the same.

The total height of the play top is also given and is equal to $4.2cm$. The height is actually the sum of the radius of the hemisphere and height of the cone.

The radius $r$ of the hemisphere$ = \dfrac{{Diameter}}{2}$

$r = \dfrac{{3.6}}{2} = 1.8cm\,\,\,\,\,\,\,\,\,\, \to \left( 1 \right)$

Also, we need to calculate the height of the cone.

$

Total\,height = Height\,of\,cone + r \\

Height\,of\,cone(h) = 4.2 - 1.8 = 2.4cm\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \left( 2 \right) \\

$

Since we have the height as well as the radius of the cone so we can calculate length the curved region that is hypotenuse using Pythagoras Theorem,

$

{(height)^2} + {(radius)^2} = {(hypotenuse)^2} \\

{(h)^2} + {(r)^2} = {(l)^2} \\

l = \sqrt {{h^2} + {r^2}} \\

l = \sqrt {{{2.4}^2} + {{1.8}^2}} = \sqrt {5.76 + 3.24} \\

l = \sqrt 9 \\

l = 3\,\,\,\,\,\,\,\,\,\,\, \to \left( 3 \right) \\

$

So the surface area of the body will be the area of the body exposed since the hemisphere and cone are joined together, the area of their bases will not be included in the total surface area of the play top.

$

\therefore Total\,surface\,area = Curved\,surface\,area\,of\,cone + Curved\,surface\,area\,of\,hemisphere \\

$

Substituting the values from equations (1) , (2) and (3) we get,

$

Total\,surface\,area = \pi rl + 2\pi {r^2} \\

Total\,surface\,area = \pi (1.8)(3) + 2\pi (1.8)(1.8) \\

Total\,surface\,area = 5.4\pi + 6.48\pi \\

Total\,surface\,area = 11.88\pi c{m^2} \\

$