Question

answer.............................................​

Answer 1

Answer:

$\Large{\boxed{\underline{\overline{\mathfrak{\star \: QuEsTiOn:- \: \star}}}}}$

A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere of the radius of the base of the cone is 21 cm and its volume is of the volume o f the hemisphere.calculate the height of the cone and surface area of the toy.

$\huge\underline{\overline{\mid{\bold{\pink{ANSWER-}}\mid}}}$

Given :-

◆ Toy is in the form of hemisphere surmounted by a right circular cone

◆ Base radius of hemisphere = Base radius of Cone = r = 21 cm

◆ Volume of hemisphere = volume of cone

To find :-

height of cone = h

surface area of toy

Formula required :-

❥ Formula of volume of cone

$\boxed{\rm{Volume\:of\:cone=\frac{1}{3}\pi r^2h}}$

(where , r is the radius of base of cone , h is the height of cone)

❥ Formula of volume of Hemisphere

$\boxed{\rm{Volume\:of\:Hemisphere=\frac{2}{3}\pi r^3}}$

(where r is the radius of hemisphere )

❥ Formula for CSA of cone

$\boxed{\rm{CSA \:of\:cone=\pi r l }}$

(where r is the radius of cone , l is the slant height of cone )

❥ Formula for CSA of hemisphere

$\boxed{\rm{CSA\:of\:hemishere=2\pi r^2}}$

(where r is the radius of hemisphere)

❥ Know Pythagoras theorem for finding the slant height of cone

In a right angled triangle

$\boxed{\rm{(Hypotenuse)^2=(height)^2+(base)^2}}$

$\Large{\underline{\underline{\bf{SoLuTion:-}}}}$

Finding the height of cone

Given that

volume of cone = volume of hemisphere

$\rm{\frac{1}{3}\pi r^2h=\frac{2}{3}\pi r^3}\\\\\rm{r^2h=2r^3}\\\\\rm{h=2r=2(21\;cm)=42\;cm}\\\\\boxed{\pink{\rm{h=42\:cm}}}$

Finding the slant height of cone

Let,

slant height of cone = l

By Pythagoras theorem

(slant height)² = (radius)² + (height)²

(slant height)² = (21)² + (42)²

$\boxed{\pink{\rm{slant\;height=l=21\sqrt{5}\;cm}}}$

Finding the Surface area of toy

S.A of toy =CSA of cone + CSA of hemisphere

$\rm{S.A\:of\:toy=\pi r l + 2\pi r^2}\\\\\rm{S.A\:of\:toy=\frac{22}{7}\times(21)\times(21\sqrt{5})+(2\times\frac{22}{7}\times(21)^2)}\\\\\rm{S.A\:of\:toy=1386\sqrt{5}\;+2772}\\\\\boxed{\boxed{\purple{\rm{S.A\:of\:toy=1386(\sqrt{5}\;+2)\;\;cm^2}}}}$