Question

A toy is made in the shape of a cylinder with one hemisphere stuck to one end

and a cone to the other end, as shown in the figure, the length of the cylindrical

part of the toy is 20cm and its diameter is 10cm. If the slant height of the cone

is 13cm. Find the surface area of the toy.​

Answer 1

Answer:

Height of the cylindrical part =13cm

Radius of cone , cylinder and hemi sphere=5cm

r=5cm for hemisphere cylinder and cone.

Height of cone h=30−5−13=12

The area of canvas required=Surface area of hemishphere,cylinder and cone parts of tent

A=2πr

2

+2πrH+πr(

h

2

+r

2

)

A=2π×5×5+2π×5×13+π×5(

5

2

+12

2

)

A=770cm

2

Answer 2

Answer:

$2990 {cm \: }^{2} \: \:is \: the \: anwer$

step by step explanation..

[i]. for Corbitat portion

r= 10√2 25cm l=13

[ii]. For cylindrical portion

r= 10/2=25cm

[iii]. For hemisphere portion

r = 10/2=5cm

Surface area of toy

= CSA of hemisphere+COSA of cylinder+ CSA

= 2π r2+2π that + πr

= πr (2r+2h+1)

= 22/7×5(2×5+2×20+13)

= 22/7×5×9 =

$2990 {cm}^{2}$

$surface \: area \: of \: toy \: is \: 2990 {cm}^{2}$