Question

.A toy is in the form of a right circular cylinder with a hemisphere on one endand a cone on the other. The height and radius of the cylindrical part are 13cm and 5 cm respectively. The radii of the hemispherical and conical parts arethe same as that of the cylindrical part. Calculate the surface area of the toy ifthe height of the conical part is 12 cm brainly

Answer 1

Answer:

770

Step-by-step explanation:

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=2πr2+2πrH+πr(h2+r2)

A=2π×5×5+2π×5×13+π×5(52+122) 

A=770cm2

Answer 2

$\begin{gathered}\begin{gathered}\bf\: Given-\begin{cases} &\sf{radius_{(cylinder)},r = 5 \: cm} \\ &\sf{radius_{(cone)},r = 5 \: cm}\\ &\sf{radius_{(hemisphere)},r = 5 \: cm}\\ &\sf{height_{(cylinder)},h = 13 \: cm}\\ &\sf{height_{(cone)},H = 12 \: cm} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{Surface Area _{(toy)}}\end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \sf \: CSA_{(cylinder)} = \:2\pi \: rh }$

$\boxed{ \sf \: CSA_{(cone)} \: \pi \: rl}$

where,

$\boxed{ \sf \: {l}^{2} \: = {r}^{2} + {h}^{2} }$

$\boxed{ \sf \: CSA_{(hemi \: sphere)} \: = \: {2\pi \: r}^{2} }$

$\large\underline{\sf{Solution-}}$

Given that,

• Radius of cone, r = 5 cm

• Height of cone, H = 12 cm

So,

• Slant height (l) of cone is

$\rm :\longmapsto\: {l}^{2} = {r}^{2} + {H}^{2}$

$\rm :\longmapsto \: {l}^{2} = {5}^{2} + {12}^{2}$

$\rm :\longmapsto\: {l}^{2} = 25 + 144$

$\rm :\longmapsto\: {l}^{2} = 169$

$\rm :\longmapsto\:l \: = \: 13 \: cm$

Now,

$\sf \: Surface Area _{(toy)} = CSA_{(cone)} + CSA_{(cylinder)} + CSA_{(hemisphere)}$

$\: \: \: = \: \sf \: \: \pi \: rl + 2\pi \: rh + {2\pi \: r}^{2}$

$\: \: \: = \: \sf\: \pi \: r \: (l + 2h + 2r)$

$\: \: \: = \sf \: \dfrac{22}{7} \times 5 \times (13 + 2 \times 13+ 2 \times5)$

$\: \: \: = \sf \: \dfrac{110}{7} \times (13 + 26 + 10)$

$\: \: \: = \sf \: \dfrac{110}{7} \times 49$

$\: \: \: = \sf \: 770 \: {cm}^{2}$

$\bf\implies \:Surface Area _{(toy)} = 770 \: {cm}^{2}$

Additional Information :-

Perimeter of rectangle = 2(length× breadth)

Diagonal of rectangle = √(length ²+breadth ²)

Area of square = side²

Perimeter of square = 4× side

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²