- Plane Figure
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- Solid Figure
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- Perimeter
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- Surface Area
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- Volume
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- Surface Area and Volume of Cube and Cuboid
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- Surface Area and Volume of Right circular cylinder
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- Surface Area and Volume of Right circular cone
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- Surface Area and Volume of sphere and hemisphere
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- How the Surface area and Volume are determined

- surfacearea_volume Formative Assignment
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- surfacearea_volume Assignments
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- surfacearea_volume Worksheet

Given below are the **Class 9 Maths** Important Questions for Surface area and volume

a) Concepts questions

b) Calculation problems

c) Multiple choice questions

d) Long answer questions

e) Fill in the blank's

a) Concepts questions

b) Calculation problems

c) Multiple choice questions

d) Long answer questions

e) Fill in the blank's

(a) Remain same

CSA= 2πrh

So if r is doubled, h is halved,CSA will remain same

(b) 616 cm

(c) 1728 cm

(d) 956 cm

(e) 1:1 In this height of cylinder will be 2r and radius of cylinder will r. So it is equal

(f) 512 m

a) A cylinder, hemisphere and cone stand on equal base and same height, the Volume ratio is 3:2:1

b) The radius of a solid sphere is 24 cm. 8 spheres can be made from it of 12cm radius

c) radius of the cone is doubled and height is halved, the volume will be halved

d) A river 10m deep and 40m wide is flowing at the rate of 2m per min. 48000m

e) A cylinder radius is halved and height is doubled, the volume will become halved

f) The side of the cube is 4 cm, the diagonal length is cm

- True.. All of them will have height as r as hemisphere height can be r only.

Hemisphere=(2/3) πr

Cone=(1/3) πr

So ratio is 3:2:1

b) True Sphere volume =(4/3) πr

c) False. Volume will be doubled Volume is given by =(1/3)πr

d) True. It is equal to the volume of the cuboid 10m,40m,and 120 m

e) True.

- False

The length, breadth and height of a room are 12 m, 10 m, and 9m respectively. Find the area of our walls of room?

a) 636 m

b) 516 m

c) 800 m

d) 456m

Area of the walls is given by =2(BH+LH) +LB=2(90+108)+120=516 m

The plastic paint in a Asian paint container is sufficient to paint an area equal to 93.75m

a) 100

b) 800

c) 940

d) 1000

Total surface area of one block = 2(lb + bh + lh)

= [2(22.5 ×10 + 10 × 7.5 + 22.5 × 7.5)] cm

= 2(225 + 75 + 168.75) cm

= (2 × 468.75) cm

= 937.5 cm

Let n blocks can be painted out by the paint of the container.

Area of n bricks = (n ×937.5) cm

Area that can be painted by the paint of the container = 93.75 m

937500 = 937.5n

n = 1000

Therefore, 100 blocks can be painted out by the paint of the container

Curved surface area of a cone is 308 cm

a) Radius of the cone is 7cm

b) total surface area is 462 cm

c) Height of the cone is (147)

d) None of the above

Curved surface area=πrl

So r=7 cm

Now total surface area=πr

Sita had to make a model of cylindrical kaleidoscope for her science project. She wanted to use black chart paper to make the curved surface of the kaleidoscope. What would be the area of chart paper required by her, if she wanted to make a kaleidoscope of length 30cm with a 2.7 cm radius? (Use p=22/7)

a) 1320 cm

b) 1400 cm

c) 986 cm

d) None of these

Curved surface of cylinder=2πrh

The radii of two cones are in the ratio of 2:3 and their heights are in the ratio of 7:3. The ratio of their volumes is

a) 20:9

b) 28:9

c) 29:29

d) None of these

Volume =(1/3)πr

Find the maximum length of the rod that can be kept in cuboidal box of sides 30cm, 20cm and 10cm.

a)√1400

b)2√400

c)2√300

d) None of these

Diagonal is the longest length in the cuboid so

D=(L

=√1400

A box is made entirely of glass panes (including base) held together with tape. It is 3 cm long, 2.5 cm wide and 2.5 cm high. How much of tape is needed for all the 12 edges?

a) 30cm

b) 32cm

c) 40 cm

d) None of these

Length of tape=4( L+B+H)=32 cm

The curved surface area of a right circular cylinder of height 14 cm is 88 cm

a) 144 cm

b) 180 cm

c) 176 cm

d) None of the above

Curved surface area of Cone |
3πr^{2} |

Curved surface of Hemisphere |
2πrH |

Curved surface area of Cylinder |
2πr^{2} |

Total surface area of Hemisphere |
$\pi r^{2}+\pi r\sqrt{r^{2}+H^{2}}$ |

Total surface area of cone |
$\pi r\sqrt{r^{2}+H^{2}}$ |

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