# Question-1

**Find the longest pole length that can be put in a room of 10 m x 10 m x 5 m dimensions.**

**Solution:**

Longest pole will be = longest diagonal of the room =

=

= 15 m

# Question-2

**A 3 cm thick iron open box is made whose external dimensions are 78 cm x 54 cm x 12 cm. Find the weight of the box if its density is 60 gm/cm**

^{3}.**Solution:**

Thickness of the iron sheet = 3 cm

External dimensions at the box = 78 cm x 54 cm x 12 cm

Therefore internal dimensions of the box = 72 m x 48 cm x 9 cm

Volume of the iron used in making the box

= External volume of the box - Internal volume

= 78 x 54 x 12 - 72 x 48 x 9

= 50544 - 31104 = 19440 cm

^{3}

Weight = Volume x density

= 19440 x 60 gm

= 1166.4 kg

# Question-3

**Find the surface area of a cube whose volume is given:**

(i) 216 cm

(i) 216 cm

^{3}(ii) 1000 cm^{3}**Solution:**

Let the edge of the cube= a cm

Then volume of the cube = a

^{3}cu. cm and surface area = 6a

^{2}sq. cm.

(i) a

^{3 }= 216 â‡’ a = 6 cm

âˆ´Total surface area = 6a

^{2}= 6 x (6)

^{2 }= 216 cm

^{2}

(ii) a

^{3}=1000 â‡’ a=10 cm

Total S.A.= 6a

^{2}= 6(10)

^{2}= 600 sq. cm

# Question-4

**Two equal cubes with sides 6 cm are placed one above the other, forming a cuboid. Find the total surface area of the cuboid thus formed.**

**Solution:**

Side of the cube = 6 cm

.

^{.}. Length of the cuboid = 6 + 6 = 12 cm

Breadth of the cuboid = 6 cm

and height = 6 cm

Total surface area = 2(lb + bh + hl)

= 2(12 x 6 + 6 x 6 + 6 x 12)

= 360 cm

^{2}

# Question-5

**Two cubes each of volume 64 cm ^{3} are joined end-to-end. Find the total surface area of the resulting cuboid.**

**Solution:**

Volume of the cube = 64 cm

^{3}

.

^{.}. a

^{3 }= 64

â‡’ a = 4 cm

.

^{.}. Length of resulting cuboid = 4 + 4 = 8 cm.

breadth = 4 cm and height = 4 cm.

Total surface area of the cuboid = 2(lb + bh + hl)

= 2(8 x 4 + 4 x 4 + 4 x 8)

= 160 cm

^{2}

# Question-6

**The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume be of the volume of the given cone, at what height above the base is the section made.**

**Solution:**

Let the height of the original cone (H) = 30 cm

âˆ´Volume of the original cone = R

^{2}H

= R

^{2 }x 30 = 10R

^{2}

Volume of the small cone( r

^{2}h) = of Volume of the original cone | h = height of small cone

â‡’ r

^{2}h = R

^{2 }â‡’ h = ........(i)

Since Î”ADE ~ Î”ABC (BC Il DE)

...........(ii)

From Eqs (i) and (ii) we get

h =

h = 10 cm.

.

^{.}. Height above the base = 30 - 10 = 20 cm.

# Question-7

**A circus tent consists of cylindrical base surmounted by a conical roof. The radius of the cylinder is 20 m. The height of the tent is 63 m and that of the cone is 21 m. Find the volume of the tent and the area of the canvas used for making it.**

**Solution:**

Height of the tent = height of the cylinder + height of the cone.

Given, height of the cone (H) = 21 m

height of cylindrical base(h) = 63 - 21 = 42 m

Radius of the cylinder (r)= radius of the cone(r) = 20 m

âˆ´ Volume of the circus tent = Volume of the cylinder + volume of the cone.

=Ï€r

^{2}h +

^{ }Ï€r

^{2 }Î— =

^{ }Ï€r

^{2 }(h+)

= x 400(42 + 7)

= 61600 m

^{3}

Slant height of the cone = = 29

Now, area of the canvas in making the tent

= curved surface area of the cylinder + Curved surface area of the cone

= 2Ï€rh + Ï€r

*l*

= Ï€r(2h +

*l*)

= x 20(84 + 29)

= 7102.85 m

^{2}.

# Question-8

**Find the volume and the curved surface area of a sphere of radius 5.6 cm.**

**Solution:**

Given, radius of the sphere=5.6 cm

Volume of the sphere =Ï€r

^{3 }

= x (5.6)

^{3}

= 735.91 cm

^{3}

Surface area of the sphere = 4Ï€r

^{2}

= 4x (5.6)

^{2}

= 394.24 cm

^{2}

# Question-9

**The volume of two hemispheres are in the ratio 8 : 27. Find the ratio of their radii.**

**Solution:**

Let the volumes of two hemispheres be V

_{1}and V

_{2}and the radii be r

_{1}and r

_{2}respectively.

.

^{.}. Ratio of the volumes of two hemispheres =

Hence, r

_{1 }: r

_{2 }= 2 : 3

# Question-10

**A 20 m deep well with diameter 14 m is dug up and the earth from digging is evenly spread to form a platform 22 mx14 m. Find the height of the platform.**

**Solution:**

Given, Depth of the well = 20 m

Diameter of the well = 14 m

hence radius of the well = 7 m

Earth taken out from the well = Ï€r

^{2}h

= x (7)

^{2 }x 20

= 3080 cm

^{3}

Volume of the platform = Volume of the earth

Area of the base x height = 3080

22 x 14 x h = 3080

Hence h=10 m.

# Question-11

**A rectangular reservoir is 120 m long and 75 m wide. Find the speed of water flowing into it through a square pipe of 20 cm side such that the water rises by 1.2 m in 9 hours.**

**Solution:**

Volume of the water flown in 9 hours = (120 x 75 x 1.2)m

^{3 }

Volume of water flown in one hour = m

^{3}

Area of the mouth of the square pipe = = 0.04m

^{2}

Speed of water per hour= = 30,000 m

.

^{.}. Hence speed of water flowing in the reservoir = 30 km/hour

# Question-12

**Find the volume of a cube whose total surface area is given:**

(i) 486 cm(i) 486 cm

^{2}(ii) 54 cm^{2}(iii) 150 cm^{2}**Solution:**

Let the edge of the cube= a cm

Then total surface area=6a

^{2}cm

^{2}

and volume of the cube=a

^{3}cu. cm.

(i) 6a

^{2 }= 486 â‡’ a=9 cm

Hence volume=(9)

^{3}=729 cm

^{3}

(ii) 6a

^{2}=54 â‡’ a = 3 cm

Hence volume=(3)

^{3}=27 cu. cm.

(iii) 6a

^{2}=150 â‡’ a=5 cm

Hence volume=(5)

^{3}=125 cu. cm.

# Question-13

**A closed cuboid water tank is made of steel sheet that is 2.5 cm thick. The outer dimensions of the tank are 2.15 m Ã— 1.5 m Ã— 1.1 m. Find the inteernal dimensions and total surface area of the tank.**

**Solution:**

External dimensions of the cube are

*l*= 215 cm,

*b*= 150 cm,

*h*= 110 cm

Since the sheet is 2.5 cm thick, the internal dimensions are

*l*= (215 - 5) cm = 210 cm, b =(150 - 5) cm = 145 cm,

*h*= (110 - 5) cm = 105 cm

âˆ´ Total surface area of the tank = 2(

*lb + bh + lh*)

= 2 Ã— (2.15 Ã— 1.5 + 1.5 Ã— 1.1 + 2.15 Ã— 1.1) m

^{3}

= 2 Ã— (3.225 + 1.65 + 2.365) m

^{3}

= 14.48 m

^{3}

# Question-14

**A rectangular piece of paper of width 20 cm and length 44 cm is rolled along its width to form a cylinder. What is the curved surface area of the cylinder so formed?****Solution:**

The length of the rectangle becomes the circumference of the base of the cylinder.

âˆ´ 2Ï€ r = 44, where r is the radius of the cylinder.

.

^{.}. r = = 7 cm

The width of the rectangle becomes the height of the cylinder,

i.e. h = 20 cm.

.

^{.}. Curved surface area of the cylinder = 2

**Ï€**rh

= 2 (22/7)Ã— 7 Ã— 20 cm

^{2}

= 880 cm

^{2}

# Question-15

**Find the curved surface area and total surface of a cone whose height is 28 cm and diameter is 42 cm.****Solution:**

Given, the diameter of the cone = 42 cm

âˆ´Radius = 42 cm Ã· 2 = 21 cm, h = 28 cm

slant height

*l*= âˆš(h

^{2}- r

^{2}) = âˆš(28

^{2}- 21

^{2}) = âˆš7Ã—49 = 7âˆš7cm

Curved surface of cone = Ï€ r

*l*

= Ã— 21 Ã— 7âˆš7 = 462âˆš7cm

^{2 }

âˆ´Total surface area of cone = Ï€ r(

*l*+ r)

= Ã— 21 Ã— (7âˆš7 + 21)

= 462(âˆš7 + 3)cm

^{2}

# Question-16

**A tent is cylindrical in shape with a conical top above it. The radius of the base of the tent is 7m. The height of the cylindrical part is 20 m and the height of the conical part is 4 m. Find the area of canvas cloth needed to construct the tent.****Solution:**

Curved surface area of cylindrical part = 2Ï€ rh

= 2 Ã—Ã— 7 Ã— 20

= 880 m

^{2 }

For finding the surface area of the curved part of the cone we have to find

*l*, the slant height.

^{ }

*l*

*= cm = = 8.06 cm*

.

^{.}. surface area of curved part of cone = Ï€ r

*l*

= Ã— 7 Ã— 8.06 m

^{2}

= 177.32 m

^{2 }

âˆ´ Total area of canvas needed = (880 + 177.32) m

^{2}

= 1057.32m

^{2}

# Question-17

**A reservoir in the form of a rectangular parallelopiped is 20 m long. If 18 kl of water is removed from the reservoir, the water level goes down by 15 cm. Find the width of the reservoir.****Solution:**

Let the width of the reservoir =

*x*m

Volume of water flowing from the reservoir = l Ã— b Ã— h

= 20 Ã—

*x*Ã— 0.15

= 3

*x*m

^{3}

Volume of the water removed = 18 m

^{3 }(1m

^{3 }= 1000 litre)

Hence 3

*x*= 18 â‡’

*x*= 6m

âˆ´ Width of the reservoir = 6m

# Question-18

**If***l, b, h*, S and V are length, breadth, height, surface area and volume respectively of a cuboid, prove that**Solution:**

Surface area of the cuboid = S = 2(

*lb + bh + hl*)

Volume of cuboid =

*l x b x h*

Now in RHS=

= LHS

# Question-19

**The length of a cinema hall is 20 m and breadth is 16 m. The sum of the areas of floor and flat roof is equal to the area of four walls. Find the height and volume of the hall.****Solution:**

Let the height of the cinema hall =

*h*m

length = 20 m and width = 16 m

Area of the floor = Area of the roof=

*l*x

*b*

= 20 x 16= 320 m

^{2}

Total area of the floor and the roof of the hall = 2 x 320 = 640 m

^{2}

Area of the four walls = 2(

*l*+

*b*) x

*h*= 2(20 + 16) x

*h*

= 72

*h*m

^{2}

Area of the four walls = Total area of the floor and the roof.

Ãž 72

*h*= 640

Ãž

*h*= 8.9 m.

âˆ´volume =

*l*x

*b*x

*h*= 20 x 16 x 8.9 = 2848 m

^{3}

# Question-20

**A cone is made from a metal sheet in the form of a sector of a circle. If the radius of the circle is 14 cm and the sector angle is 60Â°, find the radius of the base, slant height and the height of the cone formed.**

**Solution:**

Slant height of cone formed = radius of sector(R) = 14 cm

Now, circumference of base of cone = length of arc

âˆ´ 2 Ã— Ã— r =

âˆ´ r = cm = 2.33 cm (approx.)

Height of cone(h) = =cm

^{2}

= cm

^{2 }âˆ´ h = 13.8 cm

^{2 }(approx.)

# Question-21

**Find the surface area of a sphere whose radius is 2.1 cm.****Solution:**

Surface area = 4Ï€ r

^{2}

= 4 Ã—Ã— 2.1 Ã— 2.1 cm

^{2}

= 55.44 cm

^{2}

# Question-22

**How many metalboxes of size 8 cm Ã— 5 cm Ã— 3 cm can be packed in a cardboard box of size 40cm Ã— 40cm Ã— 21cm.****Solution:**

Volume of cardboard box = 40 cm Ã— 40 cm Ã— 21 cm

= 33600 cm

^{3}

Volume of 1 metalbox = 8 cm Ã— 5 cm Ã— 3 cm

= 120 cm

^{3 }

âˆ´ Number of metalboxes that can fit in the cardboard box = 33600 Ã· 120

= 280

# Question-23

**A closed cuboid water tank is made of steel sheet that is 2.5 cm thick. The outer dimensions of the tank are****2.15 m Ã— 1.5 m Ã— 1.1 m. Find the capacity of the tank, and the volume of steel used to construct it. Also find the total surface area of the ta****nk.****Solution:**

External dimensions of the cube are

*l*= 215 cm,

*b*= 150 cm,

*h*= 110 cm

Since the sheet is 2.5 cm thick, the internal dimensions are

*l*= (215 - 5) cm = 210 cm,

*b*=(150 - 5) cm = 145 cm,

h = (110 - 5) cm = 105 cm

âˆ´ External volume = (215 Ã— 150 Ã— 110) cm

^{3}

= 3547500 cm

^{3}

Internal volume = (210 Ã— 145 Ã— 105) cm

^{3}

= 3197250 cm

^{3}

Capacity of tank = internal volume

= 3197250 cm

^{3}

= 3.19725 m

^{3}

Volume of steel used = External volume - Internal volume

= (3547500 - 3197250) cm

^{3}

= 350350 cm

^{3}

= 0.35025 m

^{3}

Total surface area = 2(lb + bh + lh)

= 2 Ã— (2.15 Ã— 1.5 + 1.5 Ã— 1.1 + 2.15 Ã— 1.1) m

^{3}

= 2 Ã— (3.225 + 1.65 + 2.365) m

^{3}

= 14.48 m

^{3}

# Question-24

**A rectangular piece of paper of width 20 cm and length 44 cm is rolled along its width to form a cylinder. What is the volume of the cylinder so formed?**

**Solution:**

The length of the rectangle becomes the circumference of the base of the cylinder.

âˆ´ 2Ï€ r = 44, where r is the radius of the cylinder.

.

^{.}. r = = 7 cm

The width of the rectangle becomes the height of the cylinder,

i.e. h = 20 cm.

.

^{.}. Volume = Ï€ r

^{2}h =Ã— 7 Ã— 7 Ã— 20 cm

^{3}= 3080 cm

^{3 }

# Question-25

**Find the Total surface area and volume of a cone whose height is 28 cm and diameter is 42 cm.****Solution:**

Radius = 42 cm Ã· 2 = 21 cm, h = 28 cm

Total surface area = Ï€ r (

*l*+ r) = Ã— 21 (35 + 21) cm

^{2}= 3696 cm

^{2 }

Volume of cone = Ï€ r

^{2}h = Ã— Ã— 21 Ã— 21 Ã— 28 = 12,936 cm

^{2}

# Question-26

**A tent is cylindrical in shape with a conical top above it. The radius of the base of the tent is 7m. The height of the cylindrical part is 20 m and the height of the conical part is 4 m. Find the volume of air in the tent.**

**Solution:**

Volume of cylindrical part = Ï€ r

^{2}h

= Ã— 7 Ã— 7 Ã— 20 m

^{3}

= 3080 m

^{3 }

Volume of conical part =Ï€ r

^{2}h

= x x 7 x 7 x 4m

^{3}

^{ }= 205.3 m

^{3 }

.

^{.}.Volume of air in tent = volume of cylindrical part + volume of conical part

= (3080 + 205.3) m

^{3}

= 3285.3 m

^{3}

# Question-27

**Find the volume of a sphere whose radius is 2.1 cm.****Solution:**

Given, r = 2.1 cm

Volume of a sphere = Ï€ r

^{3 }

**=**Ã—Ã— 2.1 Ã— 2.1 Ã— 2.1 cm

^{3 }

= 38.81cm

^{3}(approx.)

# Question-28

**The surface area of a sphere and the curved surface area of a cylinder are in the ratio 2 : 1. Find the ratio of their volumes, if their radii are equal.**

**Solution:**

Given the ratio of surface area of sphere

and curved surface area of cylinder = 4r

^{2 }: 2rh = 2 : 1

(r is the common radius, and h is the height of the cylinder)

.

^{.}. i.e. r = h

The ratio of the Volume of sphere and

Volume of cylinder = (since h = r)

=

.

^{.}. Ratio of their volumes = 4 : 3

# Question-29

**Find the volume of the greatest sphere that can be cut of a cube of side 14 cm. Find also the volume of the remaining piece.**

**Solution:**

The greatest sphere will be such that

diameter of sphere = side of cube

âˆ´ radius of sphere = 14cm Ã· 2 = 7 cm

Volume of sphere = Ï€ r

^{3 }=Ã—

**Ã— 7 Ã— 7 Ã— 7 cm**

^{3 }= 1437.3 cm

^{3}

Volume of cube = (14 Ã— 14 Ã— 14) cm

^{3 }= 2744 cm

^{3 }

.

^{.}. Volume of remaining piece = (2744 - 1437.3) cm

^{3}= 1306.7 cm

^{3}

# Question-30

**The volume of two hemispheres are in the ratio 8 : 27. Find the ratio of their radii.****Solution:**

Let the volumes of two hemispheres be V

_{1}and V

_{2 }and the radii be r

_{1}and r

_{2}respectively.

Therefore the required ratio of their radii=

Hence,the ratio of the radii, r

_{1 }: r

_{2 }= 2 : 3

# Question-31

**A small indoor greenhouse (herbarium) is made entirely of glass panes (Including base) held together with tape. It is 30cm long, 25cm wide and 25 cm high.**

(i) What is the area of the glass?

(ii) How much of tape is needed for all the 12 edges?

(i) What is the area of the glass?

(ii) How much of tape is needed for all the 12 edges?

**Solution:**

**(i) For herbarium**

*l*= 30 cm,

*b*= 25 cm,

*h*= 25 cm

âˆ´ Area of the glass = 2(

*lb*+

*bh*+

*hl*)

= 2[(30)(25) + (25)(25) + (25)(30)]

= 2[750 + 625 + 750] = 4250 cm

^{2}.

(ii) The tape needed for all the 12 edges

= 4(

*l*+

*b*+

*h*)

= 4(30 + 25 + 25) = 320 cm.

# Question-32

**A Jokerâ€™s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.**

**Solution:**

Base radius (r) = 7 cm

Height (h) = 24 cm

âˆ´ Slant height (

*l*) = = = = = 25 cm

âˆ´ Curved surface area of a cap = Ï€

*rl*= = 550 cm

^{2 }

âˆ´ Curved surface area of 10 caps = 550 Ã— 10 = 5500 cm

^{2 }

Hence the area of the sheet required to make 10 such caps is 5500 cm

^{2}.