RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev

RD Sharma Solutions for Class 9 Mathematics

Created by: Abhishek Kapoor

Class 9 : RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev

The document RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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Q1) A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many liters of water can it hold?

Solution:

Given data:

Length (l) = 6m

Breadth (b) = 5m

Height (h) = 4.5m

Volume of the tank = l*b*h

= 6*5*4.5

= 135m3

It is given that,

1m3  = 1000 liters

Therefore,135m3 = (135∗1000)liters

= 135000 liters

The tank can hold 1,35,000 liters of water.

 

Q2) A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic meters of a liquid?

Solution:

Given that

Length of the vessel (l) = 10 m

Width of the Cuboidal vessel = 8 m

Let ‘h’ be the height of the cuboidal vessel.

Volume of the vessel = 380m3

Therefore,l∗b∗h = 380m3

⇒ 10∗8∗h = 380

⇒ h = RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev

⇒ h = 4.75m

Therefore, height of the vessel should be 4.75 m.

 

Q3) Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs 30 per m3

Solution:

Given,

Length of the cuboidal pit (l) = 8 m

Breadth of the cuboidal pit (b) = 6 m

Depth of the cuboidal pit (h) = 3 m

Volume of the Cuboidal pit = l*b*h

= 8*6*3

= 144 m3

Cost of digging 1m3  = Rs. 30

Cost of digging 144m3  = 144*30 = Rs.4320

 

Q4) If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then prove that RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev

Solution:

Given Data:

Length of the cube (l) = a

Breadth of the cube (b) = b

Height of the cube (h) = c

Volume of the cube (V) = l*b*h

= a*b*c

= abc

Surface area of the cube (S) = 2 (lb+bh+hl)

= 2 (ab+bc+ca)

Now, RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev

Therefore,RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev

Therefore,RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev

Hence Proved.

 

Q5) The areas of three adjacent faces of a cuboid are x, y and z.. If the volume is V, Prove that V2 = xyz.

Solution:

Let a, b and d be the length, breadth, and height of the cuboid.

Then, x = ab

y = bc

z = ca

and V = abc [V = l*b*h]

= xyz = ab*bc*ca =  (abc)2

and V = abc

V2 = (abc)2

Therefore,V2 = (xyz)

 

Q6) If the areas of three adjacent face of a cuboid are 8cm3, 18cm3 and 25cm3. Find the volume of the cuboid.

Solution:

WKT, if x, y, z denote the areas of three adjacent faces of a cuboid.

= x = l*b, y = b*h, z = l*h

Volume (V) is given by

V = l*b*h

Now, xyz = lb*bh*hl =  V2

Here x = 8

y = 18

z = 25

Therefore,V2 = 8∗18∗25 = 3600

⇒ V = 60cm3

 

Q7) The breadth of a room is twice its height, one half of its length and the volume of the room is 512 cu. dm. Find its dimensions.

Solution:

Consider l, b and h are the length, breadth and height of the room.

So, b = 2h and b = RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev

RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev = 2h

⇒ l = 4h

⇒ l = 4h,b = 2h

Now, Volume = 512dm3

⇒ 4h∗2h∗h = 512

⇒ h3 = 64

⇒ h = 4

So, Length of the room (l) = 4h = 4*4 = 16 dm

Breadth of the room (b) = 2h = 2*4 = 8 dm

And Height of the room (h) = 4 dm.

 

Q8) A river 3m deep and 40m wide is flowing at the rate of 2km per hour. How much water will fall into the sea in a minute?

Solution

Radius of the water flow = 2km per hour = RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev m/min

= RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev m/min

Depth of the river (h) = 3m

Width of the river (b) = 40m

Volume of the water flowing in 1 min =  RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev∗40∗3 = 40003

Thus, 1 minute 4000m3  = 4000000 litres of water will fall in the sea.

 

Q9) Water in a canal 30dm wide and 12dm deep, is flowing with a velocity of 100km every hour. What much area will it irrigate in 30 minutes if 8cm of standing water is desired?

Solution:

Given that,

Water in the canal forms a cuboid of Width (b) = 30dm = 3m

Height (h) = 12dm = 1.2m

Cuboid length is equal to the distance traveled in 30 min with the speed of 100 km per hour.

Therefore, Length of the cuboid =  100∗RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev = 60km  = 50000metres

So, volume of water to be used for irrigation = 5000*3*1.2 m3

Water accumulated in the field forms a cuboid of base area equal to the area of the field and height equal to 8100metres

Therefore,Area of field∗RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev = 50000∗3∗1.2

⇒ Area of field = RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev

⇒ 22,50,000metres.

 

Q10) Three metal cubes with edges 6cm, 8cm, 10cm respectively are melted together and formed into a single cube. Find the volume, surface area and diagonal of the new cube.

Solution:

Let ‘a’ be the length of each edge of the new cube.

Then a3 = (63+83+103)cm3

⇒ a3 = 1728

⇒ a = 12

Therefore,Volume of thenewcube = a3 = 1728cm3

Surface area of the new cube =  6a2 = 6∗(12)2 = 864cm2

Diagonal of the newly formed cube =  RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev

 

Q11) Two cubes, each of volume 512cm3 are joined end to end. Find the surface area of the resulting cuboid.

Solution:

Given that,

Volume of the cube = 512cm3

⇒ side= 512

⇒ side= 83

⇒ side = 8cm

Dimensions of the new cuboid formed

Length (l) = 8+8 = 16cm, Breadth (b) = 8cm, Height (h) = 8cm

Surface area = 2(lb+bh+hl)

= 2 (16*8+8*8+16*8)

= 640 cm2

Therefore, Surface area is 640 cm2.

 

Q12) Half cubic meter of gold-sheet is extended by hammering so as to cover an area of 1 hectare. Find the thickness of the gold-sheet.

Solution:

Given that, Volume of gold-sheet = 0.5m3

Area of the gold-sheet = 1 hectare = 1*10000 = 10000 m2

Therefore,Thickness of gold sheet = RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev

RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev

RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev

RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev

Therefore,Thickness of silve rsheet = RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRevcm

 

Q13) A metal cube of edge 12cm is melted and formed into three smaller cubes. If the edges of the two smaller cubes are 6cm and 8cm, find the edge of the third smaller cube.

Solution:

Volume of the large cube =  v1+v2+v3

Let the edge of the third cube be ‘x’ cm

123 = 63+83+a3 [Volume of cube = side3]

1728 = 216+512+x3

⇒ x3 = 1728−728 = 1000

⇒ x = 10cm

Therefore,Side of thirdside = 10cm

 

Q14) The dimensions of a cinema hall are 100m, 50m, 18m. How many persons can sit in the hall, if each person requires 150m3  of air?

Solution:

Given that

Volume of cinema hall = 100*50*18 m3

Volume of air required by each person = 150 m3

Number of persons who sit in the hall = RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev

 RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev = 600 [Since,V = l×b×h]

Therefore, number of persons who can sit in the hall = 600 members.

 

Q15) Given that 1 cubic cm of marble weighs 0.25kg, the weight of marble block 28cm in width and 5cm thick is 112kg. Find the length of the block.

Solution:

Let the length of the marble block be ‘l’cm

Volume of the marble block = l*b*h cm3

= l*28*5 cm3

Therefore, weight of the marble square = 140l*0.25 kg

As mentioned in the question, weight of the marble = 112 kgs

Therefore,

= 112 = 140l*0.25

= ⇒ l = RD Sharma Solutions -Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math Class 9 Notes | EduRev = 3.2cm.

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