Ex-18.2 (Part - 1), Surface Area And Volume Of A Cuboid And Cube, Class 9, Math RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

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 =

⇒ 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 

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, 

Therefore,

Therefore,

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 =

⇒  = 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 =  m/min

=  m/min

Depth of the river (h) = 3m

Width of the river (b) = 40m

Volume of the water flowing in 1 min =  ∗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∗ = 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∗ = 50000∗3∗1.2

⇒ Area of field =

⇒ 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 =  

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

⇒ side³ = 512

⇒ side³ = 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 =

Therefore,Thickness of silve rsheet = cm

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 =

  = 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 =  = 3.2cm.