Class 9 RD Sharma Solutions- Chapter 18 Surface Area and Volume of a Cuboid and Cube – Exercise 18.2 | Set 1

Question 1. 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, the length of water tank = 6 m

The breadth of water tank = 5 m

The height of water tank = 4.5 m

The quantity of water that tank can hold = Volume of Cuboid

= length × breadth × height

= 6 × 5 × 4.5 = 135 m³

As we know 1 m³ = 1000 liters

So, 135 m³ = 135 × 1000 = 135000 liters

Hence. the water tank can hold 1,35,000 liters of water.

Question 2. 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, the length of the cuboidal vessel = 10 m

The breadth of cuboidal vessel = 8 m

The volume of cuboidal vessel = 380 cubic meters

Volume of Cuboid = length × breadth × height

380 = 10 × 8 × height

So, height is 4.75 m

Hence, the vessel should be 4.75 m high in order to hold 380 cubic meters of a liquid

Question 3. 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, the length of the cuboidal pit = 8 m

The breadth of cuboidal pit = 6 m

The height of cuboidal pit = 3 m

We know Volume of Cuboid = length × breadth × height

= 8 × 6 × 3 = 144 m³

Also, the cost of digging 1 m³ = Rs 30

So, cost of digging 144 m³ = Rs 30 × 144

= Rs 4320

Hence, the cost of digging the cuboidal pit is Rs 4320

Question 4. If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then prove that 1/V = 2/S(1/a + 1/b + 1/c)

Solution:

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

Surface area of cuboid = 2(a × b + b × c + c × a)

S = 2(a × b + b × c + c × a)

Volume of cuboid = a × b × c

V = a × b × c

We need to prove 1/V = 2/S(1/a + 1/b + 1/c)

Taking L.H.S, 2/S(1/a + 1/b + 1/c)

= 2/aS + 2/bS + 2/aS = 2(bc+ca+ab)/S × abc

= 2(bc+ca+ab)/(ab + bc + ac) × abc

= 2/abc = 2/V

So, L.H.S = R.H.S

Hence proved

Question 5. 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 be the length of cuboid, b be the breadth of cuboid and c be the height

Given, the areas of three adjacent faces of a cuboid are x, y and z

So, x = ab

y = bc

z = ac

We know, Volume of cuboid = a × b × c

Multiplying x, y and z we get,

xyz = ab X bc X ac

xyz = (abc)²

So, xyz = (V)²

Hence, proved

Question 6. If the areas of three adjacent faces of a cuboid are 8 cm², 18 cm² and 25 cm². Find the volume of the cuboid.

Solution:

Let x, y, and z be the area of adjacent faces of cuboid

So, x = 8 cm²

y = 18 cm²

z = 25 cm²

We know that, product of area of adjacent faces of cuboid will be equal to square of volume of cuboid

It means, xyz = V²

So, V² = 8 × 18 × 25 = 3600

V = 60 cm³

Hence, the volume of cuboid is 60 cm³

Question 7. 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:

Given, volume of room is 512 cu.dm

Let l be the length of room, and h be the height of room

Also, b = 2 × h and b = l/2

So, we can say h = b/2 and l = 2 × b

Volume of cuboid = l × b × h

So, 512 = 2b × b × b/2

So, b = 8 cm

l = 2b = 16 cm

h = b/2 = 4 cm

Hence, the length, breadth and height of room are 16cm, 8 cm and 4 cm respectively

Question 8. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?

Solution:

Given, the depth of the river is 3 m

The width of the river is 40 m

The water flowing in the river at the rate of 2 km/hour = 100/3 m/minute

So, we can derive the volume of water flowing in 1 min = 100/3 × 40 × 3

= 4000 m³

Or 4000000 liters

Hence, 4000000 liters of water will fall into the sea in 1 minute

Question 9. Water in a canal 30 dm wide and 12 dm deep, is flowing with a velocity of 100 km every hour. What much area will it irrigate in 30 minutes if 8 cm of standing water is desired?

Solution:

Given, the width of canal = 30 dm = 3 m

The depth of canal = 12 dm = 1.2 m

And, the length of the canal will be given as the distance travelled with the speed of 100 km per hour in 30 minutes

So, length = 100 × 30/60 = 50000 m

We can say, the volume of water for irrigation = l × b × h

= 50000 × 3 × 1.2 m³

As we know the water in the field will form a cuboid, whose area will be equal to the area of field and height will be 8/100 m

So, we can conclude Area of field × 8/100 = 50000 × 3 × 1.2

Area of field = 2250000 m³

Hence, the desired area is 2250000 m³

Question 10. 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 edge of the new cube

So, the volume of cube = a³

Also, we know that sum of volumes of given three cubes will be equal to the volume of the new cube 

So, 6³ + 8³ + 10³ = a³ 

a = 12

So, the volume of the new cube = a³ = 12³

= 1728 cm³

Also, the surface area of the new cube = 6 × (a)²

= 6 × 12² = 864 cm²

And, the diagonal of the new cube = √3a = √3 × 12 

= 12√3 cm

Hence, the volume, surface area and diagonals of the new cube is 1728 cm³, 864 cm² and 12√3 cm

Question 11. Two cubes, each of volume 512 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Solution:

Let a be the edge of the cube

Given the volume of the cube is 512 cm³

So, a³ = 512 cm³

a = 8 cm

So we can say the edge of each cube is 8 cm

Now the dimensions of the resulting cuboid will be,

Length is 8 + 8 = 16 cm

Breadth is 8 cm

Height is 8 cm

So, the surface area of the cuboid = 2 (l × b + b × h + h × l)

= 2 (16 × 8 + 8 × 8 + 8 × 12)

= 640 cm²

Hence, the surface area of the resulting cube is 640 cm²

Question 12. 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, the volume of the gold-sheet is 1/2 m³

Also, the area of the gold-sheet is 1 hectare = 10000 m²

So the thickness of the gold-sheet is given by the Volume of the gold-sheet/ Area of the gold-sheet

= 0.5/10000 = 1/20000 m = 1/200 cm

Hence, the thickness of the gold-sheet is 1/200 cm

Question 13. A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of the two smaller cubes are 6 cm and 8 cm, find the edge of the third smaller cube.

Solution:

Let a be the edge of the third cube

Also, we know that sum of volumes of given three cubes will be equal to the volume of the new cube 

So, 6³ + 8³ + a³ = 12³ 

216 + 512 + a³ = 1728

a = 10 cm

Hence, the edge of the third cube is 10 cm.