**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 Ã— breadth Ã— height

= 6 Ã— 5 Ã— 4.5 = 135 m

^{3}As we know 1 m

^{3}= 1000 litersSo, 135 m

^{3}= 135 Ã— 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 Ã— breadth Ã— height

380 = 10 Ã— 8 Ã— height

So, height is 4.75 m

Hence, the vessel should be 4.75 m high in orderto 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 m**^{3}.

^{3}.

**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 Ã— breadth Ã— height

= 8 Ã— 6 Ã— 3 = 144 m

^{3}Also, the cost of digging 1 m

^{3}= Rs 30So, cost of digging 144 m

^{3}= Rs 30 Ã— 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

abe the length of cuboid,bbe the breadth of cuboid andcbe the heightSurface area of cuboid = 2(a Ã— b + b Ã— c + c Ã— a)

S = 2(a Ã— b + b Ã— c + c Ã— a)

Volume of cuboid = a Ã— b Ã— c

V = a Ã— b Ã— 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 Ã— abc

= 2(bc+ca+ab)/(ab + bc + ac) Ã— 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 V**^{2} = xyz.

^{2}= xyz.

**Solution:**

Let

abe the length of cuboid,bbe the breadth of cuboid andcbe the heightGiven, 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 Ã— b Ã— 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 cm**^{2}, 18 cm^{2} and 25 cm^{2}. Find the volume of the cuboid.

^{2}, 18 cm

^{2}and 25 cm

^{2}. Find the volume of the cuboid.

**Solution:**

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

So, x = 8 cm

^{2}y = 18 cm

^{2}z = 25 cm

^{2}We know that, product of area of adjacent faces of cuboid will be equal to square of volume of cuboid

It means, xyz = V

^{2}So, V

^{2}= 8 Ã— 18 Ã— 25 = 3600V = 60 cm

^{3}

Hence, the volume of cuboid is 60 cm^{3}

**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 Ã— h and b = l/2

So, we can say h = b/2 and l = 2 Ã— b

Volume of cuboid = l Ã— b Ã— h

So, 512 = 2b Ã— b Ã— 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 Ã— 40 Ã— 3

= 4000 m

^{3}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 Ã— 30/60 = 50000 m

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

= 50000 Ã— 3 Ã— 1.2 m

^{3}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 Ã— 8/100 = 50000 Ã— 3 Ã— 1.2

Area of field = 2250000 m

^{3}

Hence, the desired area is 2250000 m^{3}

**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

abe the edge of the new cubeSo, the volume of cube = a

^{3}Also, we know that sum of volumes of given three cubes will be equal to the volume of the new cube

So, 6

^{3}+ 8^{3}+ 10^{3}= a^{3}a = 12

So, the volume of the new cube = a

^{3}= 12^{3}= 1728 cm

^{3}Also, the surface area of the new cube = 6 Ã— (a)

^{2}= 6 Ã— 12

^{2}= 864 cm^{2}And, the diagonal of the new cube = âˆš3a = âˆš3 Ã— 12

= 12âˆš3 cm

Hence, the volume, surface area and diagonals of the new cube is 1728 cm^{3}, 864 cm^{2}and 12âˆš3 cm

**Question 11. Two cubes, each of volume 512 cm**^{3} are joined end to end. Find the surface area of the resulting cuboid.

^{3}are joined end to end. Find the surface area of the resulting cuboid.

**Solution:**

Let

abe the edge of the cubeGiven the volume of the cube is 512 cm

^{3}So, a

^{3}= 512 cm^{3}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 Ã— b + b Ã— h + h Ã— l)

= 2 (16 Ã— 8 + 8 Ã— 8 + 8 Ã— 12)

= 640 cm

^{2}

Hence, the surface area of the resulting cube is 640 cm^{2}

**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 m

^{3}Also, the area of the gold-sheet is 1 hectare = 10000 m

^{2}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

abe the edge of the third cubeAlso, we know that sum of volumes of given three cubes will be equal to the volume of the new cube

So, 6

^{3}+ 8^{3}+ a^{3}= 12^{3}216 + 512 + a

^{3}= 1728a = 10 cm

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