## RD Sharma Class 8 Solutions Chapter 21 Mensuration II (Volumes and Surface Areas of a Cubiod and a Cube) Ex 21.4

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4

**Other Exercises**

- RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.1
- RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.2
- RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.3
- RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4

**Question 1.**

**Find the length of the longest rod that can be placed in a room 12 m long, 9 m broad and 8 m high.**

**Solution:**

Length of room (l) = 12m

Breadth (b) = 9 m

Height (h) = 8 m

Longest rod to be kept in the room

**Question 2.**

**If V is the volume of the cuboid of dimensions a, b, c and S its the surface area then prove that**

**Solution:**

∵ a, b, c are the dimensions of a cuboid

∴ Volume (V) = abc

Surface area (S) = 2(ab + bc + ca)

Now

**Question 3.**

**The areas of three adjacent faces of a cuboid are .v, y and z. If the volume is V ^{1} prove that V^{2} = xyz.**

**Solution:**

Let length of cuboid = l

Breadth = b

and height = h

Volume = Ibh

∴ x = lb,y = bh and z = hl

Now x.y.z = lb.bh.hl

= l

^{2}b

^{2}h

^{2}= (Ibh)

^{2}= V

^{2}

∴ V

^{2}= xyz Hence proved

**Question 4.**

**A rectangular water reservoir contains 105 m3 of water. Find the depth of the water in the reservoir if its base measures 12 m by 3.5 m.**

**Solution:**

Volume of the water in reservoir = 105 m^{2}

Length (l)= 12 m

and breadth (b) = 3.5 m

**Question 5.**

**Cubes A, B, C having edges 18 cm, 24 cm and 30 cm respectively are melted and moulded into a new cube D. Find the edge of the bigger cube D.**

**Solution:**

Edge of cube A = 18 cm

∴ Volume = a^{2} = (18)3 cm^{3} = 5832 cm^{3}

Edge of cube B = 24 cm

∴ Volume = (24)^{3} = 13824 cm^{3}

Edge of cube C = 30 cm

∴Volume = (30)^{3} = 27000 cm^{3}

Volume of A, B, C cubes

= 5832+ 138-24+ 27000 = 46656 cm^{3}

Volume of cube D = 46656 cm^{3}

**Question 6.**

**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:**

Volume of room = 512 cu.dm

Let height of the room (h) = x

Then breadth (b) = 2x

and length (l) = 2x x 2 = 4x.

∴ Volume = l x b x h = 4x x 2x x x = 8×3

**Question 7.**

**A closed iron tank 12 m long, 9 m wide and 4 m deep is to be made. Determine the cost of iron sheet used at the rate of Rs 5 per metre sheet, sheet being 2 m wide.**

**Solution:**

Length of iron tank (l) = 12 m

Breadth (b) = 9 m

Depth (h) = 4 m

∴ Surface area of the tank = 2(l x b + b x h + h x l)

= 2(12 x 9 + 9 x 4 + 4 x 12) m^{2}

= 2(108 + 36 + 48) = 2 x 192 m^{2}

= 384 m^{2}

Width of sheet used = 2 m

**Question 8.**

**A tank open at the top is made of iron sheet 4 m wide. If the dimensions of the tank are 12 m x 8 m x 6 m, find the cost of iron sheet at Rs 17.50 per metre.**

**Solution:**

Dimensions of the open iron tank = 12mx 8m.x 6m

∴ Surface area (without top)

= 2(1 x b) x h + lb

= 2(12 + 8) x 6+12 x 8m^{2}

= 2 x 20 x 6 + 96 = 240 + 96 m^{2} = 336 m^{2}

Width of sheet used = 4 m

∴ Length of sheet = \(\frac { Area }{ b }\) = \(\frac { 336 }{ 4 }\) m = 84 m b 4

Rate of sheet = Rs 17.50 per m.

∴ Total cost = Rs 17.50 x 84 = Rs 1470

**Question 9.**

**Three equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.**

**Solution:**

Let edge of each equal cubes = x

Then, surface area of one cube = 6x^{2}

and surface area of three cubes = 3 x 6x^{2} = 18x^{2}

By placing the cubes in a row,

The length of newly formed cuboid (l) = 3x

Breadth (b) = x

and height (h) = x

∴ Surface area of the cuboid so formed

**Question 10.**

**The dimensions of a room are 12.5 m by 9 m by 7 m. There are 2 doors and 4 windows in the room; each door measures 2.5 m by 1.2 m and each window 1.5 m by 1 m. Find the cost of painting the walls at Rs 3.50 per square metre.**

**Solution:**

Dimensions of a room = 12.5 m x 9 m x 7 m

∴ Total surface area of the walls = 2(1 + b) x h = 2(12.5 + 9) x 7 m^{2}

= 2 x 21.5 x 7 = 301 0 m^{2}

Area of 2 doors = 2 x (2.5 x 1.2) m^{2} = 2 x 3.00 = 6 m^{2}

Area of 4 windows = 4 x (1.5 x 1) m^{2}

4 x 1.5 = 6 m^{2}

∴ Remaining area of the walls = 301 -(6 + 6) m^{2}

= 301 – 12 = 289 m^{2}

∴ Rate of painting the walls = Rs 3.50 per m^{2}

∴ Total cost = Rs 3.50 x 289 = Rs 1011.50

**Question 11.**

**A field is 150 m long and 100 m wide. A plot (outside the field) 50 m long and 30 m wide is dug to a depth of 8 m and the earth taken out from the plot is spread evenly in the field. By how much the level of field is raised ?**

**Solution:**

Length of the plot (l) = 50 m

Width (b) = 30 m

and depth (h) = 8 m

∴ Volume of the earth dug out = l x b x h = 50 x 30 x 8 = 12000 m^{3}

Length of the field = 150 m

and breadth = 100 m

∴ Height of the earth spread out on the field

**Question 12.**

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

**Solution:**

Volume of each cube = 512 cm

^{3}

Now by joining the two equal cubes of side 8 cm, the length of so formed cuboid (l)

= 2 x 8 = 16 cm

Breadth (b) = 8 cm

and height (h) = 8 cm

∴ Surface area = 2( l x b + b x h + h x l)

= 2(16 X 8 + 8 X 8 + 8X16) cm

^{2}

= 2(128 + 64 + 128) cm

^{2}

= 2 x 320 = 640 cm

^{2}

**Question 13.**

**Three cubes whose edges measure 3 cm, 4 cm and 5 cm respectively are melted to form a new cube. Find the surface area of the new cube formed.**

**Solution:**

Edge of first cube = 3 cm

∴ Volume = a^{3} = (3)^{3} 27 cm^{3}

Edge of second cube = 4 cm

∴Volume = a^{3} = (4)^{3} = 64 cm^{3}

Edge of third cube = 5 cm

∴ Volume = a^{3} = (5)^{3} = 125 cm^{3}

Volume of three cubes together = 27 + 64+ 125 = 216 cm^{3}

∴ Volume of the new cube = 216 cm^{3}

**Question 14.**

**The cost of preparing the’walls of a room 12 m long at the rate of Rs 1.35 per square metre is Rs 340.20 and the cost of matting the floor at 85 paise per square metre is Rs 91.80. Find the height of the room.**

**Solution:**

Length of the room (l) = 12 m

Rate of matting the floor = 85 paise per m^{2}

Total cost of matting = Rs 91.80

**Question 15.**

**The length of a hall is 18 m and width 12 m. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the wall.**

**Solution:**

Length of hall (l) = 18 m

and breadth (b) = 12 m

∴ Area of floor = l x b = 18 x 12 = 216 m^{2}

and area of roof = 216 m^{2}

Total area of floor and roof

= (216 + 216) m^{2} = 432 m^{2}

∴ Area of four walls = 432 m^{2}

But area of 4 walls = 2(l + b) x h

∴ 2h (l + b) = 432

⇒ 2h (18 + 12) = 432

⇒ 2h x 30 = 432 432

⇒ h = \(\frac { 432 }{ 60 }\) = 7.2m

∴ Height of the wall = 7.2 m

**Question 16.**

**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:**

Edge of metal bigger cube = 12 cm

∴ Volume = (12)^{3} = 1728 cm^{3}

∴ Sum of volumes of 3 smaller cubes = 1728 cm^{3}

Edge of first smaller cube = 6 cm

∴ Volume = (6)^{3} = 216 cm^{3}

Edge of second smaller cube = 8 cm

∴ Volume = (8)^{3} = 512 cm^{3}

Sum of volumes of two smaller cubes = 216+ 512 = 728 cm^{3}

∴ Volume of third smaller cube = 1728-728 cm^{3} = 1000 cm^{3}

**Question 17.**

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

**Solution:**

Length of cinema hall (l) = 100 m

Breadth (b) = 50 m

and height (h) = 18 m

∴ Volume of air of the hall = l x b x h

= 100 x 50 x 18 m^{3}

= 90000 m^{3}

Each person requires air = 150 m^{3}

∴ Number of persons = \(\frac { 90000 }{ 150 }\)= 600

**Question 18.**

**The external dimensions of a closed wooden box are 48 cm, 36 cm, 30 cm. The box is made of 1.5 cm thick wood. How many bricks of size 6 cm x 3 cm x 0.75 cm can be put in this box ?**

**Solution:**

Outer dimensions of a closed wooden box = 48 cm x 36 cm x 30 cm

Thickness of wood = 1.5 cm.

∴ Inner length (l) = 48 – 2 x 1.5 cm = 48 – 3 = 45 cm

Breadth(b) = 36-2 x 1.5 = 36-3 = 33 cm

Height (h) = 30 – 2 x 1.5 = 30 – 3 = 27 cm

∴ Volume of inner box = l x b x h = 45 x 33 x 27 cm^{3} = 40095 cm^{3}

Volume of one brick of size 6 cm x 3 cm x 0.75 cm

= 6 x 3 x 0.75 = 6 x 3 x \(\frac { 3 }{ 4 }\) cm^{3} = \(\frac { 27 }{ 2 }\) cm^{3}

∴ Number of bricks = \(\frac { 40095 x 2 }{ 27 }\)

= 1485 x 2 = 2970 bricks

**Question 19.**

**The dimensions of a rectangular box are in the ratio of 2 : 3 : 4 and the difference between the cost of covering it with sheet of paper at the rates of Rs 8 and Rs 9.50 per m2 is Rs 1,248. Find the dimensions of the box.**

**Solution:**

Ratio in the dimensions of a box =2:3:4

Difference in total cost = Rs 1,248

Difference in rates = Rs 9.50 – Rs 8 = Rs 1.50

Let length (l) = 2x

Then breadth (b) = 3x

and height (h) = 4x

∴ Surface area = 2 (l x b + b x h + h x l)

= 2(2x 3x + 3x x 4x + 4x x 2x)

= 2(6x^{2} + 12x^{2} + 8 x^{2}) = 2 x 26x^{2} = 52x^{2}

First rate of paper = Rs 9.50 per m^{2}

and second rate = 8.00 per m^{2}

∴ First cost = Rs 52x^{2} x 9.50

and second cost = Rs 52x^{2} x 8

∴ 52x^{2} x 9.50 – 52x^{2} x 8= 1248

⇒ 52x^{2} (9.50 – 8) = 1248

⇒ 52x^{2}(1.50) = 1248

Hope given RD Sharma Class 8 Solutions Chapter 21 Mensuration II Ex 21.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.