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Class 9 RD Sharma Solutions – Chapter 18 Surface Area and Volume of a Cuboid and Cube – Exercise 18.1
  • Last Updated : 01 Dec, 2020
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Question 1: Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm, and height 20 cm.

Solution:

Given, dimensions of cuboid are:

Length(l) = 80 cm 

Breadth(b) = 40 cm 

Height(h) = 20 cm



Formula for total surface area of cuboid:

TSA(Cuboid) = 2[lb + bh + hl]

Now, substituting the given values of in the formula,

= 2[(80)(40) + (40)(20) + (20)(80)]

= 2[3200 + 800 + 1600]

= 2[5600]

= 11200

Therefore, Total Surface Area = 11200 cm2

Now, Formula for Lateral Surface area of cuboid:

LSA(Cuboid) = 2[l + b] * h

= 2[80 + 40]20

= 2[120]20

= 40[120]

= 4800

Thus, Lateral Surface Area is 4800 cm2.

Question 2: Find the lateral surface area and total surface area of a cube of edge 10 cm.

Solution:

Given,

Side of the Cube = 10 cm



Formula for Lateral Surface Area of Cube :

LSA(cube) = 4 side2

= 4(10 × 10)

= 400 cm2

Formula for Total Surface Area of Cube :

TSA(cube) = 6 side2

= 6(10 × 10)

= 6(100)

= 600 cm2

Question 3: Find the ratio of the total surface area and lateral surface area of a cube.

Solution:

Formula for Lateral Surface Area of Cube :

LSA(cube) = 4 side2

Formula for Total Surface Area of Cube :

TSA(cube) = 6 side2

Therefore, 

Ratio of TSA and LSA = (6 side2)/(4 side2) = 6/4 = 3/2 or 3:2

Question 4: Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with colored paper with a picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth, and height as 80 cm, 40 cm, and 20 cm respectively. How many square sheets of paper of side 40 cm would she require?

Solution:

Given the dimensions of the wooden block are:

Length(l) = 80 cm

Breadth(b) = 40 cm 

Height(h) = 20 cm

Surface area of the wooden box = 2[lb + bh + hl]

= 2[80 × 40 + 40 × 20 + 20 × 80]

= 2[5600]

= 11200

Therefore, Surface Area of wooden box is 11200 cm2 .

Now, the area of each sheet of paper = 40 × 40 = 1600 cm2.

So, the total number of sheets required = (Surface area of the box)/(Area of one sheet of paper)

= 11200/1600

= 7

Therefore, Mary would require 7 sheets.

Question 5: The length, breadth, and height of a room are 5m, 4m and, 3m respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of Rs 7.50 m2.

Solution:

Total area to be whitewashed = lb + 2(l + b)h  —-(1)

Given,

Length(l) = 5m

Breadth(b) = 4m

Height(h) = 3m

Total area to be whitewashed = (5 × 4) + 2(5 + 4)3

= 20 + 54

= 74 

Total area to be whitewashed is 74 m2.

Now, cost of white washing1m2 is Rs. 7.50 (given)

Therefore, the cost of white washing 74 m2 = 74 × 7.50

= Rs. 555

Question 6: 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 three cubes.

Solution:

Let the breadth of the cuboid = a

Then, length of the new cuboid = 3a and

Height of the new cuboid = a

Now, 

Total surface Area of the new cuboid (TSA) = 2(lb + bh + hl)

= 2(3a × a + a × a + a × 3a)

= 14 a2

Again,

Total Surface area of three cubes = 3 × (6 side2)

= 3 × 6a2

= 18a2

Therefore, ratio of a total surface area of the new cuboid to that of the sum of the areas of three cubes = 14a2/18a2

= 7/9 or 7:9

Question 7: A 4cm cube is cut into 1 cm cubes. Calculate the total surface area of small cubes.

Solution: 

Given,

Edge of the cube = 4 cm

We know that,

Volume of the cube = Side3

= 43

= 4 × 4 × 4 = 64

Therefore, volume of the cube is 64cm3

Now, 

Edge of the cube = 1 cm3

So, Total number of small cubes = 64cm3/1cm3 = 64

And, the total surface area of all the cubes = 64 × 6 × 1 = 384 cm2.

Question 8: The length of a hall is 18m and the width 12m. 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 hall.

Solution:

Given, dimensions of the hall are:

Length(l) = 18 m

Breadth(b)= 12m

Let the height of the walls be h

From the given statement,

Area of floor and flat roof = sum of area of four walls           —-(1)

On applying respective formulas,

Area of floor and flat roof = 2lb = 2 × 18 × 12 = 432 sq/ft    —-(2)

Sum of area of four walls = (2 × 18h + 2 × 12h)                      —-(3)

On using equations (2) and (3) in (1), we get 

432 =2 × 18h + 2 × 12h

18h + 12h = 216

or h = 7.2

Therefore, height of the hall is 7.2 m.

Question 9: Hameed has built a cubical water tank with a lid for his house, with each other edge 1.5 m long. He gets the outer surface area of the tank excluding the base, covered with square tiles of a side 25 cm. Find how much he would spend for the tiles if the cost of tiles is Rs. 360 per dozen.

Solution:

Edge of the cubical tank = 1.5m or 150 cm

Surface area of the cubical tank(5 faces)= 5 x Area of one Face

= 5 × (150 × 150)cm2                   —-(1)

Find area of each square tile:

Side of tile = 25 cm(Given)

Area of one tile = 25 × 25 cm2      —–(2)

Now,

Number of tiles required = (Surface Area of Tank)/(Area of each Tile)

= (5 × 150 × 150) / 25 × 25

= 180

Find cost of tiles:

Cost of 1 dozen tiles, i.e., cost of 12 tiles = Rs.  360

Therefore, cost of one tile = Rs.360/12 = Rs. 30

So, the cost of 180 tiles = 180 × 30 = Rs. 5400

Question 10: Each edge of a cube is increased by 50 %. Find the percentage increase in the surface area of the cube.

Solution:

Let ‘a’ be the edge of a cube.

Surface area of the cube having edge ‘a’ = 6a2    —-(1)

After increasing an edge by 50%, we get,

The new edge = a + 50a/100

= 3a/2

Surface area of the cube having edge ‘3a/ 2’ = 6 × (3a/2)2 = (27/2) a2   —-(2)

Subtracting equation(1) from (2) to find the increase in the Surface Area :

Increase in the Surface Area = (27/2) a2 – 6a2

= (15/2)a2

Now, Percentage increase in the surface area = ((15/2)a2/6a2) × 100

= 15/12 × 100

= 125%

= Therefore, percentage increase in the surface area of a cube is 125.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

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