# RBSE Solutions for Class 9 Maths Chapter 12 Surface Area and Volume of Cube and Cuboid Ex 12.1

RBSE Solutions for Class 9 Maths Chapter 12 Surface Area and Volume of Cube and Cuboid Ex 12.1 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 12 Surface Area and Volume of Cube and Cuboid Exercise 12.1.

 Board RBSE Textbook SIERT, Rajasthan Class Class 9 Subject Maths Chapter Chapter 12 Chapter Name Surface Area and Volume of Cube and Cuboid Exercise Ex 12.1 Number of Questions Solved 8 Category RBSE Solutions

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 12 Surface Area and Volume of Cube and Cuboid Ex 12.1

Question 1.
The length, breadth and height of (RBSESolutions.com) a closed wooden box are 1 m, 60 cm and 40 cm respectively. Find the external surface area of the box.
Solution.
External surface area = 2 (lb + bh + hl)
= 2[100 x 60 + 60 x 40 + 40 x 100]
= 2[6000 + 2400 + 4000]
= 2[12400]
= 24800 cm2
= 2.48 m2

Question 2.
How many square centimetre will be required to cover (RBSESolutions.com) the box whose dimensions are 40 cm, 30 cm and 20 cm respectively?
Solution.
Let x cm2 area will covers the whole box.
x = 2 (lb + bh + hl)
= 2[40 x 30 + 30 x 20 + 20 x 40]
= 2[1200 + 600 + 800]
= 5200 cm2
Thus, to cover the whole box 5200 cm2 is required.

Question 3.
A room is of length 5 metres, breadth 3.5 metres and height 4 metres. Find the (RBSESolutions.com) cost of white washing the four walls and the roof at the rate of ₹ 15 per square metre.
Solution.
Area of four walls = 2(l + b) x h = 2[5 + 3.5] x 4 = 8.5 x 8 = 68.0 sq. m
Area of roof = l x b = 5 x 3.5 = 17.5 m2
Total area = 17.5 + 68 = 85.5 m2
Cost of expenditure per square metre is Rs. 15
Cost of expenditure of 85.5 sq. m = 85.5 x 15 = Rs. 1282.5

Question 4.
The edge of a cubical chalk box is 4 cm, then find the total surface area (RBSESolutions.com) of the chalk box and also find the length of its diagonal.
Solution.
Surface area of cubical chalk box = 6a2
where a = edge of cubical box
Here a = 4 cm
Surface Area = 6 x (4)2 = 96 cm2
Diagonal of a cube = √ 3 a
Diagonal = √ 3 x 4 = 4√ 3 cm

Question 5.
The total surface area of a cube is 1014 sq. m. Find (RBSESolutions.com) the length of its edge.
Solution.
Total surface area of a cube = 6 (side)2
1014 = 6 (side)2
(side)2 = $$\frac { 1014 }{ 6 }$$
⇒ edge or side = √169 = 13 m

Question 6.
The inner dimensions of a closed wooden box are 1 metre, 65 cm and 55 cm. Thickness of (RBSESolutions.com) the wood is 2.5 cm. Find the cost of painting the outer surface at the rate of Rs. 15 per square metre.
Solution.
The inner length of the wooden box (l) = 1 m = 100 cm
Inner breadth (b) = 65 cm
Inner height (h) = 55 cm
Thickness of the wood = 2.5 cm
Outer length (L) = 100 + 2 x 2.5 = 105 cm = 1.05 m
Outer breadth (B) = 65 + 2 x 2.5 = 70 cm = 0.70 m
Outer height (H) = 55 + 2 x 2.5 = 60 cm = 0.60 m
Total outer surface area = 2(LB + BH + HL)
= 2(1.05 x 0.70 + 0.70 x 0.60 + 0.60 x 1.05)
= 2(0.735 + 0.42 + 0.630)
= 2(1.785)
= 3.570 sq. m
Cost of painting per sq. m is Rs. 15.
Cost of painting 3.570 sq. m is 15 x 3.570 = Rs. 53.55

Question 7.
Each surface of a cube is 100 square cm. If it is cut parallel to the base by (RBSESolutions.com) a plane in two equal parts. Find the total surface area of each part.
Solution.
Here each surface of a cube is given 100 sq. cm.

If surface of a cube is cut by plane parallel to base into two equal parts, then area of upper and lower surface of a cubes remain same, whereas area of remaining four faces becomes half.
Therefore area of each part of the separated portion is equal to = 100 + 100 + (50 + 50 + 50 + 50) sq. cm = (200 + 200) sq. cm = 400 sq. cm
Hence, area of each separated portion is equal to 400 sq. cm.

Question 8.
An open box is made of wood 3 cm thick. Its external length, breadth and (RBSESolutions.com) height are 146 cm, 116 cm and 83 cm. Find the inside surface area.
Solution.
Outer length of the box (L) = 146 cm
Outer breadth (B) = 116 cm
Outer height (H) = 83 cm
Thickness of the wood = 3 cm
Inner length (l) = 146 – 3 x 2 = 140 cm
Inner breadth (b) = 116 – 2 x 3 = 110 cm
Inner height (h) = 83 – 3 x 1 (box is open)
Inner area of four walls = 2 (l + b) x h = 2 (140 + 110) x 80 = 250 x 160 = 40,000 sq. cm
and area of its bottom surface = l x b = 140 x 110 = 15400 sq. cm
Total surface area = 40000 + 15400 = 55400 sq. cm.

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