# KSEEB Solutions for Class 8 Maths Chapter 16 Mensuration Ex 16.1

In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 16 Mensuration Ex 16.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 8 Maths Chapter 16 Mensuration Ex 16.1 pdf, free KSEEB SSLC Class 8 Maths Chapter 16 Mensuration Ex 16.1 pdf download. Now you will get step by step solution to each question.

### Karnataka State Syllabus Class 8 Maths Chapter 16 Mensuration Ex 16.1

Question 1.
Find the total surface area of a cuboid with l= 4m, b = 3m and h = 1.5 m
T.S.A = 2 (lb + bh + lh)
= 2(4 × 3 + 3 × 1.5 + 1.5 × 4)
= 2(12 + 4.5 + 6)
= 2(22.5)
T.S.A = 45 m²

Question 2.
Find the area of four walls of a room whose length is 3.5m, breath 2.5m and height is 3m.
L = 3.5m, b = 2.5m, h = 3m
Area of the 4 walls = L.S.A of cuboid
= 2h(l + b)
= 2 × 3(3.5 + 2.5)
= 6(6) = 36m²

Question 3.
The dimensions of a room are l = 8m, b = 5m and h = 4m. Find the cost of distempering its four walls at the rate of 40/m2.
L.S.A of cuboid = 2h (l + b)
= 2 × 4 (8 + 5) = 8 (13)
L.S.A = 104 m²
The cost of distempering 1 m² = Rs 40
The cost of distempering 104m² = 104 × 40 = Rs. 4160.

Question 4.
A room is 4.8m long, 3.6m broad and 2m high. Find the cost of laying tiles on its floor and its four walls at the rate of 100/m²
L = 4.8m, b = 3.6m , h = 2m
Required area = Area of the floor + L.S.A of cuboid
= lb + 2h(l + b)
= 4.8 × 3.6 + 2 × 2(4.8 + 3.6)
= 17.28 + 33.6 = 50.88 m²
The cost of laying tiles for 1m² = Rs 100
The cost of laying tiles for 50.88 m² = 50.88 × 100 = Rs. 5088

Question 5.
A closed box is 40 cm long, 50 cm wide and 60 deep. Find the area of the foil needed for covering it.
L = 40 cm, b = 50cm, h = 60 cm
Area of the foil = T.S. A of cuboid
= 2 (lb + bh + lh)
= 2(40 × 50 + 50 × 60 + 60 × 40)
= 2(7400)= 14,800 cm²
Area of the foil required 14,800 cm²

Question 6.
The total surface area of a cube is 384 cm2 calculate the side of the cube.
T.S.A of cube = 384 cm²
61² = 384
l² = x=3846
l² = 64
l = √64
l = 8cm

Question 7.
The L.S.A of a cube is 64m², calculate the side of the cube.
L.S.A of cube = 41²
64 = 41²
x=644 = l²
∴ l = √16 = 4m
The side of the cube is 4m.

Question 8.
Find the cost of whitewashing the four walls of a cubical room of side 4m at the rate of Rs.20/m2.
l = 4m
Area to be whitewashed
= L.S.A of the cube
= 4l² = 4 × 4² = 4 × 16 = 64 m²
cost of whitewashing 1m² = 20
∴ The cost of whitewashing 64m² = 64 × 20 = Rs. 1280.

Question 9.
A cubical box has edge 10cm and another cuboidal box is 12.5 cm long, 10cm wide and 8cm high
i. Which box has a smaller total surface area?
ii. If each edge of the cube is doubled, how many times will its T.S.A increase?
T.S.A of cube 61²= 6 × 10² = 6 ×100 = 600 cm²
T.S.A of a cuboid 61² = 6 × 102 = 6 x 100 = 600 cm²
T.S.A of a cuboid = 2(lb + bh + Ih)
= 2[12.5 x 10 +10 x 8 + 8 x 12.5]
= 2[125 + 80 + 100]
= 2[305]
t = 610 cm²
1. The cubical box has smaller total surface area.
2. If the side of the cube is doubled its side will be 20cm.
T.S.A 6l² = 6 × 202 = 6 × 400 = 2400 cm²
original area = 600cm²
2400600 = 4
∴ Its area will increase by 4 times

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