NCERT Solutions: Mensuration (Exercise 10.2 & 10.3)

# NCERT Solutions for Class 8 Maths Chapter 9 - Mensuration (Exercise 10.2 and 10.3)

## Exercise 10.2

Q1. Find the areas of the following figures by counting squares:

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

(k)

(l)

(m)

(n)

Ans:
(a) Number of filled square = 9
∴ Area covered by squares = 9 x 1 = 9 sq. units

(b) Number of filled squares = 5
∴ Area covered by filled squares = 5 x 1 = 5 sq. units

(c) Number of full filled squares = 2
Number of half-filled squares = 4
∴ Area covered by full filled squares = 2 x 1 = 2 sq. units
And Area covered by half-filled squares  =
∴ Total area = 2 + 2 = 4 sq. units

(d) Number of filled squares = 8
∴ Area covered by filled squares = 8 x 1 = 8 sq. units

(e) Number of filled squares = 10
∴ Area covered by filled squares = 10 x 1 = 10 sq. units

(f) Number of full filled squares = 2
Number of half-filled squares = 4
∴ Area covered by full filled squares = 2 x 1 = 2 sq. units
And Area covered by half-filled squares  =
∴ Total area = 2 + 2 = 4 sq. units

(g) Number of full filled squares = 4
Number of half-filled squares = 4
∴ Area covered by full filled squares = 4 x 1 = 4 sq. units
And Area covered by half-filled squares  =
∴ Total area = 4 + 2 = 6 sq. units

(h) Number of filled squares = 5
∴ Area covered by filled squares = 5 x 1 = 5 sq. units

(i) Number of filled squares = 9
∴ Area covered by filled squares = 9 x 1 = 9 sq. units

(j) Number of full filled squares = 2
Number of half-filled squares = 4
∴ Area covered by full filled squares = 2 x 1 = 2 sq. units
And Area covered by half-filled squares  =
∴ Total area = 2 + 2 = 4 sq. units

(k) Number of full filled squares = 4
Number of half-filled squares = 2
∴ Area covered by full filled squares = 4 x 1 = 4 sq. units
And Area covered by half-filled squares
∴ Total area = 4 + 1 = 5 sq. units

(l) Number of full filled squares = 3
Number of half-filled squares = 10
∴ Area covered by full filled squares = 3 x 1 = 3 sq. units
And Area covered by half-filled squares
∴ Total area = 3 + 5 = 8 sq. units

(m) Number of full filled squares = 7
Number of half-filled squares = 14
∴ Area covered by full filled squares = 7 x 1 = 7 sq. units
And Area covered by half-filled squares
∴ Total area = 7 + 7 = 14 sq. units

(n) Number of full filled squares = 10
Number of half-filled squares = 16
∴ Area covered by full filled squares = 10 x 1 = 10 sq. units
And Area covered by half-filled squares
∴ Total area = 10 + 8 = 18 sq. units

## Exercise 10.3

Q1. Find the areas of the rectangles whose sides are:
(a) 3 cm and 4 cm
(b) 12 m and 21 m
(c) 2 km and 3 km
(d) 2 m and 70 cm
Ans:
(a) Area of rectangle = length x breadth
= 3 cm x 4 cm = 12 cm
(b) Area of rectangle = length x breadth
= 12 m x 21 m = 252 m

(c) Area of rectangle = length x breadth
= 2 km x 3 km = 6 km

(d) Area of rectangle = length x breadth
= 2 m x 70 cm
= 2 m x 0.7 m = 1.4 m

Q2. Find the areas of the squares whose sides are:
(a) 10 cm
(b) 14 cm
(c) 5 m
Ans:
(a) Area of square = side x side = 10 cm x 10 cm = 100 cm2
(b) Area of square = side x side = 14 cm x 14 cm = 196 cm2
(c) Area of square = side x side = 5 m x 5 m = 25 m2

Q3. The length and breadth of three rectangles are as given below:
(a) 9 m and 6 m
(b) 17 m and 3 m
(c) 4 m and 14 m
Which one has the largest area and which one has the smallest?
Ans:
(a) Area of rectangle = length x breadth = 9 m x 6 m = 54 m
(b) Area of rectangle = length x breadth= 3 m x 17 m = 51 m
(c) Area of rectangle = length x breadth= 4 m x 14 m = 56 m
Thus, the rectangle (c) has largest area, and rectangle (b) has smallest area.

Q4. The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
Ans:Length of rectangle = 50 m and Area of rectangle = 300 m2
Since, Area of rectangle = length x breadth
Therefore,
Thus, the breadth of the garden is 6 m.

Q5. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of Rs 8 per hundred sq m?
Ans: Length of land = 500 m and Breadth of land = 200 m
Area of land = length x breadth = 500 m x 200 m = 1,00,000 m2
Cost of tilling 100 sq. m of land = 8
∴ Cost of tilling 1,00,000 sq. m of land =   =  8000

Q6. A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
Ans: Length of table = 2 m
Breadth of table = 1 m 50 cm = 1.50 m
Area of table = length x breadth
= 2 m x 1.50 m = 3 m2

Q7. A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room?
Ans: Length of room = 4 m
Breadth of room = 3 m 50 cm = 3.50 m
Area of carpet = length x breadth
= 4 x 3.50 = 14m2

Q8: A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Ans: Length of floor = 5 m and breadth of floor = 4 m
Area of floor = length x breadth
= 5 m x 4 m = 20 m2 Now, Side of square carpet = 3 m
Area of square carpet = side x side = 3 x 3 = 9 m2
Area of floor that is not carpeted = 20 m2 – 9 m2 = 11 m2

Q9. Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?
Ans: Side of square bed = 1 m
Area of square bed = side x side = 1 m x 1 m = 1 m2
∴ Area of 5 square beds = 1 x 5 = 5 m2
Now, Length of land = 5 m
Breadth of land = 4 m
∴ Area of land = length x breadth = 5 m x 4 m = 20 m2
Area of remaining part = Area of land – Area of 5 flower beds
= 20 m2 – 5 m2 = 15 m2

Q10. By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

Ans:
(a)

Area of HKLM = 3 x 3 = 9 cm2
Area of IJGH = 1 x 2 = 2 cm2
Area of FEDG = 3 x 3 = 9 cm
Area of ABCD = 2 x 4 = 8 cm2
Total area of the figure = 9 + 2 + 9 + 8 = 28 cm
(b)

Area of ABCD = 3 x 1 = 3 cm2
Area of BDEF = 3 x 1 = 3 cm2
Area of FGHI = 3 x 1 = 3 cm2
Total area of the figure = 3 + 3 + 3 = 9 cm2

Q11. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)

Ans:
(a)

Area of rectangle ABCD = 2 x 10 = 20 cm2
Area of rectangle DEFG = 10 x 2 = 20 cm2
Total area of the figure = 20 + 20 = 40 cm2
(b)

There are 5 squares.
Each side is 7 cm
Area of 5 squares = 5 × 72
= 245 cm2
(c)

Area of rectangle ABCD = 5 x 1 = 5 cm
Area of rectangle EFGH = 4 x 1 = 4 cm
Total area of the figure = 5 + 4 cm

Q12. How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively:
(a) 100 cm and 144 cm
(b) 70 cm and 36 cm
Ans:
(a) Area of region = 100 cm x 144 cm = 14400 cm2
Area of one tile = 5 cm x 12 cm = 60 cm2

Thus, 240 tiles are required.
(b) Area of region = 70 cm x 36 cm = 2520 cm2
Area of one tile = 5 cm x 12 cm = 60 cm2

Thus, 42 tiles are required.

The document NCERT Solutions for Class 8 Maths Chapter 9 - Mensuration (Exercise 10.2 and 10.3) is a part of the Class 6 Course NCERT Textbooks & Solutions for Class 6.
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## FAQs on NCERT Solutions for Class 8 Maths Chapter 9 - Mensuration (Exercise 10.2 and 10.3)

 1. What is mensuration?
Ans. Mensuration is the branch of mathematics that deals with the measurement of geometric figures such as length, area, volume, and perimeter.
 2. What are the different formulas used in mensuration?
Ans. There are several formulas used in mensuration depending on the type of geometric figure being measured. Some of the commonly used formulas include the formula for the area of a circle, which is πr², the formula for the volume of a cylinder, which is πr²h, and the formula for the perimeter of a rectangle, which is 2(l+b).
 3. What is the importance of mensuration in real life?
Ans. Mensuration has several practical applications in real life, such as in construction, architecture, and engineering. It is used to calculate the amount of materials required for building structures, as well as to measure the dimensions of objects and spaces.
 4. How can I improve my skills in mensuration?
Ans. To improve your skills in mensuration, you can practice solving problems using different formulas and techniques. You can also use online resources such as tutorials and practice exercises to enhance your understanding of the subject.
 5. What are some common mistakes to avoid when solving mensuration problems?
Ans. Some common mistakes to avoid when solving mensuration problems include not using the correct formula for the given figure, not converting units properly, and not being careful with calculations. It is important to double-check your work and ensure that you have used the correct formula and units.

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