NCERT Solutions: Mensuration

# NCERT Solutions for Class 8 Maths Chapter 9 - Mensuration

## Exercise 10.1

Q1. Find the perimeter of each of the following figures:

 (a) (b) (c) (d) (e) (f)

Ans:
(a) Perimeter
= Sum of all the sides
= 4 cm + 2 cm + 1 cm + 5 cm
= 12 cm

(b) Perimeter
= Sum of all the sides
= 23 cm + 35 cm + 40 cm + 35 cm
= 133 cm

(c) Perimeter
= Sum of all the sides
= 15 cm + 15 cm + 15 cm + 15 cm
= 60 cm

(d) Perimeter
= Sum of all the sides
= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm
= 20 cm

(e) Perimeter
= Sum of all the sides
= 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm
= 15 cm

(f) Perimeter
= Sum of all the sides
= 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm
= 52 cm

Q2. The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?

Ans: Total length of tape required = Perimeter of rectangle
= 2 (40 + 10)
= 2 x 50
= 100 cm
= 1 m
Thus, the total length of tape required is 100 cm or 1 m.

Q3. A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?

Ans: Length of table top = 2 m 25 cm = 2.25 m
Breadth of table top = 1 m 50 cm = 1.50 m
Perimeter of table top = 2 x (length + breadth)
= 2 x (2.25 + 1.50)
= 2 x 3.75
= 7.50 m
Thus, the perimeter of table top is 7.5 m.

Q4. What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?

Ans: Length of wooden strip = Perimeter of photograph
Perimeter of photograph = 2 x (length + breadth)
= 2 (32 + 21)
= 2 x 53 cm
= 106 cm
Thus, the length of the wooden strip required is equal to 106 cm.

Q5. A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?

Ans: Since the 4 rows of wires are needed.
Therefore the total length of wires is equal to 4 times the perimeter of rectangle.
Perimeter of field = 2 x (length + breadth)
= 2 x (0.7 + 0.5)
= 2 x 1.2
= 2.4 km
= 2.4 x 1000 m
= 2400 m
Thus, the length of wire = 4 x 2400 = 9600 m = 9.6 km

Q6. Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm and 5 cm.
(b) An equilateral triangle of side 9 cm.
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm
Ans: (a) Perimeter of ΔABC
= AB + BC + CA
= 3 cm + 5 cm + 4 cm
= 12 cm

(b) Perimeter of equilateral ABC
= 3 x side
= 3 x 9 cm
= 27 cm

(c) Perimeter of ΔABC
= AB + BC + CA
= 8 cm + 6 cm + 8 cm
= 22 cm

Q7. Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Ans: Perimeter of triangle = Sum of all three sides
= 10 cm + 14 cm + 15 cm = 39 cm
Thus, the perimeter of triangle is 39 cm

Q8. Find the perimeter of a regular hexagon with each side measuring 8 m.
Ans: Perimeter of Hexagon = 6 x length of one side
= 6 x 8 m = 48 m
Thus, the perimeter of hexagon is 48 m.

Q9. Find the side of the square whose perimeter is 20 m.
Ans: Perimeter of square = 4 x side

Thus, the side of square is 5 cm.

Q10. The perimeter of a regular pentagon is 100 cm. How long is its each side?
Ans: Perimeter of regular pentagon = 100 cm

Thus, the side of regular pentagon is 20 cm.

Q11. A piece of string is 30 cm long. What will be the length of each side if the string is used to form:
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?
Ans: Length of string = Perimeter of each figure
(a) Perimeter of square = 30 cm

Thus, the length of each side of square is 7.5 cm.
(b) Perimeter of equilateral triangle = 30 cm

Thus, the length of each side of equilateral triangle is 10 cm.
(c) Perimeter of hexagon = 30 cm

Thus, the side of each side of hexagon is 5 cm.

Q12. Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Ans: Let the length of third side be x cm.
Length of other two side are 12 cm and 14 cm.
Now, Perimeter of triangle = 36 cm
⇒ 12 +14 + x = 36
⇒ 26 + x= 36
⇒ x = 36- 26
⇒ x = 10 cm
Thus, the length of third side is 10 cm

Q13. Find the cost of fencing a square park of side 250 m at the rate of Rs 20 per metre.
Ans: Side of square = 250 m
Perimeter of square = 4 x side
= 4 x 250
= 1000 m
Since, cost of fencing of per meter = Rs 20
Therefore, the cost of fencing of 1000 meters = 20 x 1000 = Rs20,000

Q14. Find the cost of fencing a rectangular park of length 175 cm and breadth 125 m at the rate of Rs 12 per metre.
Ans: Length of rectangular park = 175 cm
Breadth of rectangular park = 125 m
Perimeter of park = 2 x (length + breadth)
= 2 x (175 + 125)
= 2 x 300 = 600 m
Since, the cost of fencing park per meter = Rs 12
Therefore, the cost of fencing park of 600 m = 12 x 600 = Rs 7,200

Q15. Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?
Ans: Distance covered by Sweety = Perimeter of square park
Perimeter of square = 4 x side
= 4 x 75 = 300 m
Thus, distance covered by Sweety is 300 m.
Now, distance covered by Bulbul = Perimeter of rectangular park
Perimeter of rectangular park = 2 x (length + breadth)
= 2 x (60 + 45)
= 2 x 105 = 210 m
Thus, Bulbul covers the distance of 210 m and Bulbul covers less distance.

Q16. What is the perimeter of each of the following figures? What do you infer from the answers?

Ans:
(a) Perimeter of square
= 4 x side
= 4 x 25 = 100 cm
(b) Perimeter of rectangle
= 2 x (length + breadth) = 2 x (40 + 10) = 2 x 50 = 100 cm
(c) Perimeter of rectangle
= 2 x (length + breadth)
= 2 x (30 + 20) = 2 x 50 = 100 cm
(d) Perimeter of triangle = Sum of all sides
= 30 cm + 30 cm + 40 cm = 100 cm
Thus, all the figures have same perimeter.

Q17. Avneet buys 9 square paving slabs, each with a side of 1/2 m. He lays them in the form of a square.

(a) What is the perimeter of his arrangement [Fig (i)]?
Ans: The arrangement so formed in Fig (i) is a square of side equal to 3 × side of the square slab
∴ Perimeter of his arrangement in Fig (i)
= 4 × ( 3 × side of the square slab)
= 4 × ( 3 × 0.5) m
= 4 × 1.5 m
= 6 m

(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [(Fig (ii)]?
Ans: The cross arrangement so formed in Fig (ii) by 9 square slabs, has perimeter, which consists of 20 length segments each equal to side of a square slab i.e. 0.5 m
∴ Perimeter of her arrangement in Fig (ii)
= 20 × ( side of the square slab)
= 20 × 0.5 m
= 10 m

(c) Which has greater perimeter?
Ans: The Cross arrangement has greater perimeter

(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e. they cannot be broken.)
Ans: No, there is no way of getting perimeter greater than 10 m from any other possible arrangement formed by 9 square slabs when placed edge by edge completely.

## 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 is a part of the Class 6 Course Mathematics (Maths) Class 6.
All you need of Class 6 at this link: Class 6

## Mathematics (Maths) Class 6

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## FAQs on NCERT Solutions for Class 8 Maths Chapter 9 - Mensuration

 1. What is mensuration and why is it important?
Ans. Mensuration is the branch of mathematics that deals with the measurement of geometric figures such as length, area, volume, and angles. It is important because it has practical applications in various fields such as construction, engineering, architecture, and science.
 2. What are the different formulas used in mensuration?
Ans. Some of the common formulas used in mensuration are: - Area of a rectangle = Length x Breadth - Area of a square = side x side - Area of a triangle = 1/2 x base x height - Circumference of a circle = 2 x π x radius - Area of a circle = π x radius^2 - Volume of a cube = side x side x side - Volume of a cylinder = π x radius^2 x height
 3. How can one improve their mensuration skills?
Ans. One can improve their mensuration skills by practicing problems from textbooks, solving sample papers and previous year question papers, and taking online quizzes. It is also important to understand the concepts and formulas thoroughly to be able to apply them correctly.
 4. What are some common mistakes to avoid while solving mensuration problems?
Ans. Some common mistakes to avoid while solving mensuration problems are: - Not using the correct formula for a given problem - Not paying attention to units of measurements - Misinterpreting the given data - Not simplifying the answer to the simplest form - Not rounding off the answer to the correct number of decimal places
 5. What are some real-life applications of mensuration?
Ans. Mensuration has various real-life applications, some of which are: - Calculating the amount of paint needed to paint a room - Calculating the area of a field to determine the amount of crops that can be grown - Calculating the volume of water required to fill a swimming pool - Calculating the amount of material needed to construct a building or a bridge - Determining the angle of inclination of a slope for construction purposes.

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