Kseeb Class 6 Mathematics Chapter 10 Mensuration Solutions
Welcome to NCTB Solutions. Here with this post we are going to help 6th class students by providing Solutions for KSEEB Class 6 Mathematics chapter 10 Mensuration. Here students can easily find all the solutions for Mensuration Exercise 10.1, 10.2 and 10.3. Also here our Expert Mathematic Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 10 solutions. Here all the solutions are based on Karnataka State Board latest syllabus.
Mensuration Exercise 10.1 Solutions :
(2) 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?
Solution :
The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape.
The length of the tape required = 2 x (Length + breadth)
The length of the tape required = 2 x ( 40 + 10 )
= 2 x 50
= 100 cm
∴ The length of the tape required is 100 cm.
(3) A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
Solution :
A table-top measures 2 m 25 cm by 1 m 50 cm.
The perimeter of the table-top = 2 x (Length + breadth)
The perimeter of the table-top
= 2 x (2 m 25 cm + 1 m 50 cm. )
= 2 x 3. 75
= 7.5 m
∴ The perimeter of the table-top is 7.5 m.
(4) What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Solution :
From the question we get,
Length 35 cm
Breadth 25 cm
∴ Required length,
= 2 (35 + 25)
= 2 × 60
= 120 cm
Therefore, the length of the wooden strip is 120 cm.
(5) 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?
Solution :
A rectangular piece of land measures 0.7 km by 0.5 km.
Each side is to be fenced with 4 rows of wires.
Length of the wire needed = 2 x (Length + breadth)
= 2 x (0.7 + 0.5)
= 2.4 km
Each side is to be fenced with 4 rows of wires.
= 2.4 km x 4
= 9.6 km
∴ Length of the wire needed is 9.6 km
(6) 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.
Solution :
(i) Perimeter = s1 + s2 + s3
Perimeter = 3 + 4 + 5
= 12 cm
Hence, perimeter of a triangle of sides 3 cm, 4 cm and 5 cm is 12 cm.
(ii) Perimeter = 3 × side
Perimeter = 3 × 9
= 27 cm.
Hence, perimeter of an equilateral triangle of side 9 cm
(iii) Perimeter = 2 × equal side + 3rd side
Perimeter = 2 × 8 + 6
= 16 + 6
= 22 cm.
Hence, perimeter of an isosceles triangle with equal sides 8 cm each and third side of 6 cm is 22 cm.
(7) Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Solution :
Perimeter of a triangle = Addition of all sides
= 10 cm + 14 cm + 15 cm
= 39 cm
Thus, the perimeter will be 39 cm.
(8) Find the perimeter of a regular hexagon with each side measuring 8 m.
Solution :
Given, Side of regular hexagon = 8 m
∴ Perimeter,
= (6 × 8)
= 48 m
So, the perimeter of a regular hexagon is 48 m.
(9) Find the side of the square whose perimeter is 20 m.
Solution :
Perimeter of square
= 4 x sides.
Perimeter of square
= 4 x 20
= 80 m
(10) The perimeter of a regular pentagon is 100 cm. How long is its each side?
Solution :
Regular pentagon has 5 sides.
The perimeter of a regular pentagon is 100 cm.
∴ Each side,
= 100/5
= 20 cm
(11) 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?
Solution :
(a) String is 30 cm long
Side = 30/4
= 7.5 cm
(b) String is 30 cm long
Side = 30/3
= 10 cm
(c) String is 30 cm long
Side = 30/6
= 5 cm
(12) Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Solution :
Two sides of a triangle are 12 cm and 14 cm.
The perimeter of the triangle is 36 cm
Perimeter of the triangle = addition of three sides.
= 36 = 12 + 14 + ?
= 36 – 26
= 10 cm.
Hence, third side will be 10 cm.
(13) Find the cost of fencing a square park of side 250 m at the rate of ₹20 per metre.
Solution :
Square Park of side 250 m
The cost of fencing a square park = 4 x side
= 4 x 250
= 1000 m.
Rate of 20 per metre.
= 1000 x 20
= Rs.20, 000
The cost of fencing a square park of side 250 m is Rs.20, 000
(14) Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of ₹12 per metre.
Solution :
∴ Perimeter,
= 2 (175 + 125)
= 2 × 300
= 600 m
∴ Cost,
= 600 × 12
= 7200 rupees
Thus, the cost of fencing a rectangular will be 7200 rupees.
(15) 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?
Solution :
Sweety runs around a square park of side 75 m.
Bulbul runs around a rectangular park with length 60 m and breadth 45 m.
Perimeter of Square park
= 4 x 75
= 300 m
Perimeter of rectangular park
= 2 x (Length + Breadth)
= 2 x ( 60 + 45)
= 210 m
∴ Bulbul covers less distance.
(16) What is the perimeter of each of the following figures? What do you infer from the answers?
Solution :
Figure – (a)
We know, Perimeter = addition of all sides.
So, 25 + 25 + 25 + 25
= 100 cm
Figure – (b)
We know, Perimeter = addition of all sides.
So, 20 + 20 + 30 + 30
= 100 cm
The perimeter of the following figure is 100 cm.
Figure – (c)
We know, Perimeter = addition of all sides.
So, 10 +10 + 40 + 40
= 100 cm
The perimeter of the following figure is 100 cm
Figure – (d)
We know, Perimeter = addition of all sides.
So, 30 + 30 + 40
= 100 cm
The perimeter of the following figure is 100 cm.
(17) Avneet buys 9 square paving slabs, each with a side of 1m. He lays them in the form of a square
(a) What is the perimeter of his arrangement [Fig (i)]?
(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)]?
(c) Which 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?
Solution :
(a) Perimeter
= 1/2 × 3 × 4
= 6 m
(b) = 1/2 × 5 × 4
= 10 m
(c) The fraction of cross has greater perimeter.
(d) No there is no such arrangement.
Mensuration Exercise 10.2 Solutions :
(1) 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
Solution :
(a) We know, Area of the rectangle = Length x Breadth
So, 3 cm x 4 cm
= 12 cm
Area of the rectangle 12 sq. cm.
(b) We know, Area of the rectangle = Length x Breadth
So, 12 x 21
= 252
Area of the rectangle = 252 sq. m.
(c) Area of the rectangle = Length x Breadth
So, 2 x 3
= 6
Area of the rectangle 6 sq. km.
(d) Area of the rectangle = Length x Breadth
So, 2 m x 0.7 m
= 1.40
Area of the rectangle 1.40 sq. m.
(2) Find the areas of the squares whose sides are :
(a) 10 cm
(b) 14 cm
(c) 5 m
Solution :
(a) Area of the square = side x side
So, 10 cm x 10 cm
= 100
Area of the square is 100 sq. cm.
(b) Area of the square = side x side
So, 14 cm x 14 cm = 196
rea of the square is 196 sq. cm.
(c) Area of the square = side x side
So, 5 m x 5 m = 25
Area of the square is 25 sq. m.
(5) What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹8 per hundred sq m.?
Solution :
Rectangular plot of land 500 m long and 200 m wide
∴ Area of the land
= (500 × 200)
= 1090000 m2
∴ The total cost,
= 100000 × 8
= 800000 rupees
Thus, the cost of tiling will be 800000 rupees.
(6) A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
Solution :
A table-top measures 2 m by 1 m 50 cm.
Area of the rectangle = Length x Breadth
2 m x 1 m 50 cm = 3 sq. m
Area of table-top is 3 sq. m.
(7) 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?
Solution :
A room is 4 m long and 3 m 50 cm wide.
Area of the rectangle = Length x Breadth
= 4 m x 3 m 50 cm
= 14 sq. m
Thus, the carpet is needed to cover the floor of the room is 14 sq. m
(8) 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.
Solution :
A floor is 5 m long and 4 m wide.
Area of the rectangle floor = Length x Breadth
= 5 x 4
= 20 sq. m
A square carpet of sides 3 m.
Area of A square carpet
= Side x side
= 3 x 3
= 9 sq. m
The area of the floor that is not carpeted = 20 sq. m – 9 sq. m = 11 sq.m.
(9) 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?
Solution :
Five square flower beds each of sides 1 m
Area of A square flower beds
= Side x side
= 1 x 1
= 1 sq. m
Five square flower beds
= 5 x 1
= 5 sq.m
Land 5 m long and 4 m wide
Area of the rectangle and = Length x Breadth
= 5 x 4
= 20 sq. m.
Area of the remaining part of the land
= 20 sq. m. – 5 sq.m
= 15 sq.m
(11) Split the following shapes into rectangles and find their areas.
Solution :
Figure – (a)
= (10 × 2) + 910 × 2)
= 40 square unit
Figure – (b)
= (7 × 7) × 5
= 245 square units
Figure – (c)
= (4 × 1) + (5 × 1)
= 9 square units
(12) 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.
Solution :
(a) 100 cm and 144 cm
Tiles whose length and breadth are 12 cm and 5 cm respectively
Area of tiles
= 12 x 5
= 60 sq. m
Area of rectangular region
= 100 x 144
= 14,400 sq. m
Tiles needed
= 14,400 sq. m / 60 sq. m
= 240
Tiles needed is 240.
(b) 70 cm and 36 cm.
Tiles whose length and breadth are 12 cm and 5 cm respectively.
Area of tiles,
= 12 x 5
= 60 sq. m
Area of rectangular region,
= 70 x 36
= 2520 sq. m
Tiles needed,
= 2520 sq. m / 60 sq. m
= 42
Tiles needed is 42.
Next Chapter Solution :
👉 Algebra
