PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 12 Perimeter and Area Ex 12.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 12 Perimeter and Area Ex 12.2

1. Find the approximate area of each of the following figures by countaing the number of squares – complete, more than half and exactly half.
PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2 1

Solution:
(i) Number of complete squares m = 7
Here we do not have any half square or more than half.
∴ n = 0, p = 0
∴ Area of plane figure = m + n + \(\frac {1}{2}\)p
= 7 + 0 + 0
= 7 sq. units

(ii) Number of complete squares m = 2
Number of more than half squares n = 4
Number of half squares p = 0
∴ Area of plane figure = m + n + \(\frac {1}{2}\)p
= 4 + 2 + 0
= 6 sq. units.

(iii) Number of complete squares m = 10
Number of more than half squares n = 0
Number of exactly half squares p = 2
∴ Area of plane figure = m + n + \(\frac {1}{2}\)p
= (10 + 0 + \(\frac {1}{2}\) × 2) sq. units
= (10 + 1) sq. units
= 11 sq. units

(iv) Number of complete squares m = 11
Number of more than half squares n = 0
Number of exactly half squares p = 2
∴ Area of plane figure = m + n + \(\frac {1}{2}\)p
= 11 + 0 + \(\frac {1}{2}\) × 2
= 11 + 1 = 12 sq. units

(v) Number of complete squares m = 10
Number of more than half squares n = 3
Number of exactly half squares p = 0
∴ Area of plane figure = m + n + \(\frac {1}{2}\)p
= 10 + 3 + 0
= 13 sq. units

PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2

2. Find the area of rectangle whose:

Question (i)
length = 12 cm, breadth = 16 cm
Solution:
Length of rectangle = 12 cm
and Breadth of rectangle = 16 cm.
∴ Area of rectangle = Length × Breadth
= 12 cm × 16 cm
= 192 sq. cm

Question (ii)
length = 25 m, breadth = 18 m
Solution:
Length of rectangle = 25 m
and Breadth of rectangle = 18 m
∴ Area of rectangle = Length × Breadth
= 25 m × 18 m
= 450 sq. m

Question (iii)
length = 2.7 m, breadth = 45 cm
Solution:
Length of rectangle = 2.7 m = 270 cm
and Breadth of rectangle = 45 cm
∴ Area of rectangle = Length × Breadth
= 270 cm × 45 cm
= 12150 sq. cm

Question (iv)
length 4.2 cm, breadth = 1.5 cm
Solution:
Length of rectangle = 4.2 cm
and Breadth of rectangle = 1.5 cm
∴ Area of rectangle = Length × Breadth
= 4.2 cm × 1.5 cm
= 6.3 sq. cm

Question (v)
length = 3.8 mm, breadth = 4 mm.
Solution:
Length of rectangle = 3.8 mm
and Breadth of rectangle = 4 mm
∴ Area of rectangle = Length × Breadth
= 3.8 mm × 4 mm
= 15.2 sq. mm

PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2

3. Find the area of the square with side:

Question (i)
19 cm
Solution:
Side of the square = 19 cm
∴ Area of the square = side × side
= 19 cm × 19 cm = 361 sq. cm.

Question (ii)
24 mm
Solution:
Side of square = 24 mm
∴ Area of square = side × side
= 24 mm × 24 mm
= 576 sq. mm

Question (iii)
3.5 cm
Solution:
Side of the square = 3.5 cm
Area of square = side × side
= 3.5 cm × 3.5 cm
= 12.25 sq. cm

Question (iv)
2.6 cm
Solution:
Side of the square = 2.6 cm
Area of square = side × side
= 2.6 cm × 2.6 cm
= 6.76 sq. cm

PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2

Question (v)
8.2 cm.
Solution:
Side of the square = 8.2 cm
Area of the square = side × side
= 8.2 cm × 8.2 cm
= 67.24 sq. cm.

4. The area of a rectangle is 216 sq. cm and its length is 12 cm. Find its breadth.
Solution:
The area of the rectangle = 216 sq. cm.
Length of the rectangle = 12 cm
Breadth of the rectangle = \(\frac{\text { Area }}{\text { Length }}\)
= \(\frac {216}{12}\) cm
= 18 cm.

5. The area of a rectangle is 225 sq. m and its breadth is 9 in. Find its length.
Solution:
The area of the rectangle = 225 sq. m
and Breadth of the rectangle = 9 m
∴ Length of the rectangle = \(\frac{\text { Area }}{\text { Breadth }}\)
= \(\frac {225}{9}\) m
= 25 m

6. The length and breadth of a ground are 32 m and 24 m. Find the cost of levelling the ground at the rate of ₹ 3 per sq. m.
Solution:
Length of ground = 32 m
and Breadth of ground = 24 m
Area of the ground = Length × Breadth
= 32 m × 24 m
= 768 sq. m.
Levelling cost of 1 sq. m = ₹ 3
Levelling cost of 768 sq. m = ₹ 3 × 768
= ₹ 2304

PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2

7. Find the perimeter of a rectangle whose area is 324 sq. cm and its one side is 36 cm.
Solution:
Area of rectangle = 324 sq. cm
One side of rectangle = 36 cm
∴ Other side of rectangle = \(\frac{\text { Area }}{\text { One side }}\)
= \(\frac {324}{36}\) = 9 cm
Perimeter of rectangle
= 2 × (1st side + 2nd side)
= 2 × (36 + 9) cm
= 2 × 45 cm
= 90 cm

8. The perimeter of a square field is 100 m. Find its area.
Solution:
The perimeter of square field = 100 m
∴ 4 × side of square = 100 m
∴ Side of a square = \(\frac {100}{4}\) = 25 m
Hence area of square field = side × side
= 25 m × 25 m
= 625 sq. m.

9. Area of a rectangle of length 20 cm is 340 sq. cm. Find its perimeter.
Solution:
Area of rectangle = 340 sq. cm
and Length of rectangle = 20 cm
Breadth of rectangle = \(\frac{\text { Area }}{\text { Length }}\)
= \(\frac {340}{20}\) cm = 17 cm
Perimeter of rectangle
= 2 × (Length + Breadth)
= 2 × (20 + 17) cm
= 2 × 37 cm
= 74 cm

PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2

10. A marble tile measure 15 cm × 20 cm. How many tiles will be required to cover a wall of size 4 m × 6 m?
Solution:
Area of the wall = 4 m × 6 m
= 400 cm × 600 cm
= 240000 sq. cm
Area of a marble tile = 15 cm × 20 cm
= 300 sq. cm
Number of tiles required to cover wall
= \(\frac{\text { Area of wall }}{\text { Area of tile }}=\frac{240000}{300}\) = 800
Hence number of tiles required to cover wall = 800

11. Find the cost of levelling the square field of side 75 m at rate of ₹ 5 per square metre.
Solution:
Side of the square field = 75 m
Area of square field = side × side
= 75 m × 75 m
= 5625 sq. m
Cost of levelling 1 sq. m = ₹ 5
Cost of levelling 5625 sq. m = ₹ 5 × 5625
= ₹ 28125

12. How many stamps of size 2 cm × 1.5 cm can be pasted on a sheet of paper of size 6 cm × 12 cm?
Solution:
Area of the sheet of paper
= 6 cm × 12 cm = 72 sq. cm
Area of one stamp = 2 cm × 1.5 cm = 3 sq cm
Number of stamps pasted on sheet of paper
= \(=\frac{\text { Area of paper sheet }}{\text { Area of stamp }}=\frac{72}{3}\) = 24.

PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2

13.

Question (i)
What will happen to the area of a square if its side is trebled (tripled)?
Solution:
Let side of the square = x cm
∴ Area of the square = (x × x) sq. m
Now, if the side is trebled, then side of new square = 3x cm
∴ Area of the new square
= [(3x) × (3x)] sq. m
= (3 × 3 × x × x) sq. m
= 9 (x × x) sq. cm
= 9 × (Area of original square)
∴ If side is trebled, then area becomes 9 times of original area.

Question (ii)
What will happen to the area of a rectangle if its length is halved and breadth is doubled?
Solution:
l cm and b cm be the length and breadth of the rectangle respectively.
∴ Area of rectangle = l × b
Now, If its length is halved and breadth is doubled.
∴ New length = \(\frac {1}{2}\) l
and new breadth = 2b
Thus area of new rectangle = length × breadth
= \(\frac {1}{2}\) × l × 2b
= \(\frac {1}{2}\) × 2 × (l × b)
= (l × b) = original area
Area will remain the same.

Question (iii)
What will happen to the area of a square if its side is halved?
Solution:
Let side of the square = x cm
∴ Area of the square = (x × x) sq. cm
Now, if side is halved, then
side of new square = \(\frac {1}{2}\)x cm
∴ Area of the new square sq.cm
PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2 2
PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2 3
Hence, if side is halved, then the area becomes one-fourth times original area.

PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2

14. Find the area of the following figures by splitting it into rectangles and squares:

Question (i)
PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2 4
Solution:
The given figure can be divided into 2 parts
PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2 5
Rectangle A of size 8 cm × 3 cm
Rectangle B of size 4 cm × 2 cm
∴ Area of rectangle A
= 8 cm × 3 cm = 24 sq. cm
and Area of rectangle B
= 4 cm × 2 cm
= 8 sq. cm
⇒ Total area of the figure
= 24 sq. cm + 8 sq. cm
= 32 sq. cm

Question (ii)
PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2 6
Solution:
The given figure can be divided into 2 parts
PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2 7
Rectangle A of size 10 cm × 2 cm
Rectangle B of size 8 cm × 2 cm
∴ Area of rectangle A
= 10 cm × 2 cm = 20 sq. cm
and Area of rectangle B
= 8 cm × 2 cm = 16 sq. cm
Total Area of the figure
= (20 + 16) sq. cm = 36 sq. cm

PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2

Question (iii)
PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2 8
Solution:
The given figure can be divided into 3 parts
PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2 9
Rectangle A of size 12 cm × 3 cm
Rectangle B of size 3 cm × 2 cm
Rectangle of size 12 cm × 3 cm
Area of rectangle A
= 12 cm × 3 cm = 36 sq. Cm
Area of rectangle B
= 3 cm × 2 cm = 6 sq. cm
Area of rectangle C
= 12 cm × 3 cm = 36 sq. cm
Total area of the figures
= (36 + 6 + 36) sq. cm
= 78 sq. cm

Question (iv)
PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2 10
Solution:
The given figure can be divided into 3 parts
PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2 11
Rectangle A of size 13 × 3 units
Rectangle B of size 5 × 3 units
Rectangle C of size 5 × 3
Area of rectangle A
= 13 × 3 = 39 sq. units
Area of rectangle B
= 5 × 3 = 15 sq. units
Area of rectangle C
= 5 × 3 = 15 sq. units
Hence total area of the figure
= (39 + 15 + 15) sq. units
= 69 sq. units Ans.

15. Fill in the blanks:

Question (i)
1 square metre = ……….. sq. cm.
Solution:
10000

Question (ii)
1 square cm = ………. sq. mm.
Solution:
100

PSEB 6th Class Maths Solutions Chapter 12 Perimeter and Area Ex 12.2

Question (iii)
Area of Rectangle = …………. × ……………
Solution:
length, breadth

Question (iv)
Length = ………….. ÷ breadth.
Solution:
Area

Question (v)
Area of square = …………. × ……………
Solution:
side × side.

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