Perimeter – Maharashtra Board Class 6 Solutions for Mathematics

Perimeter – Maharashtra Board Class 6 Solutions for Mathematics (English Medium)

MathematicsGeneral ScienceMaharashtra Board Solutions

Exercise-39

Solution 1(1):
Length (l) = 9 cm
Breadth (b) = 6 cm
Perimeter of a rectangle
= 2 (l + b)
= 2(9 + 6)
= 2(15)
= 30 cm

Solution 1(2):
Length (l) = 5.2 cm
Breadth (b) = 4 cm
Perimeter of a rectangle
= 2 (l + b)
= 2(5.2 + 4)
= 2(9.2)
= 18.4 cm

Solution 1(3):
Length (l) = 7.5 cm
Breadth (b) = 3.2 cm
Perimeter of a rectangle
= 2 (l + b)
= 2 (7.5 + 3.2)
= 2(10.7)
= 21.4 cm

Solution 2:
Side of the square (x) = 12 cm
Perimeter of a square = 4 × x
= 4 × 12
= 48 cm

Solution 3:
Sides of the triangle, a = 6 cm, b = 9 cm, c = 5 cm
Perimeter of a triangle = a + b + c
= 6 + 9 + 5
= 20 cm

Solution 4:
Sides of the triangle, a = 4.8 cm, b = 10.2 cm, c = 5.3 cm
Perimeter of a triangle = a + b + c
= 4.8 + 10.2 + 5.3
= 20.3 cm

Exercise-40

Solution 1:
Length of the rectangle (l) = 15 m
Breadth of the rectangle (b) = 10 m
Length of trimming = Perimeter of the rectangle
Perimeter of the rectangle = 2(l + b)
= 2(15 + 10)
= 2(25)
= 50 m
∴ Length of trimming required is 50 m.

Solution 2:
Side (x) of the square window = 1.5 m
The length of the wooden strip = Perimeter of the square
Perimeter of the square window = 4 x
= 4 × 1.5
= 6 m
Length of the wooden strip required is 6 m.

Solution 3:
Length of the rectangular garden (l) = 320 m.
Breadth of the rectangular garden (b) = 210 m.
The distance Sabir walks = Perimeter of the rectangle.
Perimeter of the rectangular garden
= 2(l + b)
= 2 (320 + 210)
= 2 (530)
= 1060 m
Satbir walks 1060 metres daily.

Solution 4:
Sides of the triangle, a = 30 m, b = 20 m and c = 210 m
Perimeter of the triangle
= (a + b + c)
= (30 + 20 + 25)
= 75 m
Length of wire required for 4 rounds of the fence
= 4 × perimeter
= 4 × 75
= 300 m
Total cost of wire = 2.5 × 300
= Rs. 750
∴ Total cost of the wire is Rs. 750

Solution 5:
Length of the mat (l) = 5 m 20 cm
Breadth of the mat (b) = 3 m 30 cm
Perimeter of the rectangle
= 2(l + b)
= 2(5 m 20 cm + 3 m 30 cm)
= 2(8 m and 50 cm)
= 16 m 100 cm
= 17 m
Length of the border required = Perimeter of the rectangle
∴ Length of the border required = 17 m.

Exercise-41

Solution 1:
Suppose the length of the third side is c cm.
The sides of the triangle are 15 cm, 20 cm and c cm.
Perimeter of a triangle = a + b + c
But, perimeter = 50 cm
∴ 50 = 15 + 20 + c
∴ 50 = 35 + c
∴ c = 50 – 35
∴ c = 15 cm
The length of the third side is 15 cm.

Solution 2:
Perimeter of a square = 4 × x
But, perimeter = 80 cm
∴ 4 × x = 80
But, 4 × 20 = 80
∴ x = 20 cm
∴ Side of the square is 20 cm.

Solution 3:
Perimeter of a rectangle = 2 (l + b)
∴ 2 (7 + b) = 62
But, 2 × 31 = 62
∴ 7 + b = 31
Sum of 7 and 24 is 31
∴ b = 24
∴ The breadth of the rectangle is 24 cm.

Solution 4:
Let the two equal sides be b, c and given b = c
The sides of the triangle are 15, b, c
Perimeter of a triangle = a + b + c
∴ 55 = 15 + b + b
∴ 55 = 15 + 2b
But, 15 + 40 = 55
∴ 2b = 40
∴ b = 20
∴ Length of each of remaining sides is 20 cm.

Solution 5:
Perimeter of a rectangle = 2 (l + b)
∴ 2 (l + 30) = 100
But, 2 × 50 = 100
∴ l + 30 = 50
Sum of 30 and 20 is 50
∴ l = 20
∴ The breadth of the rectangular pool is 20 m.

Solution 6:
Perimeter of a square = 4 × x
∴ 16 = 4 × x
But, 4 × 4 = 16
∴ x = 4
The length of each side of the square room = 4 m.

Solution 7:
Length of the rectangle (l) = 50 cm
Breadth of the rectangle(b) = 30 cm
Perimeter of a rectangle
= 2 (l + b)
= 2(50 + 30)
= 2(80)
= 160 cm
∴ Length of the wire = 160 cm
Now, the wire is bent into a square.
∴ Perimeter of a square = 4 × x
∴ 160 = 4 × x
But, 4 × 40 = 160
∴ x = 40
Length of each side of the square is 40 cm.