Contents

**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.