**GSEB Solutions for Class 6 Mathematics – Perimeter and Area**

GSEB SolutionsMathsScience

**Activity **

**Solution 1(1):**

(Length of each box = 1 cm)

Perimeter of the square = 5 + 5 + 5 + 5 = 20 cm

And area of the square = 5 × 5 = 25 sq. cm

Perimeter of the rectangle = 4 + 7 + 4 + 7 = 22 cm

And area of the rectangle = 4 × 7 = 28 sq. cm

**Solution 1(2):**

Area of the triangle = 8 full blocks + 2 more than half blocks + 4 half blocks

∴ Area of the triangle = 8 + 2 + 2 = 12 sq. cm

Area of the Circle = 9 full blocks + 12 more than half blocks

∴ Area of the Circle = 9 + 12 = 21 sq. cm

**Exercise**

**Solution 1(1):**

1 m = 100 cm

∴ 1 sq m = 100 cm × 100 cm = 10,000 cm

1 sq m = 10000 sq. cm

**Solution 1(2):**

1 m × 1 m = 1 sq. m

1 m = 100 cm

∴ 100 cm × 100 cm = 10,000 sq. cm

∴ 1 sq m = 10,000 sq. cm

∴ 40,000 sq. cm = 40,000 ÷ 10000 = 4 sq. m

40000 sq. cm = 4 sq. m

**Solution 1(3)****:**

Perimeter of a square = 4 × length

∴ Perimeter of the square of length 4 cm = (4 × 4) cm = 16 cm

Perimeter of a square of length 4 cm = 16 cm.

**Solution 1(4):**

Perimeter of a rectangle = 2 (length + breadth)

∴ Perimeter of a rectangle of length 3 cm and breadth 2 cm = 2 (3 + 2) m = 10 m.

Perimeter of the rectangle of length 3 m and breadth 2 m is 10 m.

**Solution 2(1):**

Length of the rectangular field = 45 m

Breadth of the rectangular field = 40 m

Area of a rectangle = length × breadth

Area of the rectangular field = 45 m × 40 m = 1800 sq. m

The cost of tiling 1 sq m field = Rs. 12

The cost of tiling 1800 sq m field = Rs. (1800 × 12)

= Rs. 21600

The cost of tiling the field is Rs. 21600.

**Solution 2(2):**

Length of the square wall = 12m

Area of a square = length × length

Area of the square wall = 12 m × 12 m = 144 sq. m

The cost of painting 1 sq m of the wall = Rs.10

∴ the cost of painting 144 sq. m wall = Rs. (144 × 10)

∴ the cost of painting 144 sq. m wall = Rs. 1440

The cost of painting the square part of wall is Rs. 1440.

**Solution 2(3)****:**

Length of the card paper = 80 cm

Breadth of the card paper = 40 cm

Area of a rectangle = length × breadth

Area of the rectangular card paper = 80 cm × 40 cm

= 3200 sq. cm

Area of a square = length × length

Area of the square to be made from card paper

= length × length

= 10 cm × 10 cm

= 100 sq. cm

Area of one square = 100 sq. cm

Number of squares made from 100 sq. cm card paper

= 1

∴ Number of squares made from 3200 sq. cm card paper = 3200 ÷ 100 = 32

Thus, 32 squares can be made from the card paper.

**Solution 2(4):**

Length of the dining hall = 15 m

Breadth of the dining hall = 12 cm

Area of a rectangle = length × breadth

Area of the rectangular dining hall = 15 m × 12 m= 180 sq. m

Area of a square = length × length

Area of the square tile = 30 cm × 30 cm = 900 sq. cm

∴ Area of one square tile = 900 sq. cm

But the area of dining hall is in the units of sq. m

Hence, convert the units of the area of the dining hall into sq. cm

Area of the rectangular dining hall = 180 sq. m

1 sq. m = 10,000 sq. cm

∴ 180 sq. m = (180 × 10,000) cm = 18,00,000 sq. cm

Area of one square tile = 900 sq. cm

Number of square tiles needed for 900 sq. cm area = 1

∴ Number of square tiles needed for 18,00,000 sq. cm area = 18,00,000 ÷ 900 = 2,000

2000 tiles are needed for the dining hall.

**solution 2(5):**

Length of the rectangular cloth = 8 m

Breadth of the rectangular cloth = 6 cm

Area of a rectangle = length × breadth

Area of the rectangular cloth = 8 m × 6 m = 48 sq. m

Area of a square = length × length

Area of the square piece of cloth = 20 cm × 20 cm = 400 sq. cm

∴ Area of one square piece of cloth = 400 sq. cm

But the unit of area of the cloth is sq. m

Hence, convert the area of the rectangular cloth into units of sq. cm

Area of rectangular cloth = 48 sq. m

1 sq. m = 10,000 sq. cm

∴ 48 sq. m = (48 × 10,000) cm = 48,00,000 sq. cm

Area of one square piece of cloth = 400 sq. cm

Number of square pieces formed from 400 sq. cm of the cloth = 1

∴ Number of square pieces formed from 48,00,000 sq. cm of the cloth = 48,00,000 ÷ 400 = 1200

Thus, 1200 square pieces of cloth can be formed.

**Practice – 1**

**Solution 1:**

Given,

Length of the drawing paper = 25 cm

Perimeter of a square = 4 × length

Hence,

perimeter of the square drawing paper = 4 × 25 = 100

Therefore, the perimeter of the square drawing paper is 100 cm.

**Solution 2:**

Given,

Length of the square = 14 cm

Perimeter of a square = 4 × length

Hence, perimeter of the square = 4 × 14 = 56 cm

Therefore, perimeter of the square having length 14 cm is 56 cm.

**Solution 3:**

Given,

Length of the square piece of cloth = 8 metres

Perimeter of the square piece of cloth = 4 × length

Hence, perimeter of the square = 4 × 8 = 32 metres

Therefore, perimeter of the square piece of the cloth is 32 metres.

**Solution 4:**

Given,

Length of the square plot = 18 metres

Perimeter of a square = 4 × length

Hence, perimeter of the square plot = 4 × 18= 72 metres

Therefore, perimeter of the square plot is 72 metres.

**Solution 5:**

Length of the square space = 9 metres (given)

Perimeter of a square = 4 × length

Hence, perimeter of the square space = 4 × 9 = 36 metres

Therefore, perimeter of the square space is 36 metres.

**Solution 6:**

Given,

Length of the square tile = 10cm

Perimeter of a square = 4 × length

Hence, perimeter of the square tile = 4 × 10 = 40 cm

Therefore, perimeter of the square tile is 40 cm.

**Practice – 2**

**Solution 1(1):**

Length of the rectangle = 18 cm

Breadth of the rectangle = 16 cm

Perimeter of a rectangle = 2 (length + breadth)

Hence,

perimeter of the rectangle = 2 (18 + 16) = 2(34) = 68 cm

Therefore, perimeter of the rectangle is 68 cm.

**Solution 1(2):**

Length of the rectangular wooden piece = 40 cm

Breadth of the rectangular wooden piece = 30 cm

Perimeter of a rectangle = 2 (length + breadth)

Hence,

perimeter of the rectangular wooden piece = 2 (40 + 30)

= 2 (70)

= 140 cm

Therefore, perimeter of the rectangular wooden piece is 140 cm.

**Solution 1(3)****:**

Length of the computer lab = 15 m

Breadth of the computer lab = 13 m

Perimeter of a rectangle = 2 (length + breadth)

Hence, perimeter of the rectangular computer lab

= 2 (15 + 13)

= 2 (28) = 56 m

Therefore, perimeter of the computer lab is 56 m.

**Solution 1(4):**

Length of the playground = 30 m

Breadth of the playground = 25 m

Perimeter of a rectangle = 2 (length + breadth)

Hence, perimeter of the playground = 2 (30 + 25)

= 2 (55)

= 110 m

Therefore, perimeter of the playground is 110 m.

**Solution 1(5)****:**

Length of the field = 45 m

Breadth of the field = 35 m

Perimeter of a rectangle = 2 (length + breadth)

Hence, perimeter of the field = 2 (45 + 35)

= 2 (80)

= 160 m

Therefore, perimeter of the field is 160 m.

**Solution 1(6):**

Length of the rectangular garden = 20 m

Breadth of the rectangular garden = 1700 cm

∴ Breadth of the rectangular garden = 1700 ÷ 100 = 17

∴ Breadth of the rectangular garden = 17 m

Perimeter of a rectangle = 2 (length + breadth)

Hence, perimeter of the field = 2 (20 + 17)

= 2 (37)

= 74 m

Therefore, perimeter of the rectangular garden is 74 m.

**Solution 2:**

(1) Perimeter of square = 4 × length = 4 × 6 = 24 m

(2) Perimeter of rectangle = 2( length + breadth)

Perimeter of rectangle = 2 ( 7 + 5) = 24 cm

(3) Perimeter of square = 4 × length = 4 × 8 = 32 cm

(4) Perimeter of rectangle = 2( length + breadth)

Perimeter of rectangle = 2 ( 8 + 7) =2 (15) = 30 cm

(5) Perimeter of square = 4 × length = 4 × 12 = 48 m

(6) Perimeter of rectangle = 2( length + breadth)

Perimeter of rectangle = 2 ( 16 + 14) = 2 (30) = 60 cm

(7) Perimeter of square = 4 × length = 4 × 15 = 60 cm

(8) Perimeter of rectangle = 2( length + breadth)

Perimeter of rectangle = 2 ( 14 + 13)= 2 (27) = 54 cm

(9) Perimeter of square = 4 × length = 4 × 17 = 68 m

(10) Perimeter of rectangle = 2( length + breadth)

Perimeter of rectangle = 2 (20 + 10)= 2 (30)= 60 m

**Practice – 3**

**Solution 1:**

To find the cost of drawing stripes of lime surrounding the garden, find the perimeter of the square garden. Length of the square garden = 35 m (given)

Perimeter of the square garden = 4 × length

Hence,

perimeter of the square garden = (4 × 35) m = 140 m

Cost of drawing 1 metre of stripes = Rs. 7

∴ Cost of drawing stripes of lime surrounding the square garden = Rs. (140 × 7) = Rs. 980

Therefore, the cost of drawing stripes of lime is Rs. 980.

**Solution 2:**

To find the cost of fixing coloured tiles on the boundary of the classroom, find the perimeter of the classroom.

Length of the square classroom = 15 m (given)

Perimeter of the square classroom = 4 × length

Hence,

perimeter of the square classroom = (4 × 15) m = 60 m

Cost of fixing coloured tiles for 1 m = Rs. 20

∴ Cost of fixing coloured tiles on the boundary of Rs. (60 × 20) =Rs. 1200

Therefore, the cost of fixing coloured tiles on the boundary is Rs. 1200.

**Solution 3:**

To find the cost of wire fencing surrounding the rectangular plot, find the perimeter of the plot.

Given,

Length of the rectangular plot = 40 m

Breadth of the rectangular plot = 30 m

Perimeter of the rectangle = 2(length + breadth)

Hence, the perimeter of the rectangular plot = 2(40 + 30) m

= 2 (70) m

= 140 m

Cost of wire fencing for 1 m = Rs. 20

∴ Cost of wire fencing surrounding the given plot

= Rs. (140 × 20)

= Rs. 2800

Therefore, the cost of wire fencing is Rs. 2800.

**Solution 4:**

To find the cost to cover the rectangular table with a stripe, find the perimeter of the table.

Given,

Length of the rectangular table = 4 m

Breadth of the rectangular table = 3 m

Perimeter of the rectangle = 2(length + breadth)

Hence, perimeter of the rectangular table = 2(4 + 3) m

= 2 (7) m

= 14 m

Cost to cover the table with a stripe for 1 m = Rs. 5

∴ Total Cost to cover the table with a stripe = Rs. (14 × 5)

= Rs. 70

Therefore, the cost to cover the table with a stripe is Rs. 70

**Solution 5:**

To find the cost of knitting a stripe surrounding the carpet, find the perimeter of the carpet.

Perimeter of the carpet = perimeter of the prayer hall.

Length of the prayer hall = 19 m (given)

Breadth of the prayer hall = 17 m (given)

Perimeter of a rectangle = 2 (length + breadth)

Hence,

perimeter of the rectangular prayer hall

= 2(19 + 17) m = 2 (36) m = 72 m

Cost of knitting 1 m stripe = Rs. 30

∴ Cost of knitting 72 m stripe = Rs. (72 × 30) = Rs. 2160

Therefore, the cost of knitting a stripe surrounding the carpet is Rs. 2160.

**Solution 6:**

To find the cost of knitting a stripe surrounding the cloth, find the perimeter of the cloth.

Given,

Length of the cloth = 8 m

Breadth of the cloth = 300 cm

But 100 cm = 1 m,

∴ 300 cm = 300÷ 100 = 3 m.

Perimeter of the rectangle = 2(length + breadth)

Hence, perimeter of the rectangular prayer hall

= 2(8 + 3) m = 2 (11) m = 22 m

Cost of knitting 1 m stripe = Rs. 30

∴ Cost of knitting 22 m stripe = Rs. (22 × 30) = Rs. 660

Therefore, the cost of knitting a stripe surrounding the cloth is Rs. 660.

**Practice – 4**

**Solution 1:**

Given,

Length of the square = 5 cm

Area of the square = length × length

∴ Area of the square = 5cm × 5 cm= 25 sq. cm

Therefore, the area of the square is 25 sq. cm

**Solution 2:**

Given,

Length of the square ground = 15

Area of the square = length × length

∴ Area of the ground = 15 m × 15 m = 225 sq. m

Therefore, the area of the square ground is 225 sq. m

**Solution 3:**

Given,

length of the square piece of cloth = 8 m

Area of the square = length × length

∴ Area of the square piece of cloth = 8 m × 8 m = 64 sq. m

Therefore, the area of the square piece of cloth is 64 sq. m

**Solution 4:**

Given,

Length of the square plot = 35 m

Area of the square = length × length

∴ Area of the square plot = 35 m × 35 m = 1225 sq. m

Therefore, the area of the square plot is 1225 sq. m

**Solution 5:**

Given,

Length of the square handkerchief = 20 cm

Area of the square = length × length

∴ Area of the square handkerchief = 20 cm × 20 cm

Area of the square handkerchief = 400 sq. cm

Therefore, the area of the square plot is 400 sq. cm

**Practice – 5**

**Solution 1(1):**

Given,

Length of the rectangle = 19 cm

Breadth of the rectangle = 17 cm

Area of the rectangle = length × breadth

∴ Area of the given rectangle = 19 cm × 17 cm= 323 sq. cm

Therefore, the area of the rectangle is 323 sq. cm

**Solution 1(2):**

Given,

Length of the top of the table = 90 cm

Breadth of the top of the table = 70 cm

Area of the rectangle = length × breadth

∴ Area of the top of the table = 90 cm × 70 cm

= 6300 sq. cm

Therefore, the area of the rectangle top of the table is 6300 sq. cm

**Solution 1(3)****:**

Length of the ground = 55 m (given)

Breadth of the ground = 50 m (given)

Area of the rectangle = length × breadth

∴ Area of the ground = 55 m × 50 m = 2750 sq. m

Therefore, the area of the rectangle ground = 2750 sq. m

**Solution 1(4):**

Given,

Length of the rectangular floor of the library = 10 m

Breadth of the floor of the library = 4 m

Area of the rectangle = length × breadth

∴ Area of the floor of the library = 10 m × 4 m = 40 sq. m

Therefore, the area of the rectangle floor of the library = 40 sq. m

**Solution 1(5)****:**

Given,

Length of the rectangular classroom = 12 m

Breadth of the classroom = 8 m

Area of a rectangle = length × breadth

∴ Area of the classroom = 12 m × 8 m = 96 sq. m

Therefore, the area of the classroom = 96 sq. m

**Solution 2:**

(1) Area of square = length × length

= 11 m × 11m

= 121 sq. m

(2) Area of rectangle = length × breadth

= 14 m × 12 m

= 168 sq. m

(3) Area of square = length × length

= 15 m × 15 m

= 225 sq. m

(4) Area of rectangle = length × breadth

= 26 cm × 24 cm

= 624 sq. cm

(5) Area of square = length × length

= 12 m × 12m

= 144 sq. m

(6) Area of rectangle = length × breadth

= 17 m × 16 m

= 272 sq. m

(7) Area of square = length × length

= 13 cm × 13 cm

= 169 sq. cm

(8) Area of rectangle = length × breadth

= 21 cm × 19 c m

= 399 sq. cm

**Practice – 6**

**Solution 1:**

1 m = 100 cm

1 sq.m = 10,000 sq. cm

∴ 5 sq.m = 50,000 sq. cm

(ii) 20 sq. m

1 m = 100 cm

1 sq. m = 10,000 sq. cm

∴ 20 sq. m = 2,00,000 sq. cm

(iii) 30 sq. m

1 m = 100 cm

1 sq. m = 10,000 sq. cm

∴ 30 sq.m = 3,00,000 sq. cm

**Solution 2:**

**Practice – 7**

**Solution 1:**

To find the cost of leveling the school ground, find its area.

Length of the school ground = 25 m (given)

Breadth of the school ground = 20 m (given)

Area of a rectangle = length × breadth

∴ Area of the school ground = 25 m × 20 m = 500 sq m

Cost of leveling 1 sq. m area = Rs. 9

Cost of leveling 500 sq. m area = Rs. (500 × 9) = Rs. 4500

Therefore, the cost of leveling the school ground is Rs. 4500.

**Solution 2:**

To find the cost of dining cloth on the table, find the area of the table.

Length of the square dining table = 4 m (given)

Area of a square = length × length

∴ Area of the given dining table = 4 m × 4 m= 16 sq. m

Cost of preparing 1 sq. m of dining cloth = Rs. 30 per sq. m

Cost of preparing 16 sq. m of dining cloth = Rs. (16 × 30)

= Rs. 480

Therefore, the cost of preparing the dining cloth on the dining table is Rs. 480.

**Solution 3:**

To find the cost of fixing tiles on the floor, find the area of the swimming pool.

Length of the swimming pool = 16 m

Breadth of the swimming pool = 4 m

Area of a rectangle = length × breadth

∴ Area of the swimming pool = 16 m × 4 m = 64 sq. m

Cost of fixing tiles on the floor for 1 sq. m area = Rs. 22

Cost of fixing tiles for 64 sq. m area = Rs. (64 × 22)

= Rs. 1408

Therefore, the cost of fixing tiles on the floor of the swimming pool is Rs. 1408.

**Solution 4:**

To find the number of boxes which can be formed in the card paper, first find the area of the card paper.

Length of the card paper = 60 cm (given)

Breadth of the card paper = 40 cm (given)

Area of a rectangle = length × breadth

∴ Area of the card paper = 60 cm × 40 cm

= 2400 sq. cm

Now, find the area of one square box.

Area of a square = length × length

Length of the square box = 5 cm

∴ Area of one square box = 5 cm × 5 cm = 25 sq. cm

One square box covers 25 sq. cm of the card paper.

∴ Number of boxes to cover 2400 sq. cm of area

= 2400 ÷ 25

∴ Number of boxes formed = 96

Therefore, 96 boxes can be formed in the card paper.

**Solution 5:**

First, we need to find the area of the floor of the prayer hall.

Area of a rectangle = length × breadth

∴ Area of the rectangular prayer hall = 14 m × 11m

∴ Area of the rectangular prayer hall = 154 sq. m

Now, let us find the area of one square tile.

Length of the square tile = 50 cm.

But the units of length of the hall are given in metre.

Hence, let us convert the area of the prayer hall in sq. cm

1 sq. m = 10,000 sq. cm

∴ 154 sq. m = 154 × 10,000 sq. cm

∴ Area of the prayer hall = 15,40,000 sq. cm

Area of the square tile = length × length

Area of the square tile = 50 × 50 = 2500 sq. cm

One square tile covers 2500 sq. cm of the prayer hall floor.

∴ Number of tiles needed to cover 15,40,000 sq. cm of area = 15,40,000 ÷ 25

Therefore, 616 tiles are needed.