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# Mathematics – Class 7 – Chapter 11 – Perimeter and Area – Exercise 11.4 – NCERT Exercise Solution

1. A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.

Solution:

Given,

Length of the garden (L) = 90 m

Breadth of the garden (B) = 75 m

Now,

Area of the garden = length × breadth

= 90 × 75

= 6750 m2

According to the figure,

After the path is included the new length and breadth of the garden is 100 m and 85 m respectively.

The new area of the garden = 100 × 85

= 8500 m2

The area of path = New area of the garden including path – Area of garden

= 8500 – 6750

= 1750 m2

Now,

For 1 hectare = 10000 m2

Hence, the area of the garden in hectare = 6750/10000

= 0.675 hectare

2. A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.

Solution:

Given,

Length of the park (L) = 125 m

Breadth of the park (B) = 65 m

Now,

Area of the park = length × breadth

= 125 × 65

= 8125 m2

According to the figure,

After the path is included the new length and breadth of the park is 131 m and 71 m respectively.

The new area of the park = 131 × 71

= 9301 m2

The area of path = New area of the park including path – Area of the park

= 9301 – 8125

= 1176 m2

3. A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.

Solution:

Given,

Length of the cardboard (L) = 8 cm

Breadth of the cardboard (B) = 5 cm

Now,

Area of the cardboard = length × breadth

= 8 × 5

= 40 cm2

According to the figure,

When margin is not included then the new length and breadth of the cardboard is 5 cm and 2 cm respectively.

The new area of the cardboard = 5 × 2

= 10 cm2

The area of margin = Area of the cardboard when the margin is including     – Area of the cardboard when the margin is not including

= 40 – 10

= 30 cm2

4. A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find:

(i) the area of the verandah.

(ii) the cost of cementing the floor of the verandah at the rate of ₹ 200 per m2.

Solution:

(i)

Given,

Length of the room (L) = 5.5 m

Breadth of the room (B) = 4 m

Now,

Area of the room = length × breadth

= 5.5 × 4

= 22 m2

According to the figure,

After verdandah is included the new length and breadth of the room is 10 m and 8.5 m respectively.

The new area of the room when verandah is included = 10 × 8.5

= 85 m2

The area of verandah = Area of the room when verandah is included – Area of the room

= 85 – 22

= 63 m2

(ii)

Given,

The cost of cementing the floor of the verandah at the rate of ₹ 200 per m2

Then the cost of cementing the 63 m2 area of the floor of the verandah = 200 × 63

= ₹ 12600

5. A path 1 m wide is built along the border and inside a square garden of side 30 m. Find:

(i) the area of the path

(ii) the cost of planting grass in the remaining portion of the garden at the rate of ₹ 40 per m2.

Solution:

(i)

Given,

Side of square garden (s) = 30 m

Now,

Area of the square garden = Side 2

= 302

= 900 m2

According to the figure,

After including the path the new side of the square garden is 28 m.

The new area of the room when verandah is included = 282

= 784 m2

The area of path = Area of the square garden when the path is included – Area of the square

when the path is not included

= 900 – 784

= 116 m2

(ii)

Given,

The cost of planting the grass in the remaining portion of the garden at the rate of

= ₹ 40 per m2

Then,

The cost of planting the grass in a 784 m2 area of the garden = 784 × 40

= ₹ 31360

6. Two cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and breadth 300 m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.

Solution:

Given,

Length of the park (L) = 700 m

Breadth of the park (B) = 300 m

Now,

Area of the park = length × breadth

= 700 × 300

= 210000 m2

Let the ABCD is the one cross road and EFGH is another cross road in the park.

The length of ABCD cross road = 700 m

The length of EFGH cross road = 300 m

Both cross road have the same width = 10 m

Now,

= 700 × 10

= 7000 m2

= 300 × 10

= 3000 m2

Area of the IJKL at center = length × breadth

= 10 × 10

= 100 m2

Area of the roads = Area of ABCD + Area of EFGH – Area of IJKL

= 7000 + 3000 – 100

= 10000 – 100

= 9900 m2

We know that, for 1 hectare = 10000 m2

So, area of roads in hectare = 9900/10000

= 0.99 hectare

Hence, Area of the park excluding roads = Area of park – Area of the roads

= 210000 – 9900

= 200100 m2

= 200100/10000 = 20.01 hectare

7. Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find

(i) the area covered by the roads.

(ii) the cost of constructing the roads at the rate of ₹ 110 per m2.

Solution:

(i)

Given,

Length of the field (L) = 90 m

Breadth of the field (B) = 60 m

Now,

Area of the field = length × breadth

= 90 × 60

= 5400 m2

Let the ABCD is the one cross road and EFGH is another cross road in the park.

The length of ABCD cross road = 90 m

The length of EFGH cross road = 60 m

Both cross road have the same width = 3 m

Now,

= 90 × 3

= 270 m2

= 60 × 3

= 180 m2

Area of the IJKL at center = length × breadth

= 3 × 3

= 9 m2

Area of the roads = Area of ABCD + Area of EFGH – Area of IJKL

= 270 + 180 – 9

= 450 – 9

= 441 m2

(ii)

Given,

The cost of constructing the roads at the rate of ₹ 110 per m2.

Then, the cost of constructing the 441 m2 roads = 441 × 110

= ₹ 48510

8. Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? (π = 3.14)

Solution:

Given,

The radius of a circular pipe = 4 cm

Side of a square = 4 cm

Now,

The perimeter of the circular pipe = 2πr

= 2 × 3.14 × 4

= 25.12 cm

The perimeter of the square = 4 × side of the square

= 4 × 4

= 16 cm

Then, the length of cord left with Pragya = Perimeter of circular pipe – Perimeter of a square

= 25.12 – 16

= 9.12 cm Yes, 9.12 cm cord is left.

9. The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find:

(i)the area of the whole land

(ii)the area of the flower bed

(iii)the area of the lawn excluding the area of the flower bed

(iv) the circumference of the flower bed.

Solution:

(i)

According to the figure,

Length of rectangular lawn = 10 m

Breadth of rectangular lawn = 5 m

Now,

Area of the rectangular lawn = Length × Breadth

= 10 × 5

= 50 m2

(ii)

According to the figure,

Radius of the flower bed = 2 m

Area of the flower bed = πr2

= 3.14 × 22

= 3.14 × 4

= 12.56 m2

(iii)

The area of the lawn excluding the area of the flower bed = Area of rectangular lawn – Area of flower bed

= 50 – 12.56

= 37.44 m2

(iv)

The circumference of the flower bed = 2πr

= 2 × 3.14 × 2

= 12.56 m

10. In the following figures, find the area of the shaded portions:

(i)

Solution:

To find the area of EFDC,

First, we have to find the area of ΔAEF, ΔEBC and rectangle ABCD

Area of ΔAEF = ½ × Base × Height

= ½ × 6 × 10

= 1 × 3 × 10

= 30 cm2

Area of ΔEBC = ½ × Base × Height

= ½ × 8 × 10

= 1 × 4 × 10

= 40 cm2

Area of rectangle ABCD = length × breadth

= 18 × 10

= 180 cm2

Now,

Area of EFDC = ABCD area – (ΔAEF + ΔEBC)

= 180 – (30 + 40)

= 180 – 70

= 110 cm2

(ii)

Solution:

To find the area of ΔQTU,

First, we have to find the area of ΔSTU, ΔTPQ, ΔQRU and square PQRS

Area of ΔSTU = ½ × Base × Height

= ½ × 10 × 10

= 1 × 5 × 10

= 50 cm2

Area of ΔTPQ = ½ × Base × Height

= ½ × 10 × 20

= 1 × 5 × 20

= 100 cm2

Area of ΔQRU = ½ × Base × Height

= ½ × 10 × 20

= 1 × 5 × 20

= 100 cm2

Area of square PQRS = Side2

= 20 × 20

= 400 cm2

Now,

Area of ΔQTU = PQRS area – (ΔSTU + ΔTPQ + ΔQRU)

= 400 – (50 + 100 + 100)

= 400 – 250

= 150 cm2

11. Find the area of the quadrilateral ABCD.

Here, AC = 22 cm, BM = 3 cm,

DN = 3 cm, and BM ⊥ AC, DN ⊥ AC

Solution:

Given,

AC = 22 cm, BM = 3 cm DN = 3 cm and BM ⊥ AC, DN ⊥ AC

To find the area of quadrilateral ABCD,

First, we have to find the area of ΔABC, and ΔADC

Area of ΔABC = ½ × Base × Height

= ½ × 22 × 3

= 1 × 11 × 3

= 33 cm2

Area of ΔADC = ½ × Base × Height

= ½ × 22 × 3

= 1 × 11 × 3

= 33 cm2

Now,

Area of quadrilateral ABCD = Area of ΔABC + Area of ΔADC

= 33 + 33

= 66 cm2

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