## Chapter 11 Perimeter and Area Exercise 11.4

**Question 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.Answer:**

Length of the garden = 90 m

Breadth of the garden = 75 m

Area of the garden = length × breadth

= 90 × 75

= 6750 m²

New length of the garden when path is included = 100 m

New breadth of the garden when path is included = 85 m

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

1 hectare = 10000 m²

Area of garden in hectare = 6750/10000

= 0.675 hectare

Therefore, the area of the path is 1750 m² and in hectares the area of garden is 0.675 hectare.

**Question 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.**

Answer:

Answer:

Length of the park = 125 m

Breadth of the park = 65 m

Area of the park = length × breadth

= 125 × 65

= 8125 m²

New length of the garden when path is included = 131 m

New breadth of the garden when path is included = 71 m

New area of the park = 131 × 71

= 9301 m²

Area of path = New area of the park including path - Area of park

= 9301 - 8125

= 1176 m²

Therefore, the area of the path is 1176 m².

**Question 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.**

Answer:

Length of the cardboard = 8 cm

Answer:

Breadth of the cardboard = 5 cm

Area of the cardboard = length × breadth

= 8 × 5

= 40 cm²

New length of the garden when path is included = 5 cm

New length of the garden when path is included = 2 cm

New area of the cardboard = length × breadth

= 5 × 2

= 10 cm²

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

= 40 - 10

= 30 cm²

**Question 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.

Answer:

(i) The area of the verandah.

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

Answer:

(i)

Length of the room = 5.5 m

Breadth of the room = 4 m

Area of the room = length × breadth

= 5.5 × 4

= 22 m²

New length of room when verandah is included = 10 m

New breadth of the room when verandah is included = 8.5 m

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)

The cost of cementing the floor of the verandah per m² = ₹ 200

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

= ₹ 12600**Question 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.Answer:**

(i)

Side of square garden = 30 m

Area of the square garden = s x s

= 30 x 30

= 900 m²

Length of new side when path is not included = 28 m

New area of the room when verandah is included = 28 x 28

= 784 m²

The area of path = Area of the square garden when path is included - Area of the square garden when path is not included

= 900 - 784

= 116 m²

(ii)

The cost of planting the grass in the remaining portion of the garden per m² = ₹ 40

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

= ₹ 31360**Question 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.Answer:**

Length of the park = 700 m

Breadth of the park = 300 m

Area of the park = length × breadth

= 700 × 300

= 210000 m²

Let ABCD be the one cross road and EFGH be 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

Area of the ABCD cross road = length × breadth

= 700 × 10

= 7000 m²

Area of the EFGH cross road = length × breadth

= 300 × 10

= 3000 m²

Area of the IJKL at centre = 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

1 hectare = 10000 m²

Therefore, area of roads in hectare = 9900/10000

= 0.99 hectare

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

= 210000 - 9900

= 200100 m²

= 200100/10000

= 20.01 hectare**Question 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.Answer:**

(i)

Length of the field = 90 m

Breadth of the field = 60 m

Area of the field = length × breadth

= 90 × 60

= 5400 m²

Let ABCD be the one cross road and EFGH be the 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

Area of the ABCD cross road = length × breadth

= 90 × 3

= 270 m²

Area of the EFGH cross road = length × breadth

= 60 × 3

= 180 m²

Area of the IJKL at centre = 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)

The cost of constructing the roads per m² = ₹ 110.

The cost of constructing the 441 m2 roads = 441 × 110

= ₹ 48510**Question 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)**

**Answer:**

Radius of a circular pipe = 4 cm

Side of a square = 4 cm

Perimeter of the circular pipe = 2πr

= 2 × 3.14 × 4

= 25.12 cm

Perimeter of the square = 4 × side of the square

= 4 × 4

= 16 cm

The length of cord left with Pragya = Perimeter of circular pipe - Perimeter of square

= 25.12 - 16

= 9.12 cm

Yes she had 9.12 cm cord left.**Question 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.**

**Answer:**

(i)

Length of rectangular lawn = 10 m

Breadth of rectangular lawn = 5 m

Area of the rectangular lawn = Length × Breadth

= 10 × 5

= 50 m²

(ii)

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**Question 10: In the following figures, find the area of the shaded portions:(i)**

**Answer:**

Area of ΔAEF = 1/2 × Base × Height

= 1/2 × 6 × 10

= 1 × 3 × 10

= 30 cm²

Area of ΔEBC = 1/2 × Base × Height

= 1/2 × 8 × 10

= 1 × 4 × 10

= 40 cm²

Area of rectangle ABCD = length × breadth

= 18 × 10

= 180 cm²

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

= 180 - (30 + 40)

= 180 - 70

= 110 cm²**(ii)**

**Answer:**

Area of ΔSTU = 1/2 × Base × Height

= 1/2 × 10 × 10

= 1 × 5 × 10

= 50 cm²

Area of ΔTPQ = 1/2 × Base × Height

= 1/2 × 10 × 20

= 1 × 5 × 20

= 100 cm²

Area of ΔQRU = 1/2 × Base × Height

= 1/2 × 10 × 20

= 1 × 5 × 20

= 100 cm²

Area of square PQRS = s x s

= 20 × 20

= 400 cm²

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

= 400 - (50 + 100 + 100)

= 400 - 250

= 150 cm²**Question 11: Find the area of the quadrilateral ABCD.Here, AC = 22 cm, BM = 3 cm,DN = 3 cm, and BM ⊥ AC, DN ⊥ AC**

**Answer:**

AC = 22 cm

BM = 3 cm

DN = 3 cm

BM ⊥ AC

DN ⊥ AC

= 1/2 × 22 × 3

= 1 × 11 × 3

= 33 cm²

Area of ΔADC = 1/2 × Base × Height

= 1/2 × 22 × 3

= 1 × 11 × 3

= 33 cm²

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

= 33 + 33

= 66 cm²

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