solution of NCERT Exercise 11.4

Perimeter and Area 7th Math

solution of NCERT 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.

Solution:

class 7th math Perimeter and area solution of ncert exercise 11.4 Q1

Let ABCD is the given rectangular garden.

In which

Length, AB = DC = 90 m

And, width, BC = AD = 75 m

A 5 m path is built outside and around the given garden.

Thus, width of the path = 5 m

Now, Length of the park along the path = 90 m + 5m +5m = 100 m

And, width of the park along the path = 75 m + 5m +5m = 85 m

Thus, area of path in hectare = ?

We know that, Area of rectangle = Length × Breadth

Thus, area of given rectangular park = 90 m × 75 m

⇒ Area of garden = 6750 m2

Now, Area of garden along with path = Length × Breadth

= 100 m × 85 m

⇒ Area of garden with path = 8500 m

Now, Area of path = Area of garden with path – Area of garden

= 8500 m – 6750 m

Thus, Area of path = 1750 m2

Area of path in Hectare

∵ We know that, 10000 m2 = 1 hectare

∴ 1 m2 = `1/10000` hectare

∴ 1750 m2 = `1/10000xx1750` hectare = 0.175 hectare

Thus, Area of path = 0.175 hectare Answer

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.

Solution :

class 7th math Perimeter and area solution of ncert exercise 11.4 Q2

Let ABCD is the given rectangular park.

In which

Length, AB = DC = 125 m

And, width, BC = AD = 65 m

A 3 m path is built outside and around the given park.

Thus, width of the path = 3 m

Now, Length of the park along the path = 125 m + 3m + 3m = 131 m

And, width of the park along the path = 65 m + 3m + 3m = 71 m

Thus, area of path = ?

We know that, Area of rectangle = Length × Breadth

Thus, area of given rectangular park = 125 m × 65 m

⇒ Area of park = 8125 m2

Now, Area of park along with path = Length × Breadth

= 131 m × 71 m

⇒ Area of park with path = 9301 m

Now, Area of path = Area of park with path – Area of park

= 9301 m – 8125 m

Thus, Area of path = 1176 m2 Answer

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.

Solution :

class 7th math Perimeter and area solution of ncert exercise 11.4 Q3

Let EFGH is the cardboard, in which

Given,

Length (HG) = 8 cm

And width (EH) = 5 cm

And margin on cardboard = 1.5 cm

Thus, Total area of margin = ?

We know that,

Area of rectangle = Length × Width

Thus, Area of given cardboard = 8 cm × 5 cm

= 40 cm2

Now, Length picture painted on the cardboard, means length of cardboard without margin

= 8 cm – 1.5 cm – 1.5 cm

= 5 cm

And, width of the cardboard without margin

= 5 cm – 1.5 cm – 1.5 cm

= 5 cm – 3 cm

= 2 cm

Thus, Area of cardboard without margin = Length × Width

= 5 cm × 2 cm

= 10 cm

Thus area of cardboard without margin = 10 cm

Now, Area of margin = Area of cardboard – Area of cardboard without margin

= 40 cm – 10 cm

= 30 cm2

Thus, Area of margin = 30 cm2 Answer

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.

Solution:

Given, Length of the room = 5.5 m

Breadth of the room = 4 m

Width of verandah along outside of the room = 2.25 m

Rate of cementing the verandah = ₹ 200 per m2

Thus, Area of verandah and cost of cementing of verandah = ?

class 7th math Perimeter and area solution of ncert exercise 11.4 Q4

(i) Calculation of the area of the verandah

We know that, Area of rectangle = Length × Breadth

Thus, area of given room = 5.5 m × 4 m

= 22 m2

Now, length of the room including verandah

= 5.5 m + 2.25 m = 7.75 m

And, width of the room including verandah

= 4 m + 2.25 m = 6.25 m

Now, Area of room including verandah = length × width

= 7.75 m × 6.25 m

= 48.437 m2

Now, Area of verandah = Area of room including verandah – Area of room

= 48.437 m2 – 22 m2

= 26.437 m2

Thus, Area of verandah = 26.437 m2

(ii) Calculation of cost of cementing of verandah

∵ Cost of 1 m2 = ₹ 200

∴ cost of 26.437 m2 = ₹ 200 × 26.437 m2

= ₹ 5287.40

Thus, cost of cementing of verandah = ₹ 5287.40

Thus, Area of verandah =26.437 m2. And cost of cementing of verandah = ₹ 5287.40 Answer

Question (5) A path 1 m wide is build 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:

class 7th math Perimeter and area solution of ncert exercise 11.4 Q5

Given, side of square garden = 30 m

And width of path inside the garden = 1 m

Thus, side of garden without path = 28 m

Rate of plantation of garden = ₹ 40 per m2

Thus,

Area of path and cost of plantation of grass in the remaining portion of garden =?

(i) Area of path

We know that, Area of a square = (side) 2

Thus, area of given garden = (30 m )2

= 900 m2

Now, Area of garden without path = (28 m)2

= 784 m2

Thus, Area of path = Area of garden – Area of garden without path

= 900 m2 – 784 m2

= 116 m2

Thus, Area of path = 116 m2

(ii) Calculation of cost of plantation in the remaining garden

∵ cost of plantation for 1 square meter = ₹ 40

∴ cost of plantation for 784 m2 = ₹ 40 × 784 m2

= ₹ 31360.00

Thus, Area of path = 116 m2 And cost of plantation of remaining of garden = ₹ 31360.00 Answer

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.

Solution :

class 7th math Perimeter and area solution of ncert exercise 11.4 Q6

Let ABCD is the rectangular park.

And road PS and road EF crosses each other at right angle.

In this according to question

Length of AB = 700 m

Length of BC = 300 m

And width of the road = 10 m

Since, roads crosses each other at right angle, thus they form a square KLMN in the middle

Since width of the road is equal to 10 m,

Thus side of the square = 10 m

Thus, Area of roads and area of park excluding roads = ?

We know that,

Area of rectangle = Length × Breadth

Thus, Area of given park = 700 m × 300 m

= 210000 m2

Calculation of area of road

Length of road parallel to length of the park = 700 m

And width of the road = 10 m

Thus, Area of road (EFGH) parallel the length of the park = 700 m × 10 m

= 7000 m2

Length of road parallel to width of the park = 300 m

Width of the road = 10 m

Thus, area of road (PQRS) parallel to width of the park = 300 m × 10 m

= 3000 m2

Calculation of area of square crosses in the middle of the road

Since, width of the road = 10 m,

Thus, side of the square (KLMN) = 10 m

We know that, area of square = side × side

Thus, area of square (KLMN) = 10 m × 10 m

= 100 m2

[Since two roads crosses each other, thus while finding the area of road, area of overlapping square is to be subtracted from total area of road.]

Now, total area of road

= area of road parallel to length (EFGH) + area of road parallel to width (PQRS) – Area of square (KLMN)

= 7000 m2 + 3000 m2 – 100 m2

= 100000 m2 – 100 m2

= 9900 m2

Thus, Area of roads = 9900 m2

Now, since, 10000 m2 = 1 hectare

Thus, 9900 m2 = `9900/10000` hectare

= 0.99 hectare

Thus, area of roads = 0.99 hectare

Now, Area of park excluding roads

= Area of park – Area of roads

= 210000 m2 – 9900 m2

= 200100 m2

Thus, area of park excluding roads = 200100 m2

Now, since 1 hectare = 10000 m2

∴ 200100m2 = 200100 × `1/10000` hectare

= 20.01 hectare

Thus, area of park excluding roads = 20.01 hectare and area of roads = 0.99 hectare Answer

Alternate Method

class 7th math Perimeter and area solution of ncert exercise 11.4 Q6-alternate method

Let assume that roads are running just along the length and width.

Therefore, Length of the park without road

= 700 m –10 m = 690 m

And, breadth of the park without road

= 300 m – 10 m = 290 m

Now, area of the park = Length × Breadth

= 700 m × 300 m

= 210000 m2

And Area of park without road = Length × Breadth

= 690 m × 290 m

= 200100 m2

Now, since 1 hectare = 10000 m2

∴ 200100m2 = 200100 × `1/10000` hectare

= 20.01 hectare

Now, area of road = Area of park – Area of park without road

= 210000 – 200100 m2

= 900 m2

Thus, area of road = 900m2

Now, since, 10000 m2 = 1 hectare

Thus, 9900 m2 = `9900/10000` hectare

= 0.99 hectare

Thus, area of roads = 0.99 hectare

Thus, area of park excluding roads = 20.01 hectare and area of roads = 0.99 hectare Answer

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