NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 with Answers

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 Perimeter and Area Class 7 Ex 11.1

Perimeter and Area Class 7 Ex 11.2
     Perimeter and Area Class 7 Ex 11.3
    Perimeter and Area Class 7 MCQ

 NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 with Answers

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 hectares.

Answer 1: 

Length of the rectangular garden = 90 metre and breadth of the rectangular garden = 75 metre

Area of the rectangular garden =( length × breadth) 

=( 90 × 75) 

= 6750 m²

Now outer length of the rectangular garden with path = (90 + 5 + 5) = 100 metre

And outer breadth of the rectangular garden with path =( 75 + 5 + 5) = 85 metre

Therefore, area of the garden with path = (100 × 85) m²

= 8500 m²

Therefore, Area of path = Area of garden with path – Area of garden without path

= 8,500 – 6,750

= 1,750 m² 

Since, 1 m² = (1/10000) hectares

Therefore, 6,750 m² =( 6750/10000) = 0.675 hectares

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 2: 

Given, the length of the rectangular park = 125 metre.  

And, the breadth of the rectangular park = 65 metre. 

Therefore, area of the rectangular park = (125 × 65) m²

= 8125 m²

Now, since width of the path = 3 metre

Therefore, length of rectangular park with path = (125 + 3 + 3 )= 131 metre

And breadth of rectangular park with path = (65 + 3 + 3 ) = 71 metre

Therefore, area of the rectangular park with path = ( 131 × 71) m²

= 9301 m²

Area of only path = Area of park with path – Area of park without path

= (9301 – 8125) 

= 1,176 m²

Thus, the area of path around the park is 1,176 m²

Question 3: 

A picture is painted on 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 3: 

Given,Length of the painted cardboard = 8 cm and breadth of the painted card = 5 cm

Area of the painted cardboard = (8 × 5) cm²

= 40cm²

Now the width of the margin is 1.5 cm long from each of its side.

Therefore reduced length of the painted cardboard = 8 – (1.5 + 1.5) = 8 – 3 = 5 cm

And reduced breadth of the painted cardboard = 5 – (1.5 + 1.5) = 5 – 3 = 2 cm

Therefore, area of the reduce painted cardboard = (5×2) cm

=10 cm²

Area of margin = Area of painted cardboard  – Area of reduced painted cardboard 

= (40 – 10) 

= 30 cm²

Thus, the total area of margin is 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 4: 

(i) The length of room = 5.5 metre and width of the room = 4 metre

Therefore area of the room = length × breadth

= (5.5 × 4) m²

= 22m²

The length of room with verandah = 5.5 + 2.25 + 2.25 = 10 metre

The width of room with verandah = 4 + 2.25 + 2.25 = 8.5 metre

Area of the room with verandah = ( 10 × 8.5) m²

= 85 m²

Therefore, area of verandah 

 = Area of room with verandah – Area of room without verandah

 = 85 – 22

 = 63 m²

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

The cost of cementing 63 m² the floor of verandah = 200 x 63 = ₹12,600

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 with Answers

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 5: 

(i) Side of the square garden with path = 30 metre

Therefore area of the square garden with path = (30 × 30) m²

= 900 m²

Now width of the path along the border inside the garden = 1 metre

Side of square garden without path = 30 – (1 + 1) = 30 – 2 = 28 metre

Therefore, area of the garden without path = (28 × 28) m²

= 784 m²

Now Area of path = Area of the garden with path – Area of the garden without path

= 900 – 784

= 116 m²

(ii) the cost of planting grass in the remaining portion of the garden at the rate 

of ₹ 40 per m²

The cost of planting grass in 784 m² of the garden = ₹40 x 784 = ₹ 31,360

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 6: 

Here, AD = BC = 10 m and AB = DC = 300 m, EF = GH = 10 m and EH = FG = 700 m

And KL = 10 m and KN = 10 m

Area of roads = Area of rectangle ABCD + Area of rectangle EFGH – Area of square  KLMN

[ KLMN is taken twice, which is to be subtracted]

= (AB × BC) + (EF x EH) – (KL x KN)  

= (300 x 10) + (10 × 700) – (10 x 10)

= 3000 + 7000 – 100

= 9,900 m²

Area of road in hectares, 1 m² = (1/10000) 

hectares

9,900 m2 = (9900/10000) 

= 0.99 hectares

Now, Area of park excluding cross roads 

= Area of park – Area of cross roads

= (PQ × QR) – 9,900

= (300 × 700) – 9,900

= 2,10,000 – 9,900

= 2,00,100 m²

= (200100/10000) hectares 

= 20.01 hectares

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 7: 

(i) Here, AB = DC = 60 m and AD = BC = 3 m, EH = FG = 90m and EF = GH = 3 m and KL = 3 m and KN = 3 m

Area of roads = Area of ABCD + Area of EFGH – Area of KLMN

[ Science, KLMN is taken twice, which is to be subtracted]

= (AB × BC) + ( EF x FG) – (KL x KN) 

= (60 x 3) + (3 × 90) – (3 x 3)

= 180 + 270 – 9

= 441 m²

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

The total cost of 441 m² constructing the roads = ₹(110 x 441) = ₹48,510

Therefore, total cost of constructing the roads = ₹48,510

NCERT class 7 Maths

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? (Take π = 3.14)   

 

Answer 8: 

Radius of pipe(r) = 4 cm

Since the cord wrapped around the pipe,

Length of the cord required = circumference of the pipe. 

Therefore, length of the cord required = 2πr

= (2 × 3.14 × 4) 

 = 25.12 cm

Now the cord wrapped around the square having side = 4 cm

Therefore, length of the cord to wrapped around the square = ( 4 × side) 

= (4 × 4) 

 = 16 cm

Remaining cord = Cord wrapped on pipe – Cord wrapped on square 

= 25.12 – 16

= 9.12 cm

Thus, she has left 9.12 cm cord.

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 9: 

(i) Length of rectangular lawn = 10 metre, 

breadth of the rectangular lawn = 5 metre

the area of the whole land = length × breadth

= 10 × 5

50 m²

(ii) The radius of the circular flower bed = 2 metre

the area of the flower bed = πr²

= (3.14 × 2 × 2) 

= 12.56 m²

(iii) the area of the lawn excluding the area of the flower bed = area of lawn with flower bed– area of flower bed

= 50 – 12.56

= 37.44 m²

(iv) The circumference of the flower bed = 2πr

= 2 x 3.14 x 2 = 12.56 m

Question 10: 

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

(i) 

(ii) 

 

Answer 10: 

(i) Here, AB = 18 cm, BC = 10 cm, AF = 6 cm, AE = 10 cm and BE = 8 cm

Area of shaded portion 

= Area of rectangle ABCD – (Area of triangle FAE + area of triangle EBC)

= (AB x BC) – { (½ x AE x AF) + (½x BE x BC)}

= (18 x 10) – {(½ x 10 x 6) + (½ x 8 x 10)}

= 180 – (30 + 40)

= 180 – 70

= 110 cm²

(ii) Here, SR = SU + UR = 10 + 10 = 20 cm, QR = 20 cm

PQ = SR = 20 cm, PT = PS – TS = 20 – 10 cm

TS = 10 cm, SU = 10 cm, QR = 20 cm and UR = 10 cm

Area of shaded region 

= Area of square PQRS – Area of triangle QPT – Area of triangle TSU – Area of triangle UQR

= (SR x QR) - {( ½ x PQ x PT) - (½ x ST x SU) – ( ½ × QR × UR) }

= 20 x 20 – {( ½ x 20 x 10) –( ½x 10 x 10) –( ½x 20 x 10) }

= 400 – (100 + 50 + 100) 

= 400 - 250

= 150 cm²

 

 NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 with Answers

Question 11: 

Find the area of the equilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm and BM perpendicularAC, DN perpendicularAC.

 

Answer 11: 

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

Area of quadrilateral ABCDF = Area of triangle ABC + Area of triangle ADC

= ( ½ x AC x BM) + (½ x AC x DN) 

= ( ½ x 22 x 3) + (½ x 22 x 3) 

= 33 + 33

= 66 cm²

Thus, the area of quadrilateral ABCD is 66 cm².