NCERT Solutions for Class 7 Maths Chapter 11: Perimeter and Area - Exercise 11.4

Exercise 11.4

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

Ans: Let us consider an imaginary ground which is $90m$ long and $75m$ broad.

(Image will be Uploaded Soon)

Length of rectangular garden = $90m$ and breadth of rectangular garden = $75m$

Outer length of rectangular garden with path = \[90 + 5 + 5 = 100m\]

Outer breadth of rectangular garden with path = \[75 + 5 + 5 = 85m\]

Outer area of rectangular garden with path = length \[ \times \] breadth = \[100 \times 85 = 8,500{m^2}\]

Inner area of garden without path = length \[ \times \] breadth = \[90 \times 75 = 6,750{m^2}\]

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

= \[8,500-6,750\]

\[ = 1,750{m^2}\]

Since, \[1{m^2}{\text{ = }}10000\]hectares 

Therefore, 

Area of the garden will be,

\[6,750{m^2} = \dfrac{{6750}}{{10000}} = 0.675\]hectares

2. A $3$ m wide path runs outside and around a rectangular park of length \[{\mathbf{125}}\] m and breadth \[{\mathbf{65}}\] m. Find the area of the path. 

Ans: Let us look at the diagram.

(Image will be Uploaded Soon)

Length of rectangular park = $125m$,

breadth of rectangular park = $65$ m and width of the path = $3m$

Length of rectangular park with path = \[125 + 3 + 3 = 131m\]

Breadth of rectangular park with path = \[65 + 3 + 3 = 71m\]

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

\[ = \left( {131 \times 71} \right)-\left( {125 \times 65} \right)\]

\[ = 9301-8125\]

\[ = 1,176{m^2}\]

Thus, the area of path around the park is \[ = 1,176{m^2}\].

3. A picture is painted on a cardboard\[\;{\mathbf{8}}\]cm long and \[{\mathbf{5}}\] cm wide such that there is a margin of \[{\mathbf{1}}.{\mathbf{5}}\]cm along each of its sides. Find the total area of the margin. 

Ans: Let us consider an imaginary park having length $8$ m and breadth$5$ m.

(Image will be Uploaded Soon)

Length of painted cardboard = $8$ cm and breadth of painted card = $5$ cm 

Since, there is a margin of $1.5$ cm long from each of its sides. 

Therefore, reduced length \[8-\left( {1.5 + 1.5} \right) = 8-3 = 5\]cm

And reduced breadth = \[5-\left( {1.5 + 1.5} \right) = 5-3 = 2\]cm

∴ Area of margin = Area of cardboard (ABCD) – Area of cardboard (EFGH)

\[ = \left( {AB \times AD} \right){\text{ - }}\left( {EF \times EH} \right)\]

\[ = \left( {8 \times 5} \right)-\left( {5 \times {\text{2}}} \right)\]

\[ = {\text{4}}0-10\]

\[ = {\text{3}}0c{m^2}\]

Thus, the total area of margin is \[{\text{3}}0c{m^2}\].

4. A verandah of width \[{\mathbf{2}}.{\mathbf{25}}\]m is constructed all along outside a room which is \[{\mathbf{5}}.{\mathbf{5}}\]m long and \[{\mathbf{4}}\] m wide. Find: 

i. The area of the verandah. 

Ans:Let us consider an imaginary park having length $5.5$ m and breadth$4$ m.

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The length of room = $5.5$ m and width of the room = $4$ m

The length of room with verandah\[ = 5.5 + 2.25 + 2.25 = 10\]m 

The width of room with verandah\[ = 4 + 2.25 + 2.25 = 8.5\]m

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

\[ = \left( {AB \times AD} \right)-\left( {EF \times EH} \right)\]

\[ = \left( {10 \times 8.5} \right)-\left( {5.5 \times 4} \right)\]

\[ = 85-22\]

\[ = 63{m^2}\]

ii. The cost of cementing the floor of the verandah at the rate of ₹\[{\mathbf{200}}\] per\[{{\mathbf{m}}^{\mathbf{2}}}\]. 

Ans: The cost of cementing $1$\[\;{m^2}\] the floor of verandah = ₹\[200\]

The cost of cementing $63$\[\;{m^2}\] the floor of verandah\[ = 200 \times 63\]

= ₹\[12,600\]

5. A path $1$ m wide is built along the border and inside a square garden on the side \[{\mathbf{30}}\] m. Find: 

(i) The area of the path. 

Ans: Let us draw a square garden having side of the square garden = \[30\] m and width of the path along the border = $1$ m 

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Side of square garden without path = \[30-\left( {1 + 1} \right) = 30-{\text{2}} = 28\]m 

Now Area of path = Area of ABCD – Area of EFGH 

\[ = \left( {AB \times AD} \right)-\left( {EF \times EH} \right)\]

\[ = \left( {30 \times 30} \right)-\left( {28 \times 28} \right)\]

\[ = 900-784\]

\[ = 116\]\[\;{m^2}\]

(ii) The cost of planting grass in the remaining portion of the garden at the rate of₹\[{\mathbf{40}}\] per \[{{\mathbf{m}}^{\mathbf{2}}}\]. 

Ans: Area of remaining portion \[ = 28 \times 28 = 784\]\[\;{m^2}\]

The cost of planting grass in 1 \[\;{m^2}\] of the garden = ₹\[{\mathbf{40}}\]

The cost of planting grass in 784 \[\;{m^2}\] of the garden =₹\[40 \times 784 = 31,360\]

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. 

Ans: Let us assume a diagram of two cross roads and name them accordingly,

(Image will be Uploaded Soon)

Now Here, \[PQ = 10\]m and \[PS = 300\]m, \[EH = 10\]m and \[EF = 700\]m and \[KL = 10\]m and \[KN = 10\]m 

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

(∵ KLMN is taken twice, which is to be subtracted)

= \[(PS \times PQ) + (EF \times EH)-(KL \times KN)\]

\[ = \left( {300 \times 10} \right) + \left( {700 \times 10} \right)-\left( {10 \times 10} \right)\]

\[ = 3000 + 7000-100\]

= \[9,900\]\[\;{m^2}\]

Area of road in hectares, \[1\;{m^2} = \dfrac{1}{{10000}}\]hectares 

\[\therefore 9,900\;{m^2} = \dfrac{{9900}}{{10000}} = 0.99\]hectares 

Now, Area of park excluding cross roads = Area of park – Area of road 

\[ = \left( {AB \times AD} \right)-9,900\]

\[ = \left( {700 \times 300} \right)-9,900\]

\[ = 2,10,000-9,900\]

\[ = 2,00,100\;\]\[\;{m^2}\]

$ = \dfrac{{2,00,100\;}}{{10000}}$ hectares 

\[ = 20.01\]hectares

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. 

Ans: Let us assume a diagram of two cross roads and name them accordingly,

(Image will be Uploaded Soon)

Now Here, \[PQ = 10\]m and \[PS = 300\]m, \[EH = 10\]m and \[EF = 700\]m and \[KL = 10\]m and \[KN = 10\]m 

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

(∵ KLMN is taken twice, which is to be subtracted)

= \[(PS \times PQ) + (EF \times EH)-(KL \times KN)\]

\[ = \left( {60 \times 3} \right) + \left( {90 \times 3} \right)-\left( {3 \times 3} \right)\]

\[ = 180 + 270-9\]

= \[441\]\[\;{m^2}\]

ii. The cost of constructing the roads at the rate of ₹\[{\mathbf{110}}\] per \[{{\mathbf{m}}^{\mathbf{2}}}\]. 

Ans: The cost of 1 \[\;{m^2}\] constructing the roads = ₹ \[110\]

The cost of \[441\]\[\;{m^2}\] constructing the roads = ₹ \[110 \times 441\]= ₹ \[48,510\]

Therefore, the cost of constructing the roads = ₹ \[48,510\]

8. Pragya wrapped a cord around a circular pipe of radius \[{\mathbf{4}}\] cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side \[{\mathbf{4}}\] cm (also shown). Did she have any cord left? (Take\[ = {\mathbf{3}}.{\mathbf{14\pi }}\])

(Image will be Uploaded Soon)

Ans: Radius of pipe $ = 4$ cm

Wrapping cord around circular pipe \[ = 2\pi r = 2 \times 3.14 \times 4 = 25.12\]cm 

Again, wrapping cord around a square \[ = 4 \times side = 4 \times 4 = 16\]cm 

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

\[ = 25.12-16\]

\[ = 9.12\]cm 

Thus, she has left a \[9.12\]cm.

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

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i. The area of the whole land.

Ans: Length of rectangular lawn$ = 10$m, breadth of the rectangular lawn $ = 5$ m

And radius of the circular flower bed \[ = 2\]m 

Area of the whole land = length $ \times $ breadth

\[ = 10 \times 5\]

\[ = 50\]m

ii.  The area of the flower bed.

Ans: Area of flower bed \[ = \pi {r^2}\]

\[ = 3.14 \times 2 \times 2\]

\[ = 12.56\]\[\;{m^2}\]

iii. The area of the lawn excluding the area of the flower bed. 

Ans: Area of lawn excluding the area of the flower bed

= area of lawn – area of flower bed

\[ = 50-12.56\]

\[ = 37.44\]\[\;{m^2}\]

iv. The circumference of the flower bed.

Ans: The circumference of the flower bed \[ = 2\pi r\]

\[ = 2 \times 3.14 \times 2\]

\[ = 12.56\]m

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

(Image will be Uploaded Soon)

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

Area of shaded portion = Area of rectangle ABCD – (Area of \[\vartriangle FAE\]+ area of\[\vartriangle EBC\]) \[ = \left( {AB \times BC} \right)-\left( {\dfrac{1}{2} \times AE \times {\text{A}}F + \dfrac{1}{2} \times BE \times BC} \right)\]

\[ = \left( {18 \times 10} \right)-\left( {\dfrac{1}{2} \times 10 \times 6 + \dfrac{1}{2} \times 8 \times 10} \right)\]

\[ = 180-\left( {30 + 40} \right)\]

\[ = 180-70\]

$ = 110c{m^2}$

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(ii) Here, \[\;SR = SU + UR = 10 + 10 = 20cm,QR = 20cm\]

\[PQ = SR = 20cm,PT = PS-TS = 20-10cm\]

\[TS = 10cm,SU = 10cm,QR = 20cm\]and \[UR = 10cm\]

Area of shaded region = Area of square PQRS – Area of $\vartriangle QPT$– Area of \[\;\vartriangle TSU\]– Area of\[\vartriangle UQR\]

\[ = \left( {SR \times QR} \right) - \dfrac{1}{2} \times PQ \times PT-\dfrac{1}{2} \times ST \times SU-\dfrac{1}{2}\]

\[ = 20 \times 20-\dfrac{1}{2} \times 20 \times 10-\dfrac{1}{2} \times 10 \times 10-\dfrac{1}{2} \times 20 \times 10\]

\[ = 400-100-50-100\]

$ = 150c{m^2}$

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

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Ans: Here, \[AC = 22cm,BM = 3cm,DN = 3cm\]

Area of quadrilateral ABCDF = Area of $\vartriangle ABC$+ Area of $\vartriangle ADC$

\[ = \dfrac{1}{2} \times AC \times BM + \dfrac{1}{2} \times AC \times DN\]

\[ = \dfrac{1}{2} \times 22 \times 3 + \dfrac{1}{2} \times 22 \times 3\]

\[ = 3 \times 11{\text{ + }}3 \times 11\]

\[ = 33 + 33\]

$ = 66c{m^2}$

Thus, the area of quadrilateral ABCD is $66c{m^2}$.

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.4

Opting for the NCERT solutions for Ex 11.4 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.4 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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