NCERT Solutions Class 7 Mathematics Chapter 11 Perimeter and Area

 

Exercise 11.1

Q.1) The length and breadth of a rectangular piece of land are 500 𝑚 and 300 𝑚 respectively.
Find:
(i) Its area.
(ii) The cost of the land, if 1𝑚² of the land costs 𝑅𝑠. 10,000.
Sol.1) Given: Length of a rectangular piece of land = 500 𝑚 and Breadth of a rectangular piece of land = 300 𝑚
(i) Area of a rectangular piece of land = 𝐿𝑒𝑛𝑔𝑡ℎ × 𝐵𝑟𝑒𝑎𝑑𝑡ℎ
= 500 × 300 = 1,50,000 𝑚²
(ii) Since, the cost of 1𝑚² land = 𝑅𝑠. 10,000
Therefore, the cost of 1,50,000 𝑚² land = 10,000 × 1,50,000
= 𝑅𝑠. 1,50,00,00,000

Q.2) Find the area of a square park whose perimeter is 320 𝑚
Sol.2) Given: Perimeter of square park = 320 𝑚
⇒ 4 × 𝑠𝑖𝑑𝑒 = 320
⇒ side = 320 4 = 80 𝑚
Now, Area of square park = 𝑠𝑖𝑑𝑒 × 𝑠𝑖𝑑𝑒
= 80 × 80 = 6400 𝑚²
Thus, the area of square park is 6400 𝑚².

Q.3) Find the breadth of a rectangular plot of land, if its area is 440 𝑚² and the length is 22 𝑚. Also find its perimeter.
Sol.3) Area of rectangular park = 440 𝑚²
⇒ length x breadth = 440 𝑚²
⇒ 22 × 𝑏𝑟𝑒𝑎𝑑𝑡ℎ = 440
⇒ breadth = 440/22 = 20 𝑚
Now, Perimeter of rectangular park = 2 (𝑙𝑒𝑛𝑔𝑡ℎ + 𝑏𝑟𝑒𝑎𝑑𝑡ℎ)
= 2 (22 + 20) = 2 × 42 = 84 𝑚
Thus, the perimeter of rectangular park is 84 𝑚.

Q.4) The perimeter of a rectangular sheet is 100 𝑐𝑚. If the length is 35 𝑐𝑚, find its breadth. Also find the area.
Sol.4) Perimeter of the rectangular sheet = 100𝑐𝑚
⇒ 2 (𝑙𝑒𝑛𝑔𝑡ℎ + 𝑏𝑟𝑒𝑎𝑑𝑡ℎ) = 100 𝑐𝑚
⇒ 2 (35 + 𝑏𝑟𝑒𝑎𝑑𝑡ℎ) = 100
⇒ 35 + 𝑏𝑟𝑒𝑎𝑑𝑡ℎ = 100/2
⇒ 35 + 𝑏𝑟𝑒𝑎𝑑𝑡ℎ = 50
⇒ breadth = 50 – 35
⇒ breadth = 15 𝑐𝑚
Now, Area of rectangular sheet = 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑏𝑟𝑒𝑎𝑑𝑡ℎ
= 35 × 15 = 525 c𝑚²
Thus, breadth and area of rectangular sheet are 15 𝑐𝑚 and 525 𝑐𝑚² respectively.

Q.5) The area of a square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 𝑐𝑚, find the breadth of the rectangular park.
Sol.5) Given: The side of the square park = 60 𝑚
The length of the rectangular park = 90 𝑚
According to the question,
Area of square park = Area of rectangular park
⇒ 𝑠𝑖𝑑𝑒 × 𝑠𝑖𝑑𝑒 = 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑏𝑟𝑒𝑎𝑑𝑡ℎ
⇒ 60 × 60 = 90 × 𝑏𝑟𝑒𝑎𝑑𝑡ℎ
⇒ breadth = 60×60/90 = 40 𝑚
Thus, the breadth of the rectangular park is 40 𝑚

Q.6) A wire is in the shape of a rectangle. Its length is 40 𝑐𝑚 and breadth is 22 𝑐𝑚. If the same wire is rebent in the shape of a square, what will be the measure of each side. Also find which shape encloses more area?
Sol.6) According to the question, Perimeter of square = Perimeter of rectangle
⇒ 4 x side = 2 (𝑙𝑒𝑛𝑔𝑡ℎ + 𝑏𝑟𝑒𝑎𝑑𝑡ℎ)
⇒ 4 x side = 2 (40 + 22)
⇒ 4 x side = 2 × 62
⇒ side = 2×62/4 = 31 𝑐𝑚
Thus, the side of the square is 31 cm.
Now, Area of rectangle = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑥 × 𝑏𝑟𝑒𝑎𝑑𝑡ℎ
= 40 × 22 = 880 𝑐𝑚²
And Area of square = 𝑠𝑖𝑑𝑒 × 𝑠𝑖𝑑𝑒
= 31 × 31 = 961 𝑐𝑚²
Therefore, on comparing, the area of square is greater than that of rectangle.

Q.7) The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also, find the area of the rectangle.
Sol.7) Perimeter of rectangle = 130 𝑐𝑚
⇒ 2 (length + breadth) = 130 𝑐𝑚
⇒ 2 (length + 30) = 130
⇒ length + 30 = 130/2
⇒ length + 30 = 65
⇒ length = 65 – 30 = 35 𝑐𝑚
Now area of rectangle = length x breadth = 35 × 30 = 1050 𝑐𝑚²
Thus, the area of rectangle is 1050 𝑐𝑚².

Q.8) A door of length 2 𝑚 and breadth 1 𝑚 is fitted in a wall. The length of the wall is 4.5 𝑚 and the breadth is 3.6 𝑚. Find the cost of white washing the wall, if the rate of white washing the wall is 𝑅𝑠. 20 𝑝𝑒𝑟 𝑚²
Sol.8) Area of rectangular door = length x breadth
= 2 𝑚 × 1 𝑚 = 2 𝑚²
Area of wall including door = length x breadth
= 4.5 𝑚 × 3.6 𝑚 = 16.2 𝑚²

Now,
Area of wall excluding door = Area of wall including door – Area of door
= 16.2 – 2 = 14.2 𝑚²
Since, The rate of white washing of 1 𝑚² the wall = 𝑅𝑠. 20
Therefore, the rate of white washing of 14.2 𝑚² the wall
= 20 × 14.2 = 𝑅𝑠. 284
Thus, the cost of white washing the wall excluding the door is 𝑅𝑠. 284.

Exercise 11.2

Q.1) Find the area of each of the following parallelogram:

Sol.1) We know that the area of parallelogram = base x height
(a) Here base = 7 cm and height = 4 cm
∴ Area of parallelogram = 7 x 4 = 28 cm²
(b) Here base = 5 cm and height = 3 cm
∴ Area of parallelogram = 5 x 3 = 15 cm²
(c) Here base = 2.5 cm and height = 3.5 cm
∴ Area of parallelogram = 2.5 x 3.5 = 8.75 cm²
(d) Here base = 5 cm and height = 4.8 cm
∴ Area of parallelogram = 5 x 4.8 = 24 cm²
(e) Here base = 2 cm and height = 4.4 cm
∴ Area of parallelogram = 2 x 4.4 = 8.8 cm²

Q.2) Find the area of each of the following triangles :
Sol.2) We know that the area of triangle = 1/2 × 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡
(a) Here, base = 4 cm and height = 3 cm
∴ Area of triangle = 1/2 × 4 × 6 = 6𝑐𝑚²
(b) Here, base = 5 cm and height = 3.2 cm
∴ Area of triangle = 1/2 × 5 × 3.2 = 8𝑐𝑚²
(c) Here, base = 3 cm and height = 4 cm
∴ Area of triangle = 1/2 × 3 × 4 = 6𝑐𝑚²
(d) Here, base = 3 cm and height = 2 cm
∴ Area of triangle = 1/2 × 3 × 2 = 3𝑐𝑚²

Q.3) Find the missing values:

Sol.3) We know that the area of parallelogram = base x height
(a) Here, base = 20 cm and area = 246𝑐𝑚2
∴ Area of parallelogram = base x height
⇒ 246 = 20 × ℎ𝑒𝑖𝑔ℎ𝑡
⇒ height = 246/20 = 12.3𝑐𝑚
(b) Here, height = 15 cm and area = 154.5 𝑐𝑚2
∴ Area of parallelogram = base x height
⇒ 154.5 = base x 15
⇒ base = 154.5/15 = 10.3 𝑐𝑚
(c) Here, height = 8.4 cm and area = 48.72 𝑐𝑚2
∴ Area of parallelogram = base x height
⇒ 48.72 = 𝑏𝑎𝑠𝑒 × 8.4
⇒ base = 48.72/8.4 = 5.8 𝑐𝑚
(d) Here, base = 15.6 cm and area = 16.38 𝑐𝑚2
∴ Area of parallelogram = base x height
⇒ 16.38 = 15.6 x height
⇒ height = 16.38/15.6 = 1.05𝑐𝑚

Sol.4) We know that the area of triangle = 1/2 × 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡
In first row, base = 15 cm and area = 87𝑐𝑚²

Q.5) PQRS is a parallelogram (Fig 11.23). QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find:
(a) the area of the parallelogram PRS (b) QN, if PS = 8 cm
Sol.5) Given: 𝑆𝑅 = 12 𝑐𝑚, 𝑄𝑀 = 7.6 𝑐𝑚, 𝑃𝑆 = 8 𝑐𝑚.
(a) Area of parallelogram = base x height
= 12 × 7.6 = 91.2 𝑐𝑚²
(b) Area of parallelogram = base x height
⇒ 91.2 = 8 × 𝑄𝑁
⇒ 𝑄𝑁 = 91.2/8 = 11.4 𝑐𝑚

Q.6) 𝐷𝐿 and 𝐵𝑀 are the heights on sides 𝐴𝐵 and 𝐴𝐷 respectively of parallelogram ABCD (Fig 11.24). If the area of the parallelogram is 1470𝑐𝑚2, 𝐴𝐵 = 35 𝑐𝑚 and 𝐴𝐷 = 49 𝑐𝑚, find the length of 𝐵𝑀 and 𝐷𝐿.
Sol.6) Given: Area of parallelogram = 1470𝑐𝑚2 Base (AB) = 35 cm
and 𝑏𝑎𝑠𝑒 (𝐴𝐷) = 49 𝑐𝑚
Since Area of parallelogram = base x height
⇒ 1470 = 35 x DL
⇒ DL = 1470/35
⇒ 𝐷𝐿 = 42 𝑐𝑚
Again, Area of parallelogram = base x height
⇒ 1470 = 49 x BM
⇒ 𝐵𝑀 = 1470/49
⇒ 𝐵𝑀 = 30 𝑐𝑚

Thus, the lengths of DL and BM are 42 cm and 30 cm respectively.

Q.7) Δ 𝐴𝐵𝐶 is right angled at 𝐴 (Fig 11.25). AD is perpendicular to BC. If 𝐴𝐵 = 5 𝑐𝑚, 𝐵𝐶 = 13 𝑐𝑚 and 𝐴𝐶 = 12 𝑐𝑚, find the area of Δ 𝐴𝐵𝐶. Also, find the length of 𝐴𝐷.
Sol.7) In right angles triangle BAC,
AB = 5 cm and AC = 12 𝑐𝑚
Area of triangle = 1/2 × 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡 
= 1/2 × 𝐴𝐵 × 𝐴𝐶 = 1/2 × 5 × 12 = 30𝑐𝑚²
Now, in Δ 𝐴𝐵𝐶,

Area of triangle ABC = 1/2 × 𝐵𝐶 × 𝐴𝐷
⇒ 30 = 1/2 × 13 × 𝐴𝐷
⇒ 𝐴𝐷 = 30 × 2/13
= 60/13 𝑐𝑚

Q.8) Δ 𝐴𝐵𝐶 is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26). The height AD from A to BC, is 6 cm. Find the area of Δ 𝐴𝐵𝐶. What will be the height from C to AB i.e., CE?
Sol.8) In Δ ABC, AD = 6 cm and BC = 9 cm
Area of triangle = 1/2 × 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡
= 1 2 × 𝐵𝐶 × 𝐴𝐷 = 1/2 × 9 × 6 = 27 𝑐𝑚²
Again, Area of triangle = 1/2 × 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡 = 1/2 × 𝐴𝐵 × 𝐶𝐸
⇒ 27 = 1/2 × 7.5 × 𝐶𝐸
⇒ 𝐶𝐸 = 27 × 2/7.5
⇒ 𝐶𝐸 = 7.2 𝑐𝑚
Thus, height from 𝐶 to 𝐴𝐵 i.e., 𝐶𝐸 is 7.2 𝑐𝑚.

Exercise 11.3

Q.1) Find the circumference of the circles with the following radius: (𝑡𝑎𝑘𝑒 𝜋 = 22/7)
(a) 14 cm (b) 28 mm (c) 21 cm
Sol.1) (a) Circumference of the circle = 2π𝑟 = 2 × 22/7 × 14 = 88 𝑐𝑚
(b) Circumference of the circle = 2π𝑟 = 2 × 22/7 × 28 = 176 𝑚𝑚
(c) Circumference of the circle = 2π𝑟 = 2 × 22/7 × 21 = 132 𝑐𝑚

Q.2) Find the area of the following circles, given that:
(a) radius = 14 mm (b) diameter = 49 m (c) radius 5 cm
Sol.2) (a) Area of circle = π𝑟² = 22/7 × 14 × 14
= 22 × 2 × 14 = 616 𝑚𝑚²
(b) Diameter = 49 𝑚 
∴ radius = 49/2 = 24.5 𝑚
∴ Area of circle = π𝑟² = 22/7 × 24.5 × 24.5
= 22 × 3.5 × 24.5 = 1886.5 𝑚²
(c) Area of circle = π𝑟² = 22/7 × 5 × 5 = 550/7 𝑐𝑚²

Q.3) If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet.
Sol.3) Circumference of the circular sheet = 154 m
⇒ 2π𝑟 = 154 𝑚
⇒ 𝑟 = 154/2π
⇒ 𝑟 = 154×7/2×22 = 24.5𝑚
Now Area of circular sheet = π𝑟² = 22/7 × 24.5 × 24.5
= 22 × 3.5 × 24.5 = 1886.5 𝑚²
Thus, the radius and area of circular sheet are 24.5 𝑚 and 1886.5 𝑚² respectively.

Q.4) A gardener wants to fence a circular garden of diameter 21 𝑚. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also, find the costs of the rope, if it cost Rs.4 per meter.
Sol.4) Diameter of the circular garden = 21 𝑚
∴ Radius of the circular garden = 21/2 𝑚
Now Circumference of circular garden = 2π𝑟 = 2 × 22/7 × 21/2 = 22 × 3 = 66 𝑚
The gardener makes 2 rounds of fence so the total length of the rope of fencing = 2 ×
2π𝑟 = 2 × 66 = 132 𝑚
Since, the cost of 1-meter rope = 𝑅𝑠. 4
Therefore, cost of 132-meter rope = 4 × 132 = 𝑅𝑠. 528

Q.5) From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take =3.14 π )
Sol.5) Radius of circular sheet (R) = 4 cm and radius of removed circle (r) = 3 cm
Area of remaining sheet = Area of circular sheet – Area of removed circle
= π𝑅² − π 𝑅² = π( 𝑅² - 𝑅²)
= π(42 − 32) = π (16 − 9)
= 3.14 × 7 = 21.98 𝑐𝑚²
Thus, the area of remaining sheet is21.98 𝑐𝑚²

Q.6) Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs 𝑅𝑠. 15.
(𝑇𝑎𝑘𝑒 = 3.14 π )
Sol.6) Diameter of the circular table cover = 1.5 m
∴ Radius of the circular table cover = 1.5 2 m
Circumference of circular table cover = 2π𝑟
= 2 × 3.14 × 1.5/2
Therefore, the length of required lace is 4.71 𝑚.
Now the cost of 1 𝑚 𝑙𝑎𝑐𝑒 = 𝑅𝑠. 15
Then the cost of 4.71 m lace = 15 × 4.71 = 𝑅𝑠. 70.65
Hence, the cost of 4.71 𝑚 lace is 𝑅𝑠. 70.65.

Q.7) Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Sol.7) Diameter = 10 𝑐𝑚
∴ Radius = 10/2 = 5𝑐𝑚
According to question,
Perimeter of figure = Circumference of semi-circle + diameter = π𝑟 + 𝐷
= 22/7 × 5 + 10 = 110/7 + 10
= 110+70/7 = 180/7 = 25.71𝑐𝑚
Thus, the perimeter of the given figure is 25.71 𝑐𝑚.

Q.8) Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is 𝑅𝑠. 15/𝑚². (𝑇𝑎𝑘𝑒 = 3.14 π )
Sol.8) Diameter of the circular table top = 1.6 m
∴ Radius of the circular table top = 1.6/2 = 0.8 𝑚
Area of circular table top = π𝑟²
= 3.14 × 0.8 × 0.8 = 2.0096 𝑚²
Now cost of 1𝑚² polishing = Rs.15
Then cost of 2.0096 𝑚² polishing = 15 × 2.0096
= 𝑅𝑠. 30.14 (𝑎𝑝𝑝𝑟𝑜𝑥. )
Thus, the cost of polishing a circular table top is 𝑅𝑠. 30.14 (𝑎𝑝𝑝𝑟𝑜𝑥.)

Q.9) Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be
the length of each of its sides? Which figure encloses more, the circle or the square?
Sol.9) Total length of the wire = 44 cm
∴ the circumference of the circle = 2π𝑟 = 44 𝑐𝑚
⇒ 2 × 22/7 × 𝑟 = 44
⇒ 𝑟 = 44 × 7/2 × 22 = 7𝑐𝑚
Now Area of the circle = π 𝑟² = 22/7 × 7 × 7 = 154 𝑐𝑚²
Now the wire is converted into square.
Then perimeter of square = 44 cm
⇒ 4 x side = 44
⇒ side = 44/4 = 11𝑐𝑚
Now area of square = 𝑠𝑖𝑑𝑒 × 𝑠𝑖𝑑𝑒 = 11 × 11 = 121 𝑐𝑚²
Therefore, on comparing,
the area of circle is greater than that of square, so the circle enclosed more area.

Q.10) From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed. (as shown in the adjoining figure).
Find the area of the remaining sheet.

Sol.10) Radius of circular sheet (R) = 14 cm and
Radius of smaller circle (r) = 3.5 cm
Length of rectangle (l) = 3 cm and
breadth of rectangle (b) = 1 cm
According to question,
Area of remaining sheet=Area of circular sheet– (Area of two smaller circle + Area of rectangle)
= 𝜋𝑅² − [2(𝜋𝑟²) + (𝑙 × 𝑏)]
= 22/7 × 14 × 14 − [(2 × 22/7 × 3.5 × 3.5) − (3 × 1)]
= 22 × 14 × 2 − [44 × 0.5 × 3.5 + 3]
= 616 − 80
= 536 𝑐𝑚²
Therefore the area of remaining sheet is 536𝑐𝑚².

Q.11) A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm.
What is the area of the left over aluminium sheet? (Take = 3.14 π )
Sol.11) Radius of circle = 2 cm and
side of aluminium square sheet = 6 𝑐𝑚
According to question,
Area of aluminium sheet left = Total area of aluminium sheet – Area of circle
= 𝑠𝑖𝑑𝑒 × 𝑠𝑖𝑑𝑒 − π𝑟²
= 6 × 6 – 22/7 × 2 × 2
= 36 – 12.56 = 23.44 𝑐𝑚²
Therefore, the area of aluminium sheet left is 23.44 𝑐𝑚²

Q.12) The circumference of a circle is 31.4 cm. Find the radius and the area of the circle. (Take =3.14 π )
Sol.12) The circumference of the circle = 31.4 cm
⇒ 2π𝑟 = 31.4
⇒ 2 × 3.14 × 𝑟 = 31.4
⇒ 𝑟 = 31.4/2×3.14 = 5 𝑐𝑚
Then area of the circle = π𝑟² = 3.14 × 5 × 5 = 78.5 𝑐𝑚²
Therefore, the radius and the area of the circle are 5 𝑐𝑚 and 78.5 𝑐𝑚² respectively.

Q.13) A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66𝑚. What is the area of this path? (Take = 3.14 π )

Sol.13) Diameter of the circular flower bed = 66 𝑚
∴ Radius of circular flower bed (𝑟) = 66/2 = 33𝑚
∴ Radius of circular flower bed with 4 m wide path (𝑅) = 33 + 4 = 37𝑚
According to the question,
Area of path = Area of bigger circle – Area of smaller circle
= π𝑅² − π𝑟² = π(𝑅² - 𝑅²)
= π[((37)² − (33)2)
= 3.14[(37 + 33)(37 − 33)]
= 3.14 × 70 × 4 = 879.20 𝑚²
Therefore, the area of the path is 879.20 𝑚²

Q.14) A circular flower garden has an area of 314𝑚². A sprinkler at the centre of the garden can cover an area that has a radius of 12 𝑚. Will the sprinkler water the entire garden? (Take = 3.14 π )
Sol.14) Circular area by the sprinkler = π𝑟²
= 3.14 × 12 × 12
= 3.14 × 144
= 452.16 𝑚²
Area of the circular flower garden = 314 𝑚²
Since Area of circular flower garden is smaller than area by sprinkler.
Therefore, the sprinkler will water the entire garden.

Q.15) Find the circumference of the inner and the outer circles, shown in the adjoining figure.
(Take = 3.14 π)

Sol.15) Radius of outer circle (𝑟) = 19 𝑚
∴ Circumference of outer circle = 2π𝑟 = 2 × 3.14 × 19 = 119.32 𝑚
Now radius of inner circle (𝑟′) = 19 – 10 = 9 𝑚
∴ Circumference of inner circle = 2 ′ π𝑟 = 2 × 3.14 × 9 = 56.52 𝑚
Therefore, the circumferences of inner and outer circles are 56.52 𝑚 and 119.32 𝑚
respectively.

Q.16) How many times a wheel of radius 28 cm must rotate to go 352 m?
Sol.16) Let wheel must be rotate 𝑛 times of its circumference.
Radius of wheel = 28 𝑐𝑚 and Total distance = 352 𝑚 = 35200 𝑐𝑚
∴ Distance covered by wheel = n x circumference of wheel
⇒ 35200 = 𝑛 × 2π𝑟
⇒ 35200 = 𝑛 × 2 × 22/7 × 28
⇒ 𝑛 = 35200 × 7/2 × 22×28
⇒ 𝑛 = 200 revolutions
Thus, wheel must rotate 200 times to go 352 𝑚

Q.17) The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (𝑇𝑎𝑘𝑒 = 3.14 π)
Sol.17) In 1 hour, minute hand completes one round means makes a circle.
Radius of the circle (𝑟) = 15 𝑐𝑚
Circumference of circular clock = 2π𝑟
= 2 × 3.14 × 15 = 94.2 𝑐𝑚
Therefore, the tip of the minute hand moves 94.2 𝑐𝑚 in 1 ℎ𝑜𝑢𝑟.

Exercise 11.4

Q.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.
Sol.1) Length of rectangular garden = 90 𝑚
and breadth of rectangular garden = 75 𝑚

Outer length of rectangular garden with path = 90 + 5 + 5 = 100 𝑚
Outer breadth of rectangular garden with path = 75 + 5 + 5 = 85 𝑚
Outer area of rectangular garden with path = length x breadth
= 100 × 85 = 8,500 𝑚²
Inner area of garden without path = length x breadth
= 90 × 75 = 6,750 𝑚²
Now, Area of path = Area of garden with path – Area of garden without path
= 8,500 – 6,750 = 1,750𝑚²
Since, 1𝑚² = (1/10000)ℎ𝑒𝑐𝑡𝑎𝑟𝑒𝑠
Therefore, 6,750 𝑚² = 6750/10000 = 0.675 ℎ𝑒𝑐𝑡𝑎𝑟𝑒𝑠.

Q.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.

Sol.2) Length of rectangular park = 125 𝑚,
Breadth of rectangular park = 65 𝑚
and Width of the path = 3 𝑚
Length of rectangular park with path = 125 + 3 + 3 = 131 𝑚
Breadth of rectangular park with path = 65 + 3 + 3 = 71 𝑚
∴ Area of path = Area of park with path – Area of park without path
= (𝐴𝐵 × 𝐴𝐷) – (𝐸𝐹 × 𝐸𝐻)
= (131 × 71) – (125 × 65)
= 9301 – 8125 = 1,176 𝑚²
Thus, area of path around the park is 1,176 𝑚².

Q.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.

Sol.3) 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 side.
Therefore reduced length = 8 – (1.5 + 1.5) = 8 – 3 = 5 𝑐𝑚
And reduced breadth = 5 – (1.5 + 1.5) = 5 – 3 = 2 𝑐𝑚
∴ Area of margin = Area of cardboard (ABCD) – Area of cardboard (EFGH)
= (𝐴𝐵 × 𝐴𝐷) – (𝐸𝐹 × 𝐸𝐻)
= (8 × 5) – (5 × 2)
= 40 – 10 = 30 𝑐𝑚2
Thus, the total area of margin is 30 𝑐𝑚2.

Q.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 𝑅𝑠. 200 𝑝𝑒𝑟 𝑚²

Sol.4) (i) 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
= (𝐴𝐵 × 𝐴𝐷) – (𝐸𝐹 × 𝐸𝐻)
= (10 × 8.5) – (5.5 × 4)
= 85 – 22 = 63 𝑚²
(ii) The cost of cementing 1𝑚² the floor of verandah = 𝑅𝑠. 200
The cost of cementing 63𝑚² the floor of verandah = 200 × 63 = 𝑅𝑠. 12,600

Q.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 𝑅𝑠. 40 𝑝𝑒𝑟 𝑚²

Sol.5) (i) Side of the square garden = 30 m
and Width of the path along the border = 1 𝑚
Side of square garden without path = 30 – (1 + 1) = 30 – 2 = 28 𝑚
Now Area of path = Area of ABCD – Area of EFGH
= (𝐴𝐵 × 𝐴𝐷) – (𝐸𝐹 × 𝐸𝐻)
= (30 × 30) – (28 × 28)
= 900 – 784 = 116 𝑚²
(ii) Area of remaining portion = 28 × 28 = 784 𝑚²
The cost of planting grass in 1𝑚² of the garden = 𝑅𝑠. 40
The cost of planting grass in 784 𝑚² of the garden = 𝑅𝑠. 40 × 784 = 𝑅𝑠. 31,360

Q.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.
Sol.6) 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 × 𝑃𝑄 + 𝐸𝐹 × 𝐸𝐻 – 𝐾𝐿 × 𝐾𝑁
= (300 × 10) + (700 × 10)– (10 × 10)
= 3000 + 7000 – 100 = 9,900 𝑚²
Area of road in hectares = 1𝑚² = (1/10000)ℎ𝑒𝑐𝑡𝑎𝑟𝑒𝑠
∴ 9,900 𝑚² = 9900/10000 = 0.99 ℎ𝑒𝑐𝑡𝑎𝑟𝑒𝑠

(ii) Now, Area of park excluding cross roads = Area of park – Area of road
= (𝐴𝐵 × 𝐴𝐷) – 9,900
= (700 × 300) – 9,900
= 2,10,000 – 9,900 = 2,00,100 𝑚²
= (200100/10000)ℎ𝑒𝑐𝑡𝑎𝑟𝑒𝑠 = 20.01 ℎ𝑒𝑐𝑡𝑎𝑟𝑒𝑠

Q.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 𝑅𝑠. 110 𝑝𝑒𝑟 𝑚².
Sol.7) Here, 𝑃𝑄 = 3 𝑚 and 𝑃𝑆 = 60 𝑚, 𝐸𝐻 = 3 𝑚 and
𝐸𝐹 = 9 and 𝐾𝐿 = 3 𝑚 and 𝐾𝑁 = 3 𝑚
Area of roads = Area of PQRS + Area of EFGH – Area of KLMN
[ KLMN is taken twice, which is to be subtracted]
= 𝑃𝑆 × 𝑃𝑄 + 𝐸𝐹 × 𝐸𝐻 – 𝐾𝐿 × 𝐾𝑁
= (60 × 3) + (90 × 3)– (3 × 3)
= 180 + 270 − 9 = 441 𝑚²

(ii) The cost of 1𝑚² constructing the roads = 𝑅𝑠. 110
The cost of 441 𝑚² constructing the roads = 𝑅𝑠. 110 × 441
= 𝑅𝑠. 48,510
Therefore, the cost of constructing the roads = 𝑅𝑠. 48,510

Q.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 = 3.14 π)

Sol.8) Radius of pipe = 4 𝑐𝑚
Wrapping cord around circular pipe = 2πr
= 2 × 3.14 × 4 = 25.12 𝑐𝑚
Again, wrapping cord around a square = 4 x side
= 4 x 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.

Q.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.

Sol.9) Length of rectangular lawn = 10 m, breadth of the rectangular lawn = 5 m and radius of the circular flower bed = 2 𝑚
(i) Area of the whole land = length x breadth = 10 × 5 = 50𝑚²
(ii) Area of flower bed = πr² = 3.14 × 2 × 2 = 12.56 𝑚²
(iii) Area of lawn excluding the area of the flower bed = area of 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

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

Sol.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 Δ 𝐹𝐴𝐸 + area of Δ 𝐸𝐵𝐶)
= (𝐴𝐵 × 𝐵𝐶) – (1/2 × 𝐴𝐸 × 𝐴𝐹 + 1/2 × 𝐵𝐸 × 𝐵𝐶)
= (18 × 10) – (1/2 × 10 × 6 + 1/2 × 8 × 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 Δ 𝑄𝑃𝑇 – Area of Δ 𝑇𝑆𝑈 – Area of Δ 𝑈𝑄𝑅
= (𝑆𝑅 × 𝑄𝑅) − 1/2 × 𝑃𝑄 × 𝑃𝑇 – 1/2 × 𝑆𝑇 × 𝑆𝑈 – 1/2
= 20 × 20 – 1 2 × 20 × 10 – 1/2 × 10 × 10 – 1/2 × 20 × 10
= 400 – 100 – 50 – 100 = 150 𝑐𝑚²

Q.11) Find the area of the equilateral 𝐴𝐵𝐶𝐷. Here, 𝐴𝐶 = 22 𝑐𝑚, 𝐵𝑀 = 3 𝑐𝑚, 𝐷𝑁 = 3 𝑐𝑚 and 𝐵𝑀 ⊥ 𝐴𝐶, 𝐷𝑁 ⊥ 𝐴𝐶.
Sol.11) Here, 𝐴𝐶 = 22 𝑐𝑚, 𝐵𝑀 = 3 𝑐𝑚, 𝐷𝑁 = 3 𝑐𝑚
Area of quadrilateral 𝐴𝐵𝐶𝐷𝐹 = Area of Δ 𝐴𝐵𝐶 + Area of Δ 𝐴𝐷𝐶
= 1/2 × 𝐴𝐶 × 𝐵𝑀 + 1/2 × 𝐴𝐶 × 𝐷𝑁
= 1/2 × 22 × 3 + 1/2 × 22 × 3
= 3 × 11 + 3 × 11
= 33 + 33 = 66 𝑐𝑚²
Thus, the area of quadrilateral ABCD is 𝑐𝑚²