Exercise 11.1
1. The Length and the 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 1 m^{2} of the land costs Rs. 10,000.
Solution:-
From the question it is given that,
Length of the rectangular piece of land = 500 m
Breadth of the rectangular piece of land = 300 m
Then,
(i) Area of rectangle = Length × Breadth
= 500 × 300
= 150000 m^{2}
(ii) Cost of the land for 1 m^{2} = Rs. 10000
Cost of the land for 150000 m^{2} = 10000 × 150000
= Rs. 1500000000
2. Find the area of a square park whose perimeter is 320m.
Solution:-
From the question it is given that,
Perimeter of the square park = 320 m
4 × Length of the side of park = 320 m
Then,
Length of the side of park = 320/4
= 80 m
So, Area of the square park = (length of the side of park)^{2}
= 80^{2}
= 6400 m^{2}
3. Find the breadth of a rectangular plot of land, if its area is 440 m^{2} and the length is 22 m. Also find its perimeter.
Solution:-
From the question it is given that,
Area of the rectangular plot = 440 m^{2}
Length of the rectangular plot = 22 m
We know that,
Area of the rectangle = Length × Breadth
440 = 22 × Breadth
Breadth = 440/22
Breadth = 20 m
Then,
Perimeter of the rectangle = 2(Length + Breadth)
= 2 (22 + 20)
= 2(42)
= 84 m
∴Perimeter of the rectangular plot is 84 m.
4. The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth.
Also find the area.
Solution:-
From the question it is given that,
Perimeter of the a rectangular sheet = 100 cm
Length of the rectangular sheet = 35 cm
We know that,
Perimeter of the rectangle = 2 (Length + Breadth)
100 = 2 (35 + Breadth)
(100/2) = 35 + Breadth
50 – 35 = Breadth
Breadth = 15 cm
Then,
Area of the rectangle = Length × Breadth
= 35 × 15
= 525 cm^{2}
∴Area of the rectangular sheet is 525 cm^{2}
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 m, find the breadth of the rectangular park.
Solution:-
From the question it is given that,
Area of a square park is the same as of a rectangular park.
Side of the square park = 60 m
Length of the rectangular park = 90 m
We know that,
Area of the square park = (one of the side of square)^{2}
= 60^{2}
= 3600 m^{2}
Area of the rectangular park = 3600 m^{2} … [∵ given]
Length × Breadth = 3600
90 × Breadth = 3600
Breadth = 3600/90
Breadth = 40 m
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?
Solution:-
By reading the question we can conclude that, perimeter of the square is same as perimeter of rectangle.
From the question it is given that,
Length of the rectangle = 40 cm
Breadth of the square = 22 cm
Then,
Perimeter of the rectangle = Perimeter of the Square
2 (Length + Breadth) = 4 × side
2 (40 + 22) = 4 × side
2 (62) = 4 × side
124 = 4 × side
Side = 124/4
Side = 31 cm
So, Area of the rectangle = (Length × Breadth)
= 40 × 22
= 880 cm^{2}
Area of square = side^{2}
= 31^{2}
= 31 × 31
= 961 cm^{2}
∴Square shaped wire encloses more area.
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.
Solution:-
From the question it is given that.
Perimeter of the rectangle = 130 cm
Breadth of the rectangle = 30
We know that,
Perimeter of rectangle = 2 (Length + Breadth)
130 = 2 (length + 30)
130/2 = length + 30
Length + 30 = 65
Length = 65 – 30
Length = 35 cm
Then,
Area of the rectangle = Length × Breadth
= 35 × 30
= 1050 cm^{2}
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 (Fig). Find the cost of white washing the wall, if the rate of white washing the wall is Rs. 20 per m^{2}.
Solution:-
From the question it is given that,
Length of the door = 2 m
Breadth of the door = 1 m
Length of the wall = 4.5 m
Breadth of the wall = 3.6 m
Then,
Area of the door = Length × Breadth
= 2 × 1
= 2 m^{2}
Area of the wall = Length × Breadth
= 4.5 × 3.6
= 16.2 m^{2}
So, Area to be white washed = 16.2 – 2 = 14.2 m^{2}
Cost of white washing 1 m^{2} area = Rs. 20
Hence cost of whit washing 14.2 m^{2} area = 14.2 × 20
= Rs. 284
Exercise 11.2
Question 1.
Find the area of each of the following parallelograms:
Solution:
(a) Area of the parallelogram
= base × altitude = 7 cm × 4 cm
= 28 cm^{2}
(b) Area of the parallelogram
= base × altitude = 5 cm × 3 cm
= 15 cm^{2}
(c) Area of the parallelogram
= base × altitude = 2.5 cm × 3.5 cm
= 8.75 cm^{2}
(d) Area of the parallelogram
= base × altitude = 5 cm × 4.8 cm
= 24.0 cm^{2}
(e) Area of the parallelogram
= base × altitude = 2 cm × 4.4 cm
= 8.8 cm^{2}
Question 2.
Find the area of each of the following triangles:
Solution:
Area of the triangle = 12 × b × h
= 12 × 4 cm × 3 cm
= 6m^{2}
(b) Area of the triangle = 12 × b × h
= 12 × 5 cm × 3.2 cm
= 8.0 cm^{2}
(c) Area of the triangle = 12 × b × l
= 12 × 3 cm × 4 cm
= 6 cm^{2}
(d) Area of the triangle = 12 × b × h
= 12 × 3 cm × 2 cm
= 3 cm^{2}
3. Find the missing values:
S.No. | Base | Height | Area of the Parallelogram |
a. | 20 cm | 246 cm^{2} | |
b. | 15 cm | 154.5 cm^{2} | |
c. | 8.4 cm | 48.72 cm^{2} | |
d. | 15.6 cm | 16.38 cm^{2} |
Solution:-
(a)
From the table,
Base of parallelogram = 20 cm
Height of parallelogram =?
Area of the parallelogram = 246 cm^{2}
Then,
Area of parallelogram = base × height
246 = 20 × height
Height = 246/20
Height = 12.3 cm
∴Height of the parallelogram is 12.3 cm.
(b)
From the table,
Base of parallelogram =?
Height of parallelogram =15 cm
Area of the parallelogram = 154.5 cm^{2}
Then,
Area of parallelogram = base × height
154.5 = base × 15
Base = 154.5/15
Base = 10.3 cm
∴Base of the parallelogram is 10.3 cm.
(c)
From the table,
Base of parallelogram =?
Height of parallelogram =8.4 cm
Area of the parallelogram = 48.72 cm^{2}
Then,
Area of parallelogram = base × height
48.72 = base × 8.4
Base = 48.72/8.4
Base = 5.8 cm
∴Base of the parallelogram is 5.8 cm.
(d)
From the table,
Base of parallelogram = 15.6 cm
Height of parallelogram =?
Area of the parallelogram = 16.38 cm^{2}
Then,
Area of parallelogram = base × height
16.38 = 15.6 × height
Height = 16.38/15.6
Height = 1.05 cm
∴Height of the parallelogram is 1.05 cm.
S.No. | Base | Height | Area of the Parallelogram |
a. | 20 cm | 12.3 cm | 246 cm^{2} |
b. | 10.3 cm | 15 cm | 154.5 cm^{2} |
c. | 5.8 cm | 8.4 cm | 48.72 cm^{2} |
d. | 15.6 cm | 1.05 | 16.38 cm^{2} |
4. Find the missing values:
Base | Height | Area of Triangle |
15 cm | 87 cm^{2} | |
31.4 mm | 1256 mm^{2} | |
22 cm | 170.5 cm^{2} |
Solution:-
(a)
From the table,
Height of triangle =?
Base of triangle = 15 cm
Area of the triangle = 16.38 cm^{2}
Then,
Area of triangle = ½ × base × height
87 = ½ × 15 × height
Height = (87 × 2)/15
Height = 174/15
Height = 11.6 cm
∴Height of the triangle is 11.6 cm.
(b)
From the table,
Height of triangle =31.4 mm
Base of triangle =?
Area of the triangle = 1256 mm^{2}
Then,
Area of triangle = ½ × base × height
1256 = ½ × base × 31.4
Base = (1256 × 2)/31.4
Base = 2512/31.4
Base = 80 mm = 8 cm
∴Base of the triangle is 80 mm or 8 cm.
(c)
From the table,
Height of triangle =?
Base of triangle = 22 cm
Area of the triangle = 170.5 cm^{2}
Then,
Area of triangle = ½ × base × height
170.5 = ½ × 22 × height
170.5 = 1 × 11 × height
Height = 170.5/11
Height = 15.5 cm
∴Height of the triangle is 15.5 cm.
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 PQRS (b) QN, if PS = 8 cm
Solution:-
From the question it is given that,
SR = 12 cm, QM = 7.6 cm
(a) We know that,
Area of the parallelogram = base × height
= SR × QM
= 12 × 7.6
= 91.2 cm^{2}
(b) Area of the parallelogram = base × height
91.2 = PS × QN
91.2 = 8 × QN
QN = 91.2/8
QN = 11.4 cm
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 1470 cm^{2}, AB = 35 cm and AD = 49 cm, find the length of BM and DL.
Solution:-
From the question it is given that,
Area of the parallelogram = 1470 cm^{2}
AB = 35 cm
AD = 49 cm
Then,
We know that,
Area of the parallelogram = base × height
1470 = AB × BM
1470 = 35 × DL
DL = 1470/35
DL = 42 cm
And,
Area of the parallelogram = base × height
1470 = AD × BM
1470 = 49 × BM
BM = 1470/49
BM = 30 cm
7. ΔABC 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 ΔABC. Also find the length of AD.
Solution:-
From the question it is given that,
AB = 5 cm, BC = 13 cm, AC = 12 cm
Then,
We know that,
Area of the ΔABC = ½ × base × height
= ½ × AB × AC
= ½ × 5 × 12
= 1 × 5 × 6
= 30 cm^{2}
Now,
Area of ΔABC = ½ × base × height
30 = ½ × AD × BC
30 = ½ × AD × 13
(30 × 2)/13 = AD
AD = 60/13
AD = 4.6 cm
8. ΔABC 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 ΔABC. What will be the height from C to AB i.e., CE?
Solution:-
From the question it is given that,
AB = AC = 7.5 cm, BC = 9 cm, AD = 6cm
Then,
Area of ΔABC = ½ × base × height
= ½ × BC × AD
= ½ × 9 × 6
= 1 × 9 × 3
= 27 cm^{2}
Now,
Area of ΔABC = ½ × base × height
27 = ½ × AB × CE
27 = ½ × 7.5 × CE
(27 × 2)/7.5 = CE
CE = 54/7.5
CE = 7.2 cm
Exercise 11.3
1. Find the circumference of the circle with the following radius: (Take π = 22/7)
(a) 14 cm
Solution:-
Given, radius of circle = 14 cm
Circumference of the circle = 2πr
= 2 × (22/7) × 14
= 2 × 22 × 2
= 88 cm
(b) 28 cm
Solution:-
Given, radius of circle = 28 cm
Circumference of the circle = 2πr
= 2 × (22/7) × 28
= 2 × 22 × 4
= 176 cm
(c) 21 cm
Solution:-
Given, radius of circle = 21 cm
Circumference of the circle = 2πr
= 2 × (22/7) × 21
= 2 × 22 × 3
= 132 cm
2. Find the area of the following circles, given that:
(a) Radius = 14 mm (Take π = 22/7)
Solution:
Given, radius of circle = 14 mm
Then,
Area of the circle = πr^{2}
= 22/7 × 14^{2}
= 22/7 × 196
= 22 × 28
= 616 mm^{2}
(b) Diameter = 49 m
Solution:
Given, diameter of circle (d) = 49 m
We know that, radius (r) = d/2
= 49/2
= 24.5 m
Then,
Area of the circle = πr^{2}
= 22/7 × (24.5)^{2}
= 22/7 × 600.25
= 22 × 85.75
= 1886.5 m^{2}
(c) Radius = 5 cm
Solution:
Given, radius of circle = 5 cm
Then,
Area of the circle = πr^{2}
= 22/7 × 5^{2}
= 22/7 × 25
= 550/7
= 78.57 cm^{2}
3. If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π = 22/7)
Solution:-
From the question it is given that,
Circumference of the circle = 154 m
Then,
We know that, Circumference of the circle = 2πr
154 = 2 × (22/7) × r
154 = 44/7 × r
r = (154 × 7)/44
r = (14 × 7)/4
r = (7 × 7)/2
r = 49/2
r = 24.5 m
Now,
Area of the circle = πr^{2}
= 22/7 × (24.5)^{2}
= 22/7 × 600.25
= 22 × 85.75
= 1886.5 m^{2}
So, the radius of circle is 24.5 and area of circle is 1886.5.
4. A gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs ₹ 4 per meter. (Take π = 22/7)
Solution:-
From the question it is given that,
Diameter of the circular garden = 21 m
We know that, radius (r) = d/2
= 21/2
= 10.5 m
Then,
Circumference of the circle = 2πr
= 2 × (22/7) × 10.5
= 462/7
= 66 m
So, the length of rope required = 2 × 66 = 132 m
Cost of 1 m rope = Rs. 4 [given]
Cost of 132 m rope = Rs. 4 × 132
=Rs. 528
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 π = 3.14)
Solution:-
From the question it is give that,
Radius of circular sheet R = 4 cm
A circle of radius to be removed r = 3 cm
Then,
The area of the remaining sheet = πR^{2 }– πr^{2}
= π (R^{2} – r^{2})
= 3.14 (4^{2} – 3^{2})
= 3.14 (16 – 9)
= 3.14 × 7
= 21.98 cm^{2}
So, the area of the remaining sheet is 21.98 cm^{2}.
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 ₹ 15. (Take π = 3.14)
Solution:-
From the question it is given that,
Diameter of the circular table = 1.5 m
We know that, radius (r) = d/2
= 1.5/2
= 0.75 m
Then,
Circumference of the circle = 2πr
= 2 × 3.14 × 0.75
= 4.71 m
So, the length of lace = 4.71 m
Cost of 1 m lace = ₹ 15 [given]
Cost of 4.71 m lace = ₹ 15 × 4.71
= ₹ 70.65
7. Find the perimeter of the adjoining figure, which is a semicircle including its diameter.
Solution:-
From the question it is given that,
Diameter of semi-circle = 10 cm
We know that, radius (r) = d/2
= 10/2
= 5 cm
Then,
Circumference of the semi-circle = πr
= (22/7) × 5
= 110/7
= 15.71 cm
Now,
Perimeter of the given figure = Circumference of the semi-circle + semi-circle diameter
= 15.71 + 10
= 25.71 cm
8. Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is Rs.15/m^{2}. (Take π = 3.14)
Solution:-
From the question it is given that,
Diameter of the circular table-top = 1.6 m
We know that, radius (r) = d/2
= 1.6/2
= 0.8 m
Then,
Area of the circular table-top = πr^{2}
= 3.14 × 0.8^{2}
= 3.14 × 0.8 ×0.8
= 2.0096 m^{2}
Cost for polishing 1 m^{2} area = ₹ 15 [given]
Cost for polishing 2.0096 m^{2} area = ₹ 15 × 2.0096
= Rs. 30.144
Hence, the Cost for polishing 2.0096 m^{2} area is Rs. 30.144.
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 area, the circle or the square? (Take π = 22/7)
Solution:-
From the question it is given that,
Length of wire that Shazli took =44 cm
Then,
If the wire is bent into a circle,
We know that, circumference of the circle = 2πr
44 = 2 × (22/7) × r
44 = 44/7 × r
(44 × 7)/44 = r
r = 7 cm
Area of the circle = πr^{2}
= 22/7 × 7^{2}
= 22/7 × 7 ×7
= 22 × 7
= 154 cm^{2}
Now,
If the wire is bent into a square,
The length of the each side of square = 44/4
= 11 cm
Area of the square = length of the side of square^{2}
= 11^{2}
= 121 cm^{2}
By comparing the two areas of the square and circle,
Clearly, circle encloses more area.
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 1cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. (Take π = 22/7)
Solution:-
From the question it is given that,
Radius of the circular card sheet = 14 cm
Radius of the two small circle = 3.5 cm
Length of the rectangle = 3 cm
Breadth of the rectangle = 1 cm
First we have to find out the area of circular card sheet, two circles and rectangle to find out the remaining area.
Now,
Area of the circular card sheet = πr^{2}
= 22/7 × 14^{2}
= 22/7 × 14 × 14
= 22 × 2 × 14
= 616 cm^{2}
Area of the 2 small circles = 2 × πr^{2}
= 2 × (22/7 × 3.5^{2})
= 2 × (22/7 × 3.5 × 3.5)
= 2 × ((22/7) × 12.25)
= 2 × 38.5
= 77 cm^{2}
Area of the rectangle = Length × Breadth
= 3 × 1
= 3 cm^{2}
Now,
The area of the remaining part = Card sheet area – (area of two small circles + rectangle
area)
= 616 – (77 + 3)
= 616 – 80
= 536 cm^{2}
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 π = 3.14)
Solution:-
From the question it is given that,
Radius of circle = 2 cm
Square sheet side = 6 cm
First we have to find out the area of square aluminium sheet and circle to find out the remaining area.
Now,
Area of the square = side^{2}
= 6^{2}
= 36 cm^{2}
Area of the circle = πr^{2}
= 3.14 × 2^{2}
= 3.14 × 2 × 2
= 3.14 × 4
= 12.56 cm^{2}
Now,
The area of the remaining part = Area of aluminum square sheet – area of circle
= 36 – 12.56
= 23.44 cm^{2}
12. The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take π = 3.14)
Solution:-
From the question it is given that,
Circumference of a circle = 31.4 cm
We know that,
Circumference of a circle = 2πr
31.4 = 2 × 3.14 × r
31.4 = 6.28 × r
31.4/6.28 = r
r = 5 cm
Then,
Area of the circle = πr^{2}
= 3.14 × 5^{2}
= 3. 14 × 25
= 78.5 cm
13. A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)
Solution:-
From the question it is given that,
Diameter of the flower bed = 66 m
Then,
Radius of the flower bed = d/2
= 66/2
= 33 m
Area of flower bed = πr^{2}
= 3.14 × 33^{2}
= 3.14 × 1089
= 3419.46 m
Now we have to find area of the flower bed and path together
So, radius of flower bed and path together = 33 + 4 = 37 m
Area of the flower bed and path together = πr^{2}
= 3.14 × 37^{2}
= 3.14 × 1369
= 4298.66 m
Finally,
Area of the path = Area of the flower bed and path together – Area of flower bed
= 4298.66 – 3419.46
= 879.20 m^{2}
14. A circular flower garden has an area of 314 m^{2}. 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 π = 3.14)
Solution:-
From the question it is given that,
Area of the circular flower garden = 314 m^{2}
Sprinkler at the centre of the garden can cover an area that has a radius = 12 m
Area of the circular flower garden = πr^{2}
314 = 3.14 × r^{2}
314/3.14 = r^{2}
r^{2} = 100
r = √100
r = 10 m
∴Radius of the circular flower garden is 10 m.
Since, the sprinkler can cover an area of radius 12 m
Hence, the sprinkler will water the whole garden.
15. Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)
Solution:-
From the figure,
Radius of inner circle = outer circle radius – 10
= 19 – 10
= 9 m
Circumference of the inner circle = 2πr
= 2 × 3.14 × 9
= 56.52 m
Then,
Radius of outer circle = 19 m
Circumference of the inner circle = 2πr
= 2 × 3.14 × 19
= 119.32 m
16. How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = 22/7)
Solution:-
From the question it is given that,
Radius of the wheel = 28 cm
Circumference of the wheel = 2πr
= 2 × 22/7 × 28
= 2 × 22 × 4
= 176 cm
Now we have to find the number of rotation of the wheel,
= Total distance to be covered/ circumference of wheel
= 352 m/176 cm
= 35200 cm/ 176 cm
= 200
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 π = 3.14)
Solution:-
From the question it is given that,
Length of the minute hand of the circular clock = 15 cm
Then,
Distance travelled by the tip of minute hand in 1 hour = circumference of the clock
= 2πr
= 2 × 3.14 × 15
= 94.2 cm
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:
Given: Length = 90 m
Breadth = 75 m
Area of the garden = l × b
= 90 m × 75 m = 6750 m^{2}
Length of the garden including path
= 90m + 5m + 5m = 100 m
Breadth of the garden including path
= 75m + 5m + 5m = 85m
Area of the garden including path
= l × b
= 100 m × 85 m = 8500 m^{2}
Area of the path = 8500 m^{2} – 6750 m^{2} = 1750 m^{2}
Hence, required area of path = 1750 m^{2} and area of the garden = 6750 m^{2} = 0.675 ha
Ex 11.4 Class 7 Maths 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:
Length of the park = 125 m
Breadth of the park = 65 m
Area of the park = l × b
= 125 m × 65 m = 8125 m^{2}
Length of the park including path
= 125 m + 3m + 3m = 131 m
Breadth of the park including path
= 65m + 3m + 3m = 71m
Area of the park including path
= 131 m × 71 m = 9301 m^{2}
∴ Area of the path
= 9301 m^{2} – 8125 m^{2} = 1176 m^{2}
Hence, the required area = 1176 m^{2}.
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:
Length = 8 cm, breadth = 5 cm
Area of the cardboard = l × b
= 8 cm × 5 cm = 40 cm^{2}
Width of the margin = 1.5 cm
Length of the inner cardboard
= 8 cm – 1.5 × 2 cm
= 8 cm – 3 cm = 5 cm
Breadth of the inner cardboard
= 5 cm – 1.5 × 2 cm
= 5 cm – 3 cm = 2 cm
Area of the inner rectangle = l × b
= 5 cm × 2 cm = 10 cm^{2} Area of the margin
= 40 cm^{2} – 10 cm^{2} = 30 cm^{2}
Hence, the required area = 30 cm^{2}.
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 Rs. 200 per m^{2}.
Solution:
Length of the room = 5.5 m
Breadth of the room = 4 m
∴ Area of the room = l × b = 5.5 m × 4 m = 22 m^{2}
Width of the verandah = 2.25 m
Length of the room including verandah
= 5.5 m + 2 × 2.25 m = 10 m
Breadth of the room including verandah
= 4 m + 2 × 2.25 m = 8.50 m^{2}
Area of the room including verandah = l × b
= 10 m × 8.50 m = 85 m^{2}
(i) Area of the verandah = 85 m^{2} – 22 m^{2}
= 63 m^{2}
(ii) Cost of cementing the floor of the verandah = Rs. 63 × 200 = Rs. 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 Rs. 40 per m^{2}.
Solution:
Area of the square garden = (Side)^{2}
= 30 m × 30 m = 900 m^{2}
Length of the garden excluding the path = 30 m – 2 × 1 m = 28 m
∴ Area of the garden excluding the path = 28 m × 28 m = 784 m^{2}
(i) Area of the path = 900 m^{2} – 784 m^{2}
= 116 m^{2}
(ii) Cost of the planting the remaining portion at the rate of Rs. 40 per m^{2}
= Rs. 40 × 784 =Rs. 31,360
Question 6.
Two cross roads, each of width 10 m, cut a 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:
Length of the road parallel to the length of the park = 700 m
Width of the road = 10 m
∴ Area of the road = l × b = 700 m × 10 m = 7000 m^{2}
Length of the road parallel to the breadth of the park = 300 m
Width of the road = 10 m Area of this road = l × b = 300 m × 10 = 3000 m^{2}
Area of the both roads
= 7000 m^{2} + 3000 m^{2} – Area of the common portion
= 10,000 m^{2} – 10 m × 10 m
= 10,000 m^{2} – 100 m^{2}
= 9900 m^{2} = 0.99 ha
Area of the park = l × b
= 700 m × 300 m = 210000 m^{2}
Area of the park excluding the roads
= 210000 m^{2} – 9900 m^{2}
= 200100 m^{2} = 20.01 ha
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 of the rate of Rs. 110 per m^{2}.
Solution:
Length of the road along the length of the field = 90 m
Breadth = 3 m
∴ Area of this road = l × b
= 90 m × 3 m = 270 m^{2}
Similarly, the area of the road parallel to the breadth of the field = l × b
= 60 m × 3 m = 180 m^{2} Area of the common portion
= 3m × 3m = 9m^{2}
(i) Area of the two roads
= 270 m^{2} + 180 m^{2} – 9 m^{2}
= 450 m^{2} – 9 m^{2} = 441 m^{2}
(ii) Cost of constructing the roads
= Rs. 110 × 441 = Rs. 48,510
Question 8.
Pragya wrapped a card around a circular pipe of radius 4 cm 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)
Solution:
Length of the cord = Circumference of the circular pipe
= 2πr = 2 × 3.14 × 4 = 25.12 cm
Perimeter of the square box
= 4 × side = 4 × 4 cm = 16 cm
Length of the cord left
= 25.12 cm – 16 cm = 9.12 cm
Yes, 9.12 cm cord is left.
Question 9.
The given 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.
Solution:
(i) Length of the lawn = 10 m
Breadth of the lawn = 5 m
Area of the lawn = l × b
= 10 m × 5 m = 50 m^{2}
(ii) Area of the circular flower bed = πr^{2}
22/7 x 2 x 2 = 88/7 m^{2 }
= 12.57 m^{2}
(iii) Area of the lawn excluding the area of the flower bed
Question 10.
In the following figures, find the area of the shaded portion.
Solution:
(i) Area of the rectangle = l × b
= 18 cm × (6 cm + 4 cm)
= 18 cm × 10 cm = 180 cm^{2}
Area of right triangle
=12×b×h=12×6×10=30 cm^{2}
Area of right ∆BCE = 12 × b × h
= 12 × 8 × 10 =40 cm^{2}
Area of the two right triangles
= 30 cm^{2} + 40 cm^{2} = 70 cm^{2}
Area of the shaded portion
= 180 cm^{2} – 70 cm^{2} = 110 cm^{2}
(ii) Area of the square PQRS = (Side)^{2}
= (20)^{2} = 400 cm^{2}
Area of the three triangles
= 50 cm^{2} + 100 cm^{2} + 100 cm^{2} = 250 cm^{2}
Area of the shaded portion
= 400 cm^{2} – 250 cm^{2} = 150 cm^{2}
Question 11.
Find the area of the quadrilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM ⊥ AC, DN ⊥ AC.
Area of the quadrilateral ABCD
= Area of ∆ABC + Area of ∆ADC
= 33 cm^{2} + 33 cm2 = 66 cm^{2}
Hence, the required area = 66 cm^{2}