Exercise 11.1 : Solutions of Questions on Page Number : 208
Q1 : 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.
Answer :
(i) Area = Length x Breadth
= 500 x 300
= 150000 m^{2}
(ii)Cost of 1 m^{2} land = Rs 10000
Cost of 150000 m^{2} land = 10000 x 150000 = Rs 1500000000
Q2 : Find the area of a square park whose perimeter is 320 m.
Answer :
Perimeter = 320 m
4 × Length of the side of park = 320
Length of the side of park =
Area = (Length of the side of park)^{2} = (80)^{2} = 6400 m^{2}
Q3 : 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.
Answer :
Area = Length × Breadth = 440 m^{2}
22 × Breadth = 440
Breadth == 20 m
Perimeter = 2 (Length + Breadth)
= 2 (22 + 20) = 2 (42) = 84 m^{2}
Q4 : The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.
Answer :
Perimeter = 2 (Length + Breadth) = 100 cm
2 (35 + Breadth) = 100
35 + B = 50
B = 50 – 35 = 15 cm
Area = Length x Breadth = 35 x 15 = 525 cm^{2}
Q5 : 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.
Answer :
Area of square park = (One of its sides)^{2} = (60)^{2} = 3600 m^{2}
Area of rectangular park = Length x Breadth = 3600
90 x Breadth = 3600
Breadth = 40 m
Q6 : 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?
Answer :
Perimeter of rectangle = Perimeter of square
2 (Length + Breadth) = 4 × Side
2 (40 + 22) = 4 × Side
2 × 62 = 4 × Side
Side == 31 cm
Area of rectangle = 40 × 22 = 880 cm^{2}
Area of square = (Side)^{2} = 31 × 31 = 961 cm^{2}
Therefore, the square-shaped wire encloses more area.
Q7 : 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.
Answer :
Perimeter = 2 (Length + Breadth) = 130
2 (Length + 30) = 130
Length + 30 = 65
Length = 65 – 30 = 35 cm
Area = Length x Breadth = 35 x 30 = 1050 cm^{2}
Q8 : 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 (see the given figure). Find the cost of white washing the wall, if the rate of white washing the wall is Rs 20 per m^{2}.
Answer :
Area of wall = 4.5 x 3.6 = 16.2 m^{2}
Area of door = 2 x 1 = 2 m^{2}
Area to be white-washed = 16.2 – 2 = 14.2 m^{2}
Cost of white-washing 1 m^{2} area = Rs 20
∴Cost of white-washing 14.2 m^{2} area = 14.2 x 20 = Rs 284
Exercise 11.2 : Solutions of Questions on Page Number : 216
Q1 : Find the area of each of the following parallelograms:
Answer :
Area of parallelogram = Base x Height
(a) Height= 4 cm
Base = 7 cm
Area of parallelogram = 7 x 4 = 28 cm2
(b) Height= 3 cm
Base = 5 cm
Area of parallelogram = 5 x 3 = 15 cm2
(c) Height= 3.5 cm
Base = 2.5 cm
Area of parallelogram = 2.5 x 3.5 = 8.75 cm2
(d) Height= 4.8 cm
Base = 5 cm
Area of parallelogram = 5 x 4.8 = 24 cm2
(e) Height= 4.4 cm
Base = 2 cm
Area of parallelogram = 2 x 4.4 = 8.8 cm2
Q3 : Find the missing values:
So No |
Base |
Height |
Area of 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} |
Answer :
Area of parallelogram = Base × Height
(a) b = 20 cm
h = ?
Area = 246 cm2
20 × h = 246
Therefore, the height of such parallelogram is 12.3 cm.
(b) b = ?
h = 15 cm
Area = 154.5 cm2
b × 15 = 154.5
b = 10.3 cm
Therefore, the base of such parallelogram is 10.3 cm.
(c) b = ?
h = 8.4 cm
Area = 48.72 cm2
b × 8.4 = 48.72
Therefore, the base of such parallelogram is 5.8 cm.
(d) b = 15.6 cm
h = ?
Area = 16.38 cm2
15.6 × h = 16.38
Therefore, the height of such parallelogram is 1.05 cm.
Q4 :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} |
Answer :
(a) b = 15 cm
h = ?
Area =
Therefore, the height of such triangle is 11.6 cm.
(b) b = ?
h = 31.4 mm
Area =
Therefore, the base of such triangle is 80 mm.
(c) b = 22 cm
h = ?
Area =
Therefore, the height of such triangle is 15.5 cm.
Q5 : PQRS is a parallelogram (see the given figure). 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
Answer :
(a) Area of parallelogram = Base × Height = SR × QM
= 7.6 × 12 = 91.2 cm^{2}
(b) Area of parallelogram = Base × Height = PS × QN = 91.2 cm^{2}
QN × 8 = 91.2
Q6 : DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (see the given figure). If the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL.
Answer :
Area of parallelogram = Base × Height = AB × DL
1470 = 35 × DL
Also, 1470 = AD × BM
1470 = 49 × BM
Q7 : ΔABC is right angled at A (see the given figure). 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.
Answer :
Area == 30 cm^{2
}
Q8 : ΔABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (see the given figure). 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?
Answer :
Exercise 11.3 : Solutions of Questions on Page Number : 223
Q12 : The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take π = 3.14)
Answer :
Circumference = 2πr = 31.4 cm
2 x 3.14 x r = 31.4
r = 5 cm
Area = 3.14 x 5 x 5 = 78.50 cm^{2}
Q14 : 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)
Answer :
Area = πr^{2} = 314 m^{2}
3.14 x r^{2} = 314
r^{2} = 100
r = 10 m
Yes, the sprinkler will water the whole garden.
Q17 : 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)
Answer :
Distance travelled by the tip of minute hand = Circumference of the clock
= 2πr = 2 x 3.14 x 15
= 94.2 cm
Exercise 11.4 : Solutions of Questions on Page Number : 226
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.
Answer :
Length (l) of garden = 90 m
Breadth (b) of garden = 75 m
Area of garden = l × b = 90 × 75 = 6750 m^{2}
From the figure, it can be observed that the new length and breadth of the garden, when path is also included, are 100m and 85m respectively.
Area of the garden including the path = 100 × 85 = 8500 m^{2}
Area of path = Area of the garden including the path − Area of garden
= 8500 − 6750 = 1750 m^{2}
1 hectare = 10000 m^{2}
Therefore, area of garden in hectare
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 :
Length (l) of park = 125 m
Breadth (b) of park = 65 m
Area of park = l × b = 125 × 65 = 8125 m^{2}
From the figure, it can be observed that the new length and breadth of the park, when path is also included, are 131 m and 71 m respectively.
Area of the park including the path = 131 × 71 = 9301 m^{2}
Area of path = Area of the park including the path − Area of park
= 9301 − 8125 = 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.
Length (l) of cardboard = 8 cm
Breadth (b) of cardboard = 5 cm
Area of cardboard including margin = l × b = 8 × 5 = 40 cm^{2}
From the figure, it can be observed that the new length and breadth of the cardboard, when margin is not included, are 5 cm and 2 cm respectively.
Area of the cardboard not including the margin = 5 × 2 = 10 cm^{2}
Area of the margin = Area of cardboard including the margin − Area of cardboard not
including the margin
= 40 − 10 = 30 cm^{2}
Question 4 :
A verandahof 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}.
Answer :
(i)
Length (l) of room = 5.5 m
Breadth (b) of room = 4 m
Area of room = l × b = 5.5 × 4 = 22 m^{2}
From the figure, it can be observed that the new length and breadth of the room, when verandah is also included, are 10 m and 8.5 m respectively.
Area of the room including the verandah = 10 × 8.5 = 85 m^{2}
Area of verandah = Area of the room including the verandah − Area of room
= 85 − 22 = 63 m^{2}
(ii) Cost of cementing 1 m2 area of the floor of the verandah = Rs 200
Cost of cementing 63 m2 area of the floor of the verandah
= 200 x 63
= 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}.
(i)
Side (a) of square garden = 30 m
Area of square garden including path = a^{2} = (30)^{2} = 900 m^{2}
From the figure, it can be observed that the side of the square garden, when path is not included, is 28 m.
Area of the square garden not including the path = (28)^{2} = 784 m^{2}
Area of path = Area of the square garden including the path − Area of square
garden not including the path
= 900 − 784 = 116 m^{2}
(ii)
Cost of planting grass in 1 m^{2} area of the garden = Rs 40
Cost of planting grass in 784 m^{2} area of the garden = 784 x 40 = Rs 31360
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 :
Length (l) of park = 700 m
Breadth (b) of park = 300 m
Area of park = 700 × 300 = 210000 m^{2}
Length of road PQRS = 700 m
Length of road ABCD = 300 m
Width of each road = 10 m
Area of the roads = ar (PQRS) + ar (ABCD) − ar (KLMN)
= (700 × 10) + (300 × 10) − (10 × 10)
= 7000 + 3000 − 100
= 10000 − 100 = 9900 m^{2} = 0.99 hectare
Area of park excluding roads = 210000 − 9900 = 200100 m^{2 }= 20.01 hectare
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 Rs 110 per m^{2}.
Answer :
Length (l) of field = 90 m
Breadth (b) of field = 60 m
Area of field = 90 × 60 = 5400 m^{2}
Length of road PQRS = 90 m
Length of road ABCD = 60 m
Width of each road = 3 m
Area of the roads = ar (PQRS) + ar (ABCD) − ar (KLMN)
= (90 × 3) + (60 × 3) − (3 × 3)
= 270 + 180 − 9 = 441 m^{2}
Cost for constructing 1 m^{2} road = Rs 110
Cost for constructing 441 m^{2} road = 110 × 441 = Rs 48510
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 :
(i) Area of whole land = Length × Breadth = 10 × 5 = 50 m^{2}
(ii) Area of flower bed = πr^{2} = 3.14 × 2 × 2 = 12.56 m^{2}
(iii) Area of lawn excluding the flower bed = Area of whole land − Area of
flower bed
= 50 − 12.56 = 37.44 m^{2}
(iv)Circumference of flower bed = 2πr = 2 × 3.14 × 2 = 12.56 m
Question 10 :
In the following figures, find the area of the shaded portions:
Answer :
(i)
Area of EFDC = ar (ABCD) − ar (BCE) − ar (AFE)
= (18 × 10) −(10 × 8) − (6 × 10)
= 180 − 40 − 30 = 110 cm^{2}
(ii)
ar (QTU) = ar (PQRS) − ar (TSU) − ar (RUQ) − ar (PQT)
= (20 × 20) −(10 × 10) − (20 × 10) −(20 × 10)
= 400 − 50 − 100 − 100 = 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
Answer :
ar (ABCD) = ar (ABC) + ar (ADC)
= (3 × 22) + (3 × 22)
= 33 + 33 = 66 cm^{2}