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Question 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 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
= 15,0000 m^{2}
(ii) Cost of the land for 1 m^{2} = Rs. 10000
Cost of the land for 150000 m^{2} = 10000 × 150000
= Rs. 15,00000000
Question 2: Find the area of a square park whose perimeter is 320 m.
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} m^{2} = 6400 m^{2}
Question 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:
Area of rectangular park = 440 m^{2}
⇒ length x breadth = 440 m^{2}
⇒ 22 x breadth = 440
⇒
Now, Perimeter of rectangular park
= 2 [length + breadth]
= 2 [22 + 20]
= 2 x 42 = 84 m
Thus, the perimeter of rectangular park is 84 m.
Question 4: The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.
Solution:
Perimeter of the rectangular sheet = 100 cm
⇒ 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 x breadth
= 35 x 15 = 525 cm^{2}
Thus, breadth and area of rectangular sheet are 15 cm and 525 cm^{2} respectively.
Question 5: The area of a square park is the same as 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.
Solution:
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 x side = length x breadth
⇒ 60 x 60 = 90 x breadth
Thus, the breadth of the rectangular park is 40 m.
Question 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:
According to the question,
Perimeter of square = Perimeter of rectangle
⇒ 4 x side = 2 (length + breadth)
⇒ 4 x side = 2 (40 + 22)
⇒ 4xside = 2x62
Thus, the side of the square is 31 cm.
Now, Area of rectangle = length x breadth = 40x22 = 880 cm^{2}
And Area of square = side x side = 31 x 31 = 961 cm^{2}
Therefore, on comparing, the area of square is greater than that of rectangle.
Question 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:
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 x 30 = 1050 cm^{2}
Thus, the area of rectangle is 1050 cm^{2}.
Question 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 m^{2}.
Solution:
Area of rectangular door = length x breadth = 2 m x 1 m = 2 m^{2}
Area of wall including door = length x breadth = 4.5 m x 3.6 m = 16.2 m^{2}
Now, Area of wall excluding door = Area of wall including door  Area of door = 16.22 = 14.2 m^{2}
Since, the rate of white washing of 1 m^{2} the wall = Rs 20
Therefore, the rate of white washing of 14.2 m^{2} the wall = 20 x 14.2 = Rs 284
Thus, the cost of white washing the wall excluding the door is Rs 284.
Solution:
We know that the area of parallelogram = base x height
(a) Here base = 7 cm and height = 4 cm
∴ Area of parallelogram = 7x4 = 28 cm^{2}
(b) Here base = 5 cm and height = 3 cm
∴ Area of parallelogram = 5x3 = 15 cm^{2}
(c) Here base = 2.5 cm and height = 3.5 cm
∴ Area of parallelogram = 2.5 x 3.5 = 8.75 cm^{2}
(d) Here base = 5 cm and height = 4.8 cm
∴ Area of parallelogram = 5 x 4.8 = 24 cm^{2}
(e) Here base = 2 cm and height = 4.4 cm
∴ Area of parallelogram = 2 x 4.4 = 8.8 cm^{2}
Question 2: Find the area of each of the following triangles:
Solution:
We know that the area of triangle
(a) Here, base = 4 cm and height = 3 cm
∴ Area of triangle
(b) Here, base = 5 cm and height = 3.2 cm
∴ Area of triangle
(c) Here, base = 3 cm and height = 4 cm
∴ Area of triangle
(d) Here, base = 3 cm and height = 2 cm
∴ Area of triangle
Question 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. 
 84 cm  48.72 cm^{2} 
d.  15.6 cm 
 16.38 cm^{2} 
Solution:
We know that the area of parallelogram = base x height
(a) Here, base = 20 cm and area = 246 cm^{2}
∴ Area of parallelogram = base x height
⇒ 246 = 20 x height
(b) Here, height = 15 cm and area = 154.5 cm^{2}
∴ Area of parallelogram = base x height
⇒ 154.5 = base x 15
(c) Here, height = 8.4 cm and area = 48.72 cm^{2}
∴ Area of parallelogram = base x height
⇒ 48.72 = base x 8.4
(d) Here, base = 15.6 cm and area = 16.38 cm^{2}
∴ Area of parallelogram = base x height
⇒ 16.38 = 15.6 x height
Thus, the missing values are:
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  84 cm  48.72 cm^{2} 
d.  15.6 cm  1.05  16.38 cm^{2} 
Question 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:
We know that the area of triangle
In first row, base = 15 cm and area = 87 cm^{2}
In second row, height = 31.4 mm and area = 1256 mm^{2}
In third row, base = 22 cm and area = 170.5 cm^{2}
Thus, the missing values are:
Base  Height  Area of triangle 
15 cm  11.6 cm  87 cm^{2} 
80 mm  31.4 mm  1256 mm^{2} 
22 cm  15.5 cm  170.5 cm^{2} 
Question 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:
Given:
SR = 12 cm, QM = 7.6 cm, PS = 8 cm,
(a) Area of parallelogram = base x height
= 12 x 7.6 = 91.2 cm^{2}
(b) Area of parallelogram = base x height
Question 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:
Given:
Area of parallelogram = 1470 cm^{2}
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.
Question 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.
Solution:
Question 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:
In ΔABC, AD = 6 cm and BC = 9 cm
Area of triangle
Again, Area of triangle
Thus, height from C to AB i.e., CE is 7.2 cm.
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