EXERCISE 11.1
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 Ã— Breadth
= 500 Ã— 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 Ã— 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 Ã— 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}
â‡’ length x breadth = 440 m^{2}
â‡’ 22 x breadth = 440
â‡’
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
â‡’ 2 (length + breadth) = 100 cm
â‡’ 2 (35 + breadth) = 100
â‡’ 35 + breadth = 100/2
â‡’ 35 + breadth = 50
â‡’ breadth = 50 - 35
â‡’ 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
â‡’ side x side = length x breadth
â‡’ 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
â‡’ 4 x side = 2 (length + breadth)
â‡’ 4 x side = 2 (40 + 22)
â‡’ 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
â‡’ 2 (length + breadth) = 130 cm
â‡’ 2 (length + 30) = 130
â‡’ length + 30 = 130/2
â‡’ length+ 30 = 65
â‡’ 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 = 16.2-2 = 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.
EXERCISE 11.2
Question 1: Find the area of each of the following parallelograms:
Solution:
We know that the area of parallelogram = base x height
(a) Here base = 7 cm and height = 4 cm
âˆ´ Area of parallelogram = 7x4 = 28 cm^{2}
(b) Here base = 5 cm and height = 3 cm
âˆ´ Area of parallelogram = 5x3 = 15 cm^{2}
(c) Here base = 2.5 cm and height = 3.5 cm
âˆ´ Area of parallelogram = 2.5 x 3.5 = 8.75 cm^{2}
(d) Here base = 5 cm and height = 4.8 cm
âˆ´ Area of parallelogram = 5 x 4.8 = 24 cm^{2}
(e) Here base = 2 cm and height = 4.4 cm
âˆ´ 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
âˆ´ Area of triangle
(b) Here, base = 5 cm and height = 3.2 cm
âˆ´ Area of triangle
(c) Here, base = 3 cm and height = 4 cm
âˆ´ Area of triangle
(d) Here, base = 3 cm and height = 2 cm
âˆ´ 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}
âˆ´ Area of parallelogram = base x height
â‡’ 246 = 20 x height
(b) Here, height = 15 cm and area = 154.5 cm^{2}
âˆ´ Area of parallelogram = base x height
â‡’ 154.5 = base x 15
(c) Here, height = 8.4 cm and area = 48.72 cm^{2}
âˆ´ Area of parallelogram = base x height
â‡’ 48.72 = base x 8.4
(d) Here, base = 15.6 cm and area = 16.38 cm^{2}
âˆ´ Area of parallelogram = base x height
â‡’ 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
â‡’ 1470 = 35 x DL
â‡’ DL = 1470/35
â‡’ DL = 42 cm
Again, Area of parallelogram = base x height
â‡’ 1470 = 49 x BM
â‡’ BM = 1470/49
â‡’ BM = 30 cm
Thus, the lengths of DL and BM are 42 cm and 30 cm respectively.
Question 7: Î” 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 Î” ABC. Also, find the length of AD.
Solution:
Question 8: Î”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 Î”ABC. What will be the height from C to AB i.e., CE?
Solution:
In Î”ABC, 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.