Exercise: Area
Questions for: Area
A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?
Answer: D
We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres?
Answer: E
Let breadth = x metres.
Then, length = (x + 20) metres.
| Perimeter = |  |
5300 |  |
m = 200 m. |
| 26.50 |
2[(x + 20) + x] = 200
2x + 20 = 100
2x = 80
x = 40.
Hence, length = x + 20 = 60 m.
The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?
Answer: B
Let original length = x and original breadth = y.
Original area = xy.
| New length = | x | . |
| 2 |
New breadth = 3y.
| New area = | |
x | x 3y | |
= | 3 | xy. |
| 2 | 2 |
| Increase % = |
|
1 | xy x | 1 | x 100 | % |
= 50%. |
| 2 | xy |
The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:
Answer: D
We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
Area = (l x b) = (63 x 40) m2 = 2520 m2.
What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?
Answer: A
Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile = (41 x 41) cm2.
| Required number of tiles = |
|
1517 x 902 | |
= 814. |
| 41 x 41 |
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