Let length of plot = L meters, then breadth = L - 20 meters
and perimeter = 2[L + L - 20] = [4L - 40] meters
[4L - 40] * 26.50 = 5300
[4L - 40] = 5300 / 26.50 = 200
4L = 240
L = 240/4= 60 meters.
Let the two parallel sides of the trapezium be a cm and b cm.
Then, a - b = 4
And,
Solving (i) and (ii), we get: a = 27, b = 23.
So, the two parallel sides are 27 cm and 23 cm.
area of a square =
We know that,
Area of trapezium = 1/2 x (Sum of parallel sides) x (Distance between Parallel sides)
= 1/2 x (12 + 10) x 14
= 22 x 14/2
= 22 x 7
= 154 sq. cm
Let the breadth of the rectangle = b mts
Then Length of the rectangle = b + 6 mts
Given perimeter = 84 mts
2(L + B) = 84 mts
2(b+6 + b) = 84
2(2b + 6) = 84
4b + 12 = 84
4b = 84 - 12
4b = 72
b = 18 mts
=> Length = b + 6 = 18 + 6 = 24 mts
Now, required Area of the rectangle = L x B = 24 x 18 = 432 sq. mts
Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile =
Required number of tiles = = 814
Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres
Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which is very important.
Hence number of poles required = 280 / 5 = 56
Perimeter of the rectangle is given by 3000/10 = 300 mts
But we know,
The Perimeter of the rectangle = 2(l + b)
Now,
2(8x + 7x) = 300
30x = 300
x = 10
Required, Area of rectangle = 8x x 7x = 56 x 100 = 5600 sq. mts.
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