Let original length = x and original breadth = y.
Decrease in area = xy - = (7/25)xy
Decrease % = (7/25)xy * (1/xy) * 100 = 28%
We have: l = 20 ft and lb = 680 sq. ft.So, b = 34 ft.
Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
Let the length of the altitude be h.
Then, h =
=
= 4.5
square and a rectangle with equal areas will satisfy the
relation p < q
100 cm is read as 102 cm
a = 100 x 100 sq.cm and b = 102 x 102 sq.cm
then, Increase in area = [ (102 x 102) - (100 x 100) ] sq.cm
percentage error = = 4.04%
We know that,
Area of rectangle A = L x W = 5 x 3 = 15 sq.inches.
Let breadth = x metres.
Then, length = (x + 20) metres.
Perimeter =5300/23.50
2[(x + 20) + x] = 200
2x + 20 = 100
2x = 80
x = 40.
Hence, length = x + 20 = 60 m.
Let the radius of the circle be r cm. Then,
r=3.
let the diagonals of the squares be 2x and 5x respectively
ratio of their areas = = 4:25
Area of a square with side a = a² sq unts
Area of a triangle with base a = (1/2) * a * h sq.unts
a² =1/2 *a *h
=> h = 2a
altitude of the triangle is 2a
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