Distance covered in one revolution = = 88m.
distance covered in 1 revolution =
= 2 x (22/7) x 20 = 880/7 cm
required no of revolutions = 17600 x (7/880) = 140
Let x and y be the length and breadth of the rectangle respectively.
Then, x - 4 = y + 3 or x - y = 7 ----(i)
Area of the rectangle =xy; Area of the square = (x - 4) (y + 3)
(x - 4) (y + 3) =xy <=> 3x - 4y = 12 ----(ii)
Solving (i) and (ii), we get x = 16 and y = 9.
Perimeter of the rectangle = 2 (x + y) = [2 (16 + 9)] cm = 50 cm.
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.
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
Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile =
Required number of tiles = = 814
let the radius of the pool be Rft
Radius of the pool including the wall = (R+4)ft
Area of the concrete wall = = sq feet
=> 8
Radius of the pool R = 20ft
Let length = x meters, then breadth = 0.6x
Given that perimeter = 800 meters
=> 2[ x + 0.6x] = 800
=> x = 250 m
Length = 250m and breadth = 0.6 x 250 = 150m
Area = 250 x 150 = 37500 sq.m
Area of the sheet =
Area used for typing =
required % = =64%
circumference of a circle = 2 r
=> 2 × 22/7 × r ? r = 37
=> 37/7 × r = 37
=> r = 7 cm.
Diameter D = 2r = 7×2 = 14 cm.
The triangle with sides 26 cm, 24 cm and 10 cm is right angled, where the hypotenuse is 26 cm.
Area of the triangle =1/2 x base x height => 1/2 x 24 x 10 = 120 sq.cm
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