a : b =(1/2 * x * H) : (1/2 * y * h)
bxH = ayh => H/h =ay/bx
Hence H:h = ay:bx
area of theplot = 110 * 65 = 7150 sq.m
area of the plot excluding the path = (110-5)* (65-5) = 6300 sq.m
area of the path = 7150 - 6300 =850 sq.m
cost of gravelling the path = 850 x (80/100) = 680 Rs
Length of the first carpet = (1.44)(6) = 8.64 cm
Area of the second carpet = 8.64(1 + 40/100) 6 (1 + 25/100)
= 51.84(1.4)(5/4) sq m = (12.96)(7) sq m
Cost of the second carpet = (45)(12.96 x 7) = 315 (13 - 0.04) = 4095 - 12.6 = Rs. 4082.40
The cow can graze an area of 2826 sq.m i.e it is able to cover area of circle of 2826 sq.m
Therefore,
=2826
=(2826×7/22) = 899.18 = 900
r = 30 m.
Therefore, the radius of the circle implies the length of the rope = 30 mts.
perimeter of the plot = 2(90+50) = 280m
no of poles =280/5 =56m
Let each side of the square be a. Then, area = .
New side =
. New area =
=
Increase in area =
Increase% =
% = 56.25%.
Take a square of side 4cm and a rectangle having l=6cm and b=2cm
then perimeter of square = perimeter of rectangle
area of square = 16 sq.cm
area of rectangle = 12 sq.cm
Hence a >b
Let l = 9 ft.
Then l + 2b = 37
=> 2b = 37 ? l = 37 ? 9 = 28
b = 28/2 = 14 ft.
Area = lb = 9 × 14 = 126 sq. ft.
Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile = (41 x 41) sq.cm
Required number of tiles = 1517 x (902/41) x 41 = 814
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