let ABC be the isosceles triangle, the AD be the altitude
Let AB = AC = x then BC= 32-2x [because parameter = 2 (side) + Base]
since in an isoceles triange the altitude bisects the base so
BD = DC = 16-x
In a triangle ADC,
BC = 32-2x = 32-20 = 12 cm
Hence, required area = = = 60 sq cm
The diameter is equal to the shortest side of the rectangle.
So radius= 14/2 = 7cm.
Therefore,
ratio =
length of wire =
= 2 x (22/7 ) x 56 = 352 cm
side of the square = 352/4 = 88cm
area of the square = 88 x 88 = 7744sq cm
let the side of the square be x meters
length of two sides = 2x meters
diagonal =
= 1.414x m
saving on 2x meters = .59x m
saving % = %
= 30% (approx)
Let original length = x metres and original breadth = y metres.
Original area = xy sq.m
Increased length = and Increased breadth =
New area =
The difference between the Original area and New area is:
Increase % = = 44%
Area of the field = Total cost/rate = (333.18/25.6) = 13.5 hectares
Let altitude = x metres and base = 3x metres.
Then,
Base = 900 m and Altitude = 300 m.
Area to be plastered =
=
Cost of plastering =
ratio =
Area of the room=(544 * 374)
size of largest square tile= H.C.F of 544 & 374 = 34 cm
Area of 1 tile = (34 x 34)
Number of tiles required== [(544 x 374) / (34 x 34)] = 176
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