Area to be plastered =
=
Cost of plastering =
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.
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
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
Let the breadth of the given rectangle be x then length is 2x.
thus area of the given rect is
after dec 5cm from length and inc 5cm breadth , new lenght becomes 2x-5 and breadth is x+5.thus new area =(2x-5)(x+5)=
since new area is 75 units greater than original area thus
5x=75+25
5x=100
therefore x=20
hence length of the rectangle is 40 cm.
Circumference = No.of revolutions * Distance covered
Distance to be covered in 1 min. = (66 X1000)/60 m = 1100 m.
Circumference of the wheel = 2 x (22/7) x 0.70 m = 4.4 m.
Number of revolutions per min. =(1100/4.4) = 250.
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