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. Then, its diagonal =
Radius of incircle =
Radius of circum circle=
Required ratio =
Let the side of the square(ABCD) be x meters.
Then, AB + BC = 2x metres.
AC = = (1.41x) m.
Saving on 2x metres = (0.59x) 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%
Other side = [(17 x 17) - (15 x 15)] = (289 - 225) = 8m
Area = 15 x 8 =120 sq. 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
We know that d=?2s
Given diagonal = 20 cm
=> s = 20/ cm
Therefore, perimeter of the square is 4s = 4 x 20/ = 40 cm.
Let breadth = x m
Then, length = (x+5)m
Area of a rectangle = Length x Breadth
x(x+5) = 750
x² + 5x - 750= 0
(x+30)(x-25)= 0
x = 25 or x = -30
Hence, breadth x = 25m
=> Length = x + 5 = 25 + 5 = 30m.
Area of the park = (60 x 40) = 2400
Area of the lawn = 2109
Area of the crossroads = (2400 - 2109) = 291
Let the width of the road be x metres. Then,
(x - 97)(x - 3) = 0
x = 3.
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