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%
let ABCD be the given parallelogram
area of parallelogram ABCD = 2 x (area of triangle ABC)
now a = 30m, b = 14m and c = 40m
s=1/2 x (30+14+40) = 42
Area of triangle ABC =
= = 168sq m
area of parallelogram ABCD = 2 x 168 = 336 sq m
(or)
Also,
= 41 + 40 = 81
(l + b) = 9.
Perimeter = 2(l + b) = 18 cm.
Let original radius = R.
New radius = =
Original area = and new area =
Decrease in area = = 75%
let length = x and breadth = y then
2(x+y) = 46 => x+y = 23
x²+y² = 17² = 289
now (x+y)² = 23²
=> x²+y²+2xy= 529
=> 289+ 2xy = 529
=> xy = 120
area = xy = 120 sq.cm
let each side of the square be a , then area =
As given that The side is increased by 25%, then
New side = 125a/100 = 5a/4
New area =
Increased area=
Increase %= % = 56.25%
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 the side of the square be x. Then, its diagonal =
Radius of incircle =
Radius of circum circle=
Required 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 =
The diameter is equal to the shortest side of the rectangle.
So radius= 14/2 = 7cm.
Therefore,
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