Area Problems
Three circles of radius 3.5cm are placed in such a way that each circle touches the other two. The area of the portion enclosed by the circles is
Let length of rectangle = 5kand breadth of rectangle = 3kAccording to the quecation,5k - 3k = 8 ? 2k =8? k = 4? Lenght = 5k = 5 x 4 = 20 mBreadth = 3k = 3 x 4 = 12 m? Required area = Lenght x Breadth = 20 x 12 = 240 sq m
Let original length = x metres and original breadth = y metres. Original area = (xy) m2. New length = ❨ 120 x ❩m = ❨ 6 x ❩m. 100 5 New breadth = ❨ 120 y ❩m = ❨ 6 y ❩m. 100 5 New Area = ❨ 6 x x 6 y ❩m2 = ❨ 36 xy ❩m2. 5 5 25 The difference between the original area = xy and new-area 36/25 xy is = (36/25)xy - xy = xy(36/25 - 1) = xy(11/25) or (11/25)xy ∴ Increase % = ❨ 11 xy x 1 x 100 ❩% = 44%
Increase in circumference of circle = 5%? Increase in radius is also 5%.Now, increase in area of circle = 2a + (a2/100) %Where, a = increase in radius= 2 x 5 + (5 x 5)/100 % = 10.25%
let the side of the square be x meterslength of two sides = 2x metersdiagonal = 2 x = 1.414x m saving on 2x meters = .59x m saving % = 0 . 59 x 2 x * 100 % = 30% (approx)
EXAMIANS