Area Problems
The diameter of the driving wheel of a bus is 140 cm. How many revolution, per minute must the wheel make in order to keep a speed of 66 kmph ?
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.
Speed = 12 x (5/18) m/sec =10/3 m/sec there4; perimeter = (10/3) x 15 x 60 m=3000 m? 2( a + 4a) = 3000 m? a = 300 mSo, length = 1200 m and breadth = 300 m ? Area = (1200 x 300 ) m2 = 360000m2
Area of the field =1215/135 = 9 hec= 90000 m2 [1 hec =10000 m2]? Side of the field = ?90000 = 300 mPerimeter of the field = 4 x 300 = 1200 mNow, cost of putting a fence around field = (1200 x 75)/100 = ? 900
In a triangleSum of two sides is always greater than 3rd side i.e., x < 25 + 15 = 40 .....(i)Difference of two sides is always less than 3rd side i.e., 25 - 15 = 10 < x ...(ii) From Eqs. (i) and (ii) , we get 10 < x < 40
Let the side of the square = 100 m So area of square = 100 x 100 = 10000.New length = 140 m, New breadth = 130 mIncrease in area = [(140 x 130) - (100 x 100)] m2= 8200 m2? Required increase percent = (8200/ 10000) x 100 % = 82%
EXAMIANS