Area Problems
The area of a rectangle is thrice that of square. Length of the rectangle is 40 cm and the breadth of the rectangle is ( 3 / 2 ) times that of the side of the square. The side of the square in cm is?
Let the side of the square = y cmThen, breadth of the rectangle = 3y/2 cm ? Area of rectangle = (40 x 3y/2) cm2= 60y cm2? 60y = 3y2? y = 20Hence, the side of the square = 20 cm
Let the side of the square(ABCD) be x metres. Then, AB + BC = 2x metres. AC = √2x = (1.41x) m. Saving on 2x metres = (0.59x) m. Saving % = ❨ 0.59x x 100 ❩% = 30%
Let original length = x and original breadth = y. Decrease in area = xy - ❨ 80 x x 90 y ❩ 100 100 = ❨ xy - 18 xy ❩ 25 = 7 xy. 25 ∴ Decrease % = ❨ 7 xy x 1 x 100 ❩% = 28%
clearly the cow will graze a circular field of area 9856 sq m and radius equal to the length of the rope.
Let the length of the rope be r mts then,
= 9856
= (9856 × 7) / 22 = 3136
r = 56 m
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.
Area of the plot = (3 x 1200) m2= 3600 m2Let breadth = y metersThen Length = 4y meters,Now area = 4y x y = 3600 m2? y2 = 900 m2? y = 30 m? Length of plot = 4y m= (4 x 30) m=120 m
EXAMIANS