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 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
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
Let the sides of trapezium be 5k and 3k, respectively According to the question, (1/2) x [(5k + 3k) x 12] = 384? 8k = (384 x 2)/12 = 64 ? k = 64/8 = 8 cmLength of smaller of the parallel sides = 8 x 3 = 24 cm
Diameter of the circle = 13 + 5 = 18 cm? Radius = Diameter/2 =18/2 = 9 cm Area of the circle = ?r2 = (22/7) x 92 = (22 x 81)/9 = 1782/7 = 254.57 sq cm= 255 sq cm
Let h be the altitude of triangle.So area of triangle = (1/2)xh area of square = x2From question area of square = area of triangle x2 = (1/2)xh ? h = (2x2)/x = 2x
EXAMIANS