Area Problems
The two parallel sides of a trapezium are 1 meters and 2 meters respectively . The perpendicular distance between them is 6 meters. The area of the trapezium is?
Original circumference = 2?r New circumference = (150 /100) x 2 ?r = 3?r 2?R = 3?r? R = 3r/2 Original area = ?r2New area = ?R2= ?9r2 / 4 = 9?r2/4Increase in area = (9?r2/4 ) - (?r2)= (5/4) ?r2Req. increase per cent = [{(5/4) ?r2} / {?r2}] x 100 = 125 %
Let length of rectangular field = 5y,so width = 4y.From question5y - 4y = 20m? x = 20m ? Length = (5 x 20)m = 100 mBreadth = (4 x 20) m = 80 m ? Perimeter = 2 (100 + 80 ) m = 360 m
Area of the field = Total cost/rate = (333.18/25.6) = 13.5 hectares 13 . 5 * 10000 m 2 = 135000 m 2 Let altitude = x metres and base = 3x metres.Then, 1 2 * 3 x * x = 135000 ? x 2 = 90000 ? x = 300 Base = 900 m and Altitude = 300 m.
EXAMIANS