Area Problems
The cross section of a canal is a trapezium in shape . If the canal is a trapezium in shape. If the canal is 10 meters wide at the top and 6 meters wide at bottom and the area of cross section is 640 sq. meters The depth of canal is?
Cross section area = 1/2 x ( a + b ) x d where a and b are the parallel sides, d is the perpendicular distance between them.? 1/2 x ( a + b ) x d = 640? d = (640 x 2) / 16 = 80m
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
Original area = ?(d/2)2= (?d2) / 4New area = ?(2d/2)2= ?d2Increase in area = (?d2 - ?d2/4)= 3?d2/4? Required increase percent = [(3?d2)/4 x 4/(?d2) x 100]%= 300%
let ABCD be the given parallelogram area of parallelogram ABCD = 2 x (area of triangle ABC) now a = 30m, b = 14m and c = 40m s=1/2 x (30+14+40) = 42 Area of triangle ABC = s s - a s - b s - c = 42 12 28 2 = 168sq m area of parallelogram ABCD = 2 x 168 = 336 sq m
EXAMIANS