Area Problems
The area of a trapezium is 384 sq.cm 2. If its parallel sides are in ratio 3 : 5 and the perpendicular distance between them be 12 cm, The smaller of parallel sides is?
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
Given that, area = 40 sq cm, base = 28 cm and height = perpendicular = ?Area = (base x perpendicular) / 2 ? 40 = (28 x perpendicular) / 2 ? perpendicular = 40/14 = 20/7 = 26/7 cm
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
We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103. Solving the two equations, we get: l = 63 and b = 40. ∴ Area = (l x b) = (63 x 40) m2 = 2520 m2
EXAMIANS