three sides of a trapezium are each equal to k cm. find the greatest possible area of the trapezium.
ans---Undetermined error: @@3/4sqrt(3)k^2@@
please help
Let the base of each triangles formed by drawing perpendicular on the longer side of the trapezium = a cm. Given hypotenuse = k cm.
So, height or perpendicular = sqrt (k2-a2)
We know area of trapezium = 1/2 (sum of parallel sides) X (perpendicular distance between them).
So area = 1/2 (a+k+k+a) x sqrt (k2-a2)
= (k+a) x sqrt (k2-a2)
= sqrt(k+a)2(k2-a2)
= sqrt (k4+2k3a-2ka3-a4)
Now, to get max are we need the place where the slope is 0.
We can do this with a derivative.
So, d/da Area = d/da (sqrt k4+2k3a-2ka3-a4)
0 = required derivative
So, we get a = k/2
Hence max area = (k+k/2) x sqrt (k2-k2/4) = 3/4 k2sqrt 3.