Two 2*8 rectangles overlap, as shown below. Find the area of the overlapping region (which is shaded)
+129845
Let the 2 right triangles contained within ABCD have legs of 2 and x
Their combined areas = (2) (1/2) ( 2x) = 2x
The shaded area is a parallelogram with a height of 2 and bases of sqrt (2^2 + x^2) = sqrt (4 + x^2)
So....these combined areas = 2 * 8 = 16
So
2x + sqrt (4 + x^2) * 2 = 16
x + sqrt ( 4 + x^2) = 8
sqrt ( 4 + x^2) = 8 - x square both sides
4 + x^2 = x^2 - 16x + 64
16x - 64 + 4 = 0
16x - 60 = 0
4x - 15 = 0
4x = 15
x = 15/4
So....the area of the parallelogram (shaded area ) =
sqrt [ 4 + (15/4)^2 ] * 2 =
sqrt [ 64 + 225 ] / 4 * 2 =
17 / 2 =
8.5 units ^2
