Two rectangles overlap, as shown below. Find the area of the overlapping region (which is shaded) if AB = BE = 2 and AD = ED = 4.
The area of a rectangle of length l and width w is given by the multiplication of the dimensions, as follows:
A = lw.
The dimensions of the right triangle as follows:
Hence the remaining leg on the overlapping region is:
4 - x, as AD = 4.
By symmetry, the other dimension of the overlapping region is also of:
4 - x.
Being also the hypotenuse of the right triangles.
The value of x can be found applying the Pythagorean Theorem as follows:
x² + 2² = (4 - x)²
x² + 4 = 16 - 8x + x²
8x = 12
x = 1.5.
Then the two dimensions of the shaded region are:
4 - 1.5 = 2.5.
Meaning that the area is of:
A = 2.5 x 2.5 = 6.25 units squared.
The shaded area is NOT a rectangle or square! The bottom triangle with base CD ==2 and its height is ==1.5. Its area ==[2 x 1.5] / 2 ==1.5 u^2
Do the same with the similar triangle at the top with the base AB ==2
Add the area of the 2 triangles ==1.5 + 1.5 ==3 u^2
The area of the vertical rectangle ==2 x 4 ==8 u^2
The shaded area ==area of the vertical rectangle - the sum of 2 triangles at the top and the bottom
== 8 - 3 == 5 u^2 - area the shaded section.