There is a pentagon ABCDE such that angles A, C, and E are all right angles and it is concave at angle C.
If a ray starting at D was aimed at C, it would go exactly 1 inch before going through C, and would eventually go through A.
If BC and AE are both 5 inches, what is the area of this pentagon?
Using Tristan's drawing, by Pythagoras, BD² = 5² + 1² = 26.
Then draw a ray from D to AB, parallel to AE, meeting at X.
So BX² = BD² - DX² = 1. Hence AB = DE + 1.
By Pythagoras, DE² + 25 = (AC + 1)², (1) and
AC² + 25 = AB², and so AC² + 25 = (DE + 1)². (2)
(1) - (2) => DE² - AC² = AC² - DE² + 2AC - 2DE.
Hence 2(DE² - AC²) = 2(DE - AC)(DE + AC) = 2(AC - DE).
Either DE - AC = 0 or DE + AC = -1, so we choose DE = AC.
Then, from (1), 2DE + 1 = 25, and DE = 12.
The area of the pentagon equals the area of right triangle ABC plus the area of right triangle ADE.
So the area equals ½AC·BC + ½AE·DE.
Area = ½·12·5 + ½·5·12 = 60 square inches.
Solution
