Hello, Singular!
I found a solution, but I hope there is a better way . . .
ABCD is a square.
Find the area of triangle AYX. Code:
q X s-q
C * - - - - * - - - - * D
| :::* |
| *:::::* 4 | s-p
| 5 :::::::::* |
| *:::::::::::* |
s | :::::::::::::::* Y
| *:::::::::::* |
| :::::::::* | p
| *:::::* 3 |
|:::* |
A * - - - - - - - - - * B
s
Let s = side of the square.
. . We see that: .s² .= .∆ + 12 .[1]
Let YB = p, then DY = s - p
Let CX = q, then XD = s - q
From right triangle YBA: .½ps = 3 . → . ps = 6---[2]
From right triangle XCA: .½qs = 5 . → . qs = 10 .[3]
From right triangle XDY: .½(s - p)(s - q) = 4 . → . s² - ps - qs + pq .= .8 .[4]
Substitute [2] and [3] into [4]: .s² - 6 - 10 + pq .= .8 . → . s² + pq .= .24 .[5]
Multiply [2] and [3]: .pqs² = 60 . → . pq = 60/s²
Substitute into [5]: .s² + 60/s² .= .24 . → . s^4 - 24s² + 60 .= .0
. . . . . . . . . - . . . - . . . - . . . . . .____________
. . . . . . . . . . . . . . . . . . . 24 ± √24² - 4(1)(60) . . . . . . . . .__
Quadratic Formula: . s² .= .------------------------- .= .12 ± 2√21
. . . . . . . . . . . . . . . . . . . . . . . . 2(1)
. . . . . . . . . . . . - . . . - . . . . . . . .__
Equate to [1]: . ∆ + 12 .= .12 ± 2√21
. . . . . . . . . . . . . . . . .__
. . Therefore: . ∆ .= .2√21