From problem diagram above, let x and b be the base and height of triangle '12'. Let x - a and y - b be the base and height of triangle 14. Then we may write:
½ ay = 36, ay = 72, and a = 72/y
½ bx = 12, bx = 24, and b = 24/ x
½ (x- a) (y - b) = 14 or (x - a) (y - b) = 28
Then:
xy - bx - ay + ab = 28
xy - 24 - 72 + ab = xy - 96 + ab = 28
xy - 124 + ab = 0
xy - 124 + (72/y) (24/x) = xy - 124 + 1728/ xy = 0
Multiply both sides by xy:
(xy) 2 - 124 xy + 1728 = 0
Solve for xy, the area of the triangle using the quadratic formula:
xy = {124 + Ö(124)2 - (4)(1)(1728)} / 2
xy = {124 + Ö(15376) - 6912)} / 2
xy = {124 + Ö(8464)} / 2
xy = 62 + 46 = 108 or 16
But since xy cannot be smaller than any of its triangles, then xy = 108
Finally, the area of the interior triangle (?) is equal to the area of the rectangle minus the area of the exterior triangles or:
A(?) = 108 - 36 - 12 - 14 = 46
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