Wednesday, August 7, 2024

Solution to Mensa Math (Geometry) Problem

 




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)-   (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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