Imagine a couple of (differently) scaled down versions of the triangle, one where the corner square becomes area 1, and one where the sloping square becomes area 1. From using similar triangles and chasing the angle ‘a’ around the figure we can see that the areas of some of the other parts of the triangle will be the same across the two figures. We can call these triangular areas x, y and z.

(I should probably point out at this point, while I’m introducing lots of new variables, that I’ve no intention of ever calculating what a, x or y evaluate to. z on the other hand…).

Each of the triangles with area x, y or z has one right angle, one angle ‘a’, and one side of length 1. You can express x y and z in terms of trig functions of a:

x = tan(a)/2 ( = sin(a)/2cos(a) )

y = cot(a)/2 ( = cos(a)/2sin(a) )

z = sin(a)cos(a)/2

If you multiply x and y together you get simply xy = 1/4

If you add the reciprocals of x and y:

1/x + 1/y = 2cos(a)/sin(a) + 2sin(a)/cos(a) = 2((cos(a))^2+(sin(a))^2)/ sin(a)cos(a)

But since (cos(a))^2+(sin(a))^2 = 1 by definition:

1/x + 1/y = 1/z

Putting these two facts together we get (x+y)/(1/4)=1/z

4(x+y)=1/z

x+y = 1/4z

Now scaling the two triangles back up so that the square areas are 363 and 336 respectively, and equating the areas of the two triangles we find that:

Area = 363(1+x+y) = 336(1+x+y+z)

Substituting x+y = 1/4z

Area = 363(1+1/4z) = 336(1+1/4z+z)

I’m now going to multiply each term by z and group terms on one side:

336zˆ2 – 27z – 27/4 = 0

Using the quadratic formula we see that:

z = (27 +/- sqrt(27^2 + 27x336))/(2x336)

z = (27+99)/672 (since z is an area and must be positive)

z = 3/16

Plugging that back into one of the area formulae:

Area = 363(1+1/4z) = 363+484 = 847, which is the final answer.