3-4-5 Square

Let $k$ be the side of the square, $l$ the bottom left side, $m$ the bottom right side, $n$ the right bottom side, $o$ the right top side.
Since, it's a square it's true that:
$l+m=k$ and $n+o=k$
$\frac{1}{2}kl=4\Rightarrow l=\frac{8}{k}$
$l+m=k\Rightarrow m=\frac{k^{2}-8}{k}$
$\frac{1}{2}mn=5\Rightarrow n=\frac{10k}{k^{2}-8}$
$n+o=k\Rightarrow o=\frac{k^{3}-18k}{k^{2}-8}$
$\frac{1}{2}ko=3\Rightarrow k\frac{k^{3}-18k}{k^{2}-8}=6\Rightarrow k^{4}-24k^{2}+48=0$
The discriminant is:
$D=24^{2}-4*48=384=64*6\Rightarrow \sqrt{D}=8\sqrt{6}$
$k^{2}=\frac{24\pm 8\sqrt{6}}{2}\Rightarrow$
$k^{2}=12\pm 4\sqrt{6}$
If we use the negative sign, we see that $k^{2}=4(3-\sqrt{6})\approx 2.2$
That can not be the case since $k^{2}$ is the area of the square which can not be less than 3.
Therefor:
$k^{2}=12+4\sqrt{6}\Rightarrow$
$k^{2}=4(3+\sqrt{6})$
So:
$\boxed{a=4}$ $\boxed{b=3}$ $\boxed{c=1}$ $\boxed{d=6}$
And the answer is $\boxed{14}$

Note: Sorry that i've skipped over some tedious algebra

Let the side of the square be 'l'. Then the other side of the triangle with area 4 must be (8 / l) , (since l x (8 / l) / 2 = 4 which is given). Similarly the other side of the triangle with area 3 must be (6 / l), (since l x (6 / l) / 2 = 3 which is given. And the area of the 3rd triangle with a given area is half of the sides multiplied together, and this area is given as 5. So (l - 8 / l) x (l - 6 / l) / 2 = 5. Multiply both sides of this equation by 2 x l^2 (one l multiplying one bracket, and the other l multiplying the other bracket. So (l^2 - 8) x (l^2 - 6) = 10 x l^2. Now note that the area of the square which I'll call A is equal to l^2. So substituting is gives (A - 8) x (A - 6) = 10 x A. Multiplying and simplifying this equation gives A^2 - 24 x A + 48 = 0. From this A = [24 (+/-) SQUROOT (24^2 - 4 x 1 x 48)] / (2 x 1). So A = 12 (+/-)SQUROOT (96). So A = 12 (+/-) 4 x SQUROOT 6. So A = 4 x [3 (+/-) SQUROOT 6]. Now SQUROOT 6 is clearly more than 2. So the negative value of this root gives A < 4 x [3 - 2]. So for the negative value of the squroot one gets that A<4. This is clearly absurd, because the areas of the 3 triangles that were given add up to 12. (ALTHOUGH I DON'T KNOW WHY THIS ROOT EXISTS FOR THIS EQUATION - I EXPECTED IT TO BE NEGATIVE. CAN SOMEONE EXPLAIN THIS TO ME?). But then the answer is the positive squroot, so the answer is A = 12 + 4 x SQUROOT 6. Which can be expressed as A = 4 x (3 + SQUROOT 6). AND WE'D BETTER NOT FORGET THE ONE TIMES THE SQUROOT!! So A = 4 x (3 + 1 x SQUROOT 6). And finally 4 + 3 + 1 + 6 = 14 which is the answer of this question. Regards, David