Volume of a ramp

Notice that since the circle is an incircle, the area of the triangle is given by $\dfrac{ar}{2} + \dfrac{br}{2} + \dfrac{hr}{2} = \dfrac{ab}{2}$ , where $a$ , $b$ , and $h$ are the side lengths of the triangle with $h$ being the hypotenuse, and $r$ is the radius of the inscribed circle. But notice that $\dfrac{ar}{2} + \dfrac{br}{2} + \dfrac{hr}{2} = \dfrac{r}{2} \left( a+b+h \right) = \dfrac{r}{2}P$ , where $P$ is the perimeter of the triangle.

So $A = \dfrac{Pr}{2} \implies r = \dfrac{2A}{P}.$ Note that the triangle is an isosceles right triangle, and so $a = b \implies h = \sqrt{a^2+a^2} = a\sqrt{2}.$

So $P = 2a + a\sqrt{2} = a \left( 2 + \sqrt{2} \right)$ . Likewise, $A = \dfrac{ab}{2} = \dfrac{a^2}{2}$ , so $r = \dfrac{a^2}{P} = \dfrac{a}{2 + \sqrt{2}}.$

But since $r = 4$ , we see that $4 = \dfrac{a}{2 + \sqrt{2}} \implies a = 4\left( 2 + \sqrt{2} \right).$

Hence, $V_{\text{ramp}} = 18 \left( \dfrac{a^2}{2} \right) = \dfrac{18 \left( 16 \left( 2 + \sqrt{2} \right)^2 \right) }{2} = 9 \left( 32 \left( 3 + 2\sqrt{2} \right) \right).$

$\therefore V_{\text{ramp}} = 288 \left( 3 + 2\sqrt{2} \right) \text{ ft}^3 \implies a+b+c+d = 288+3+2+2 = \boxed{295}.$