Truncated Right Triangular Prism

Using the pythagorean theorem we have:

$AB = \sqrt{13}a$ and $BC = \sqrt{10}a = AC$

To find the area $A_{\triangle{ABC}}$ we use Heron's formula obtaining:

$s = \dfrac{\sqrt{13} + 2\sqrt{10}}{2}a \implies A_{\triangle{ABC}} = \dfrac{3\sqrt{39}}{4}a^2$

For the area of the trapezoidal faces we have:

$A_{AEGB} = 18a^2$

$A_{BGFC} = \dfrac{33}{2}a^2$

$A_{AEFC} = \dfrac{39}{2}a^2$

and

$A_{\triangle{EGF}} = \dfrac{9\sqrt{3}}{4}a^2$

$\implies A_{s} = \dfrac{3}{4}(72 + \sqrt{39} + 3\sqrt{3})a^2$

and

$V = \dfrac{9\sqrt{3}}{4}(\dfrac{7 + 6 + 5}{3})a^3 = \dfrac{54\sqrt{3}}{4}a^3$

$\implies \dfrac{V}{A_{s}} = \dfrac{18\sqrt{3}a}{72 + \sqrt{39} + 3\sqrt{3}} =$ $72 + \sqrt{39} - 3\sqrt{3} \implies$

$a = \dfrac{72^2 + 144\sqrt{39} + 39 - 27}{18\sqrt{3}} = \dfrac{2}{3\sqrt{3}}(433 + 12\sqrt{39}) =$

$\dfrac{\alpha}{\beta\sqrt{\beta}}(\lambda + \omega\sqrt{\gamma})$ $\implies \alpha + \beta + \lambda + \omega + \gamma = \boxed{489}$ .

Volume of the Solid

$\begin{aligned} V & = \text{Base area} \times \text{Average height of the solid} \\ & = \frac {\sqrt 3(3a)^2}4 \times \frac {5a+6a+7a}3 = \frac {27\sqrt 3}2 a^3 \end{aligned}$

Surface Area of the Solid is the sum of:

• the base area $\dfrac {\sqrt 3 (3a)^2}4 = \dfrac {9\sqrt 3}4a^2$
• the area of the three vertical sides $3a\left(\dfrac {5a+6a}2 + \dfrac {6a+7a}2 + \dfrac {7a+5a}2\right) = 54a^2$
• the area of the top triangle: The top triangle is an isosceles triangle with equal sides of length $\sqrt{1^2+3^2}a = \sqrt{10}a$ and base of length $\sqrt{2^2+3^2}a = \sqrt{13}a$ . Therefore, a height $h = \sqrt{10-\dfrac {13}4} a = \dfrac {3\sqrt 3}2a$ and an area $A = \dfrac 12 \times \sqrt{13}a \times \dfrac {3\sqrt 3}2a = \dfrac {3\sqrt{39}}4a^2$

Therefore $A_s = \dfrac {216+3\sqrt{39}+9\sqrt 3}4$ .

And we have:

$\begin{aligned} \frac V{A_s} & = 72 + \sqrt{39} - 3\sqrt 3 \\ \frac {54\sqrt 3}{216+3\sqrt{39}+9\sqrt 3} a & = 72 + \sqrt{39} - 3\sqrt 3 \\ \frac {18\sqrt 3}{72+\sqrt{39}+3\sqrt 3} a & = 72 + \sqrt{39} - 3\sqrt 3 \\ \implies a & = \frac {(72 + \sqrt{39})^2 - (3\sqrt 3)^2}{18\sqrt 3} \\ & = \frac {5196 + 144\sqrt{39}}{18\sqrt 3} \\ & = \frac {866+ 24\sqrt{39}}{3\sqrt 3} \\ & = \frac 2{3\sqrt 3}(433+ 12\sqrt{39}) \end{aligned}$

Therefore $\alpha + \beta + \lambda + \omega + \gamma = 2+3+433+12+39 = \boxed{489}$ .