A Short Problem!

$A_{\triangle{BDE}} = \dfrac{1}{2}(2a)h = ah$

$\dfrac{h}{4a} = \tan(\alpha), \dfrac{a}{h} = \tan(\beta)$ and $\dfrac{3a}{h} = \tan(\alpha + \beta)$

Using the tangent formula for the sum of two angles

$\implies 3a = \dfrac{h^2 + 4a^2}{3a} \implies h^2 + 4a^2 = 9a^2 \implies h = \sqrt{5}a$ $\implies$

$A_{\triangle{BDE}} = \sqrt{5}a^2 = \dfrac{5 + 3\sqrt{5}}{2} \implies a^2 = \dfrac{5 + 3\sqrt{5}}{2\sqrt{5}} =\dfrac{3 + \sqrt{5}}{2} =$

$\dfrac{6 + 2\sqrt{5}}{4} = (\dfrac{\sqrt{5} + 1}{2})^2 \implies a = \dfrac{\sqrt{5} + 1}{2} = \dfrac{\sqrt{p} + q}{r} \implies p + q + r = \boxed{8}$ .

BDE & CDB are similar triangles.

BD / 2a = 3a / BD
BD² = 6a²
BD = √6 × a

AB = √(BD² – AD²)
= √5 × a

(5 + 3√5) / 2 = (1/2)(2a)(√5a)
5 + 3√5 = (2a)(√5a) = 2√5 × a²
a² = (3 + √5) / 2
= (6 + 2√5) / 4
a = (1 + √5) / 2

Answer = 5 + 1 + 2 = 8