Area of a Region

Using the equations of rotation to rotate the x and y axis we have:

$$

${\bf x = x' cos\theta - y' sin\theta }$

${\bf y = x' sin\theta + y' cos\theta }$

Replacing the equations of rotation in the original equation and simplifying we obtain: $$

${\bf (4 cos^2\theta + 12 cos\theta sin\theta + 9 sin^2\theta)\:x'^2 + (4 sin^2\theta - 12 cos\theta sin\theta + 9 cos^2\theta)\:y'^2 + }$

${\bf (12\sqrt{13} sin\theta + 8\sqrt{13} cos\theta)\:x' + (12\sqrt{13} cos\theta - 8\sqrt{13} sin\theta) \:y' + (5 sin(2\theta) + 12 cos(2\theta))\:x'y' }$ ${\bf - 65 = 0 }$

To eliminate the ${\bf x'y'}$ term let ${\bf 5 sin(2\theta) + 12 cos(2\theta) = 0 \implies }$

${\bf tan(2\theta) = \frac{-12}{5} \implies }$ ${\bf \frac{2 tan\theta}{1 - tan^2\theta} = \frac{-12}{5} \implies }$

${\bf 6 tan^2\theta - 5 tan\theta - 6 = 0 \implies tan\theta = \frac{5 +- 13}{12} }$

Choosing the positive value for ${\bf tan\theta \implies tan\theta = \frac{3}{2} \implies }$ ${\bf cos\theta = \frac{2}{\sqrt{13}}\: and \:sin\theta = \frac{3}{\sqrt{13}} \implies }$

${\bf x = \frac{2}{\sqrt{13}}\: x' - \frac{3}{\sqrt{13}} y' }$ ${\bf \:and \:y = \frac{3}{\sqrt{13}}\:x' + \frac{2}{\sqrt{13}}\:y' }$

$$

Replacing the equations of rotation in the original equation and simplifying we obtain: $$

${\bf 169x'^2 + 676x' - 845 = 0 \implies x'^2 + 4x' - 5 = 0 \implies (x' + 5)(x' - 1) = 0 }$

${\bf \therefore }$ we have two parallel lines ${\bf x' = 1\: and \: x' = -5 }$ in the ${\bf x'y' }$ system. $$

Since the circle ${\bf x^2 + y^2 = 36 }$ is invariant under any rotation, we can write:

${\bf x'^2 + y'^2 = 36 }$

$$

For ${\bf x' = -5 \implies y' = +- \sqrt{11} }$ and ${\bf x' = 1 \implies y' = +- \sqrt{35} }$

$$

${\bf \therefore }$ The points of intersection of the circle and the parallel lines in the x'y' system are: $$

${\bf A:(-5, -\sqrt{11}), B:(-5, \sqrt{11}), C:(1,\sqrt{35}), \:, and \: D:(1,-\sqrt{35}) }$

$$

so that, ${\bf AB = 2 * \sqrt{11}, CD = 2 * \sqrt{35} }$ and the height ${\bf h = 6 \implies }$ $$

Area of trapezoid ${\bf ABCD = 1/2 * (6) * (2 * \sqrt{35} + 2 * \sqrt{11}) = 6 * (\sqrt{35} + \sqrt{11}) = a * (\sqrt{b} + \sqrt{c}) }$

${\bf \implies a + b + c = 52. }$