Area of conic section

First we note that the semi-vertical angle is $45^{\circ}$ . If the base of the cone is placed on the $xy$ plane, with the center of the base at the origin, then the equation of the cone is

$x^2 + y^2 - (z - 10)^2 = 0$

Now, if we cut it by the plane $x = 5$ , then

$y^2 - (z - 10)^2 = -25$

or

$(z-10)^2/25 - y^2/25 = 1$

which is a hyperbola. Solving for z,

$z = 10 - \sqrt{ 25 + y^2}$

when $z = 0$ , we have $y = \pm 5 \sqrt{3}$

By direct integration, and using symmetry, we arrive at,

$\text{Area} = 2 \displaystyle \int_{ 0} ^{5 \sqrt{3}} 10 - \sqrt{ 25 + y^2 } dy$

by using trigonometric substitution, let $y = 5 \tan t$ , then $dy = 5 \sec^2 t \hspace{4pt} dt$

and the integral becomes,

$\text{Area} = 2 [ 10(5 \sqrt{3} ) - \displaystyle \int_{0}^{\dfrac{\pi}{3}} 25 \sec^3 t dt ]$

from the integration tables, we have, $\displaystyle \int \sec^3 t \hspace{4pt} dt = \frac{1}{2} \sec t \tan t + \frac{1}{2} \ln | \sec t + \tan t |$

hence,

$\text{Area} = 2 [ 50 \sqrt{3} - 25/2 ( 2 \sqrt{3} - 0 + \ln (2 + \sqrt{3} )]$

$= 100 \sqrt{3} - 50 \sqrt{3} - 25 \ln (2 + \sqrt{3} )$

$= 50 \sqrt{3} - 25 \ln(2 + \sqrt{3} )$

$= 5^2 ( 2 \sqrt{3} - \ln( 2 + \sqrt{3} ) )$

And we deduce, that $a = 5 , b = 2 , c = 3$ , which makes the answer $\boxed{10}$