Rectangular Pyramid

Let $\angle {FDC}=\angle {FAD}=α,\angle {FCD}=\angle {CBE}=β$ . Then $|\overline {BC}|=|\overline {AD}|=\sqrt {7^2+24^2}=25,|\overline {FD}|=\sqrt {25^2-15^2}=20$ , $\sin α=\dfrac{20}{25}=\dfrac{4}{5}\implies \cos α=\dfrac{3}{5}$ , $\sin β=\dfrac{24}{25}\implies \cot β=\dfrac{7}{24}$ . $|\overline {CD}|=|\overline {FD}|\times \dfrac{\sin (α+β)}{\sin β}=20\times (\cos α+\cot β\sin α)=\dfrac{50}{3}$ . From the given condition, the height of the pyramid $h$ is given by $2πh=2\left (\dfrac{50}{3}+25\right )\implies h=\dfrac{125}{3π}$ . Height of the triangle on base $\overline {AD}$ is $\sqrt {\left (\dfrac{125}{3π}\right ) ^2+\left (\dfrac{25}{3}\right) ^2}\approx 15.6636$ , and on base $\overline {CD}$ is $\sqrt {\left (\dfrac{125}{3π}\right) ^2+\left (\dfrac{25}{2}\right) ^2}\approx 18.2251$ . Area of base = $25\times \dfrac{50}{3}\approx 416.667$ . Lateral surface area is $2\times \dfrac{1}{2}\times 15.6636+2\times \dfrac{1}{2}\times 18.2251=695.3426$ . Hence the total surface area is $416.667+695.3426=\boxed {1112.0093}$ .

Let $BE = 7, EC = 24$ and $AF = 15$ .

$BC = AD = \sqrt{7^2 + 24^2} = \sqrt{625} = 25$ and $225 = 25 * AG \implies AG = 9$

$\implies GD = HC = 16 \implies h_{1} = \sqrt{15^2 - 9^2} = \sqrt{144} = 12$

$\triangle{BEC} \sim \triangle{HFC} \implies \dfrac{7}{h_{2}} - \dfrac{24}{16} \implies h_{2} = \dfrac{14}{3} \implies$

$AB = CD = 12 + \dfrac{14}{3} = \dfrac{50}{3} \implies A_{ABCD} = 25(\dfrac{50}{3}) = \dfrac{1250}{3}$

The perimeter $P = 50 + \dfrac{100}{3} = \dfrac{250}{3} = 2\pi r \implies r = \dfrac{125}{3\pi} = OP$

$OT = \dfrac{25}{3}$ and $OS = \dfrac{25}{2}$ and $OP = \dfrac{125}{3\pi} \implies$

$PT = \dfrac{25}{3\pi}\sqrt{\pi^2 + 25}$ and $PS = \dfrac{25}{6\pi}\sqrt{9\pi^2 + 100} \implies$

The total surface area $A = \dfrac{1250}{3} + \dfrac{625}{3\pi}\sqrt{\pi^2 + 25} + \dfrac{625}{9\pi}\sqrt{\pi^2 + 100} =$

$\dfrac{625}{9\pi}(6\pi + 3\sqrt{\pi^2 + 25} + \sqrt{9\pi^2 + 100}) \approx \boxed{1112.00931858}$ .