jugoslav

No explanation is needed... I hope!?! 

It would be interesting to find a triangle in which F and F' share the same point. 

My goal is to find the area of a segment BGC. Let's get started.

AJ = sqrt(AK² - JK²) = 30√3

AH = 2/3(AJ) = 20√3

AE = HJ = 1/3(AJ) = 10√3

Angle BAC = 2[arccos(AB / AE)] = 109.4712206º

Area of sector ABGC = 30²pi * (∠BAC / 360) = 859.784956 u²

Area of triangle ABC = BE * AE = 424.2640687 u²

Area of segment BGC = [ABGC] - [ABC] = 435.5208873 u²

Area of circle H = 30²pi = 2827.433388

and finally...

Shaded area = 2827.433388 - (6 * 435.5208873) = 214.3080644 u²

EB * CB = AB * DB

9 * 22 = AB * 30

AB = 12 ³/₁₁

DA = DB - AB = 17 ⁸/₁₁

x(52 - x) = 24²

x = 16,  x = 36

24² = 576

576 / 16 = 36

AG = 26 - 16 = 10

PQ cannot be 8 !!!

..........................................

PQ = 6

Radius = √22 

A = (9 * 5) + (6 * 4.5)

A triangle in the top-left corner is an isosceles triangle (50º, 65º, 65º)

Thus, angle x = 180 - 65 = 115º

This is a right trapezoidal prism because the base and the top are right-angled trapezoids.

Prism height = 14 m

Base area = [1/2(16 + 11)] * 9.5 = 128.25 m²

V = 128.25 * 14 = 1795.5 m³

[BAD] = {(160 / 15) * 1/2(9 + 15)} - 80 = 48