Triangle Geometry

Firstly, using Pythagoras' Theorem, we can say that: $\overline {BC} = \sqrt {8^2 + 15^2}$ $\overline {BC} = \sqrt {64 + 225} = \sqrt {289} = 17$ Therefore: $\triangle BCD = \frac {17\times 4}{2} = \frac {68}{2} = \boxed {34}$ Now, using right-triangle trigonometry, we know that $Sin(\theta)=\frac O H$ , therefore: $sin(53.1)=\frac 8 H$ $H = \frac 8 {sin(53.1)}=10$ Now if we call the angle at 53.1 point E, then: $\overline {BE}=10$ $\overline {AE}=\sqrt {\overline {BE}^2 - \overline {AB}^2}=\sqrt{10^2-8^2}=\sqrt{36}=6$ $\overline{CE} = 15 - \overline{AE} = 15 - 6 = 9$ $\triangle BEC=\frac{8\times 9} 2=\frac {72} 2 =\boxed {36}$ Therefore the area of the blue shade is: $\triangle BCD + \triangle BEC = 34 + 36 = 70$ Thank you to everyone who attempted my puzzle!