Can you find area of the Blue triangle? | (Fun Geometry Problem) |

[@matthieudutriaux]

Here is a simpler method :
0:00 to 2:42 same method than you to find a=b ; angle (PEQ)=angle (QER)
Let's call angle (RED)=c
Then, angle (BCE)=90°+c and angle(BAE)=90°-(c+2*b)
Law of sinuses in triangle BCE :
BC/sin(angle(QER))=BE/sin(angle(BCE))
BC/sin(b)=BE/sin(90°+c)
8*k/sin(b)=BE/cos(c)
Law of sinuses in triangle BAE :
AB/sin(angle(PEQ))=BE/sin(angle(BAE))
AB/sin(b)=BE/sin(90°-(c+2*b))
10*k/sin(b)=BE/cos(c+2*b)
(8*k/sin(b))/(10*k/sin(b))=(BE/cos(c))/(BE/cos(c+2*b))
4/5=cos(c+2*b)/cos(c)
EP=ED*cos(c+2*b)
ER=ED*cos(c)
EP/ER=cos(c+2*b)/cos(c)
EP/ER=4/5
Then, same method than you, from 10:50 to the end.
A|EPQ|=1/2*EQ*EP*sin(b)
A|EQR|=1/2*EQ*ER*sin(b)=200 cm²
A|EPQ|=EP/ER*A|EQR|=4/5*200=160 cm²

[@jimlocke9320]

Special case: Let CD = 0. C, D, and R become the same point, let's call it R. All linear dimensions are units of k. Early in the video, we find that

[@gregevgeni1864]

Nice problem!

[@MathandEngineering]

Thanks!