Challenging Geometry Puzzle: Problem 1578. Share your solution by posting it in the comment box provided.

Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.

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ReplyDeleteDenote [XYZ]= area of XYZ; p= half perimeter of triangle AED

r= incircle of triangle AED

1.Since ABCD is a tangential quadr. => AB+CD=AD+BC= 26+4=30

Since ABCD is a cyclic quard. => ^(EBC)= ^(ADE)=> triangle BEC similar to DEA ( case AA)

[ABCD]= [AOD]+[OBC]+[OCD]+[OAB]

[ABCD]= ½( 26r + 4r+30r)=30r

2. triangle BEC similar to DEA with similarity ratio= BC/AD= 2/13 and ratio of areas= (2/13)^2= 4/169

Denote Z=[BEC] we have Z/(Z+[ABCD])= 4/169

Replace [ABCD]= 30r in above we get Z=[BEC]= 8/11.r

3. So [AED]= [EBC]+[ABCD]= 338/11. r= p.r

So p= 338/11= 30.7272

Note that r= p- AD= 4.7272

So [ABCD]= 30.r= 141.816