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In triangle $ABC,$ $\angle C = 90^\circ.$ A semicircle is constructed along side $\overline{AC}$ that is tangent to $\overline{BC}$ and $\overline{AB}.$ If the radius of the semicircle is equal to $4,$ and $AB = 10$, then find the area of triangle $ABC$.

 May 1, 2024
 #1
avatar+128794 
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I don't think this is possible  if AB  =  10   ( it must be greater)

 

If we can....let us imagine this  figure ....  { AB  is  actually ≈ 13.45 }

 

 

 

Let DC = DE =  4

Let AB  = "10"

Let AE =  "x"

And BE  =  BC  =   "10 - x"

And AD  =   sqrt [ DE^2 + AE^2]  = sqrt [ 4^2 + x^2 ] 

 

Note that ...Triangle ADE ≈  Triangle ABC

 

So

 

DE / AD  =  BC / AB

 

4 / sqrt [ 4^2 + x^2  ] =  (10 - x )  /  10        

 

The only real solution to  this  is  x =   0  =   AE    which is  impossible

 

 

cool cool cool

 May 2, 2024

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