+330 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 $ABC$.
+129847 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
