+0  
 
-2
1
1
avatar+171 

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 $1$ and $BC = \sqrt{3}$, then find $AB$.

 Jun 8, 2024
 #1
avatar+129401 
+1

Construct a circle at B  with radius BC  that intercepts the semi-circle at E

Because BC = BE......then BE will be tangent to the semi-circle

Carry BE through to A ....let AD = x

OE = OC

BC = BE

BO = BO

Then triangles EOB and COB are congruent right triangles

tan angle OBC  = 1/sqrt 3  =  tan angle OBE 

 

tan ( OBC + OBE)  =  tan (ABC)....so.....

 

[ 2 * 1/sqrt 3 ] /  [1 - (1/sqrt 3)^2]  =   (2 + x) / sqrt 3

 

2/sqrt 3 * sqrt 3  =  ( 1 -1/3) (2 + x)

 

2  =  (2/3) (x + 2)

 

3 = x + 2

 

x =1

 

So

 

AB  = sqrt [ AC^2 + BC^2 ] = sqrt [ (2 + x)^2  + (sqrt3)^2 ]  = sqrt [ 3^2 + 3 ]  = sqrt 12  =  2sqrt 3

 

cool cool cool

 Jun 8, 2024
edited by CPhill  Jun 8, 2024
edited by CPhill  Jun 8, 2024

4 Online Users

avatar
avatar
avatar