+0  
 
0
7
1
avatar+1612 

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 $AB$.

 Jan 8, 2024
 #1
avatar+128732 
+1

Construct  a semicircle with a radius of 1 at O = (0,1)

The equation of this  circle  is x^2 + (y-1)^2  = 1  → x^2 + y^2 - 2y  = 0      (1)

 

Constuct another circle with a radius of sqrt 3 at B = (sqrt 3,0) 

The equation of this  circle is ( x - sqt 3)^2 + y^2  =  3 ⇒  x^2 -2sqrt (3)x + 3  + y^2  = 3   →

x^2 -2sqrt(3)x + y^2  = 0       (2)

These circles will intersect at D such that  CD = DB

To find the D, set the equations of the  circles  equal

 

x^2 + y^2 -2y =  x^2 + y^2 - 2sqrt (3)x

 

y = sqrt (3)x         sub this  value for  y back into (2) to find the x coordinate of D

 

x^2 - 2sqrt (3)x + (sqrt (3) x)^2  = 0

 

x^2 -sqrt (12 )x + 3x^2  = 0

 

4x^2  - sqrt (12)x  = 0

 

x ( 4x - sqrt (12))  = 0

 

4x -sqrt (12) = 0

 

4x - 2sqrt (3) = 0

 

4x = 2sqrt (3)

 

x = sqrt (3) / 2  = the x coordinate of D

 

y= sqrt (3) * (sqrt (3)) / 2  =  3/2  

 

So  D =  ( sqrt (3)/2 , 3/2)

 

The slope of a line through OD =   (3/2 - 1) / (sqrt (3)/2 - 0)  = (1/2)/(sqrt ((3)/2)  = 1/sqrt (3)

 

 

The line through BD  will be perpendicular to  the line through OD so its slope = - sqrt (3)

 

The equation of the line through BD is

 

y= (-sqrt (3)) ( x -sqrt (3))

 

To find the y coordinate of A

 

When x = 0 , y = 3

 

So   A= (0,3)

 

And AB  =  sqrt [ (sqrt (3))^2 + 3^2 ] = sqrt [ 3 + 9 ] = sqrt [12]  = 2sqrt (3) ≈  3.46

 

 

cool cool cool

 Jan 8, 2024

4 Online Users

avatar
avatar