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In right triangle ABC, angle B = 90 degrees, AB = 5, and BC = 12. Circle O is drawn with the center O on AB, and tangent to the hypotenuse of the triangle and BC. What is the exact radius of the circle, expressed as a decimal?

 Dec 30, 2020
 #1
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Here we go, ABC is a right triangle, since AB=5 and BC=12, AC=13 due to Pythagorean theorem, the formula is 

area/semi perimeter = radius, the area is 30, then the semi perimeter is 15, 30/15=2

 

so the radius should be 2

 Dec 30, 2020
 #2
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AC  has  the slope  (-5/12)

 

And  the  equation of the  line through AC  is

 

y = (-5/12)x  +  5

 

12y   = -5x +  60

 

5x + 12y  - 60  =   0

 

The point  on AB  will be  (0, a)

 

And we  can solve this to  find "a"         (abs   =  absolute value )

 

abs (  5(0) +12a  - 60 )  / sqrt ( 5^2 + 13^2)   =  a

 

abs  ( 12a  -60_  / sqrt (169)

 

abs ( 12a - 60)   = sqrt (169)  a

 

abs  (12a  -60)   =13a

 

Either

 

12a   - 60  =  13a

 

-60  =  a        reject

 

or

 

60   -12a    =13a

 

60 =  25a

 

60/25   = a  =  12/5  = 2.4 = a   = the radius of the  circle

 

Here's a pic  :

 

 

cool cool cool

 Dec 30, 2020
 #3
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In right triangle ABC, angle B = 90 degrees, AB = 5, and BC = 12. Circle O is drawn with the center O on AB, and tangent to the hypotenuse of the triangle and BC. What is the exact radius of the circle, expressed as a decimal?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

AB = 5       BC = 12

 

∠ACB = tan-1(5 / 12)

 

Radius   r = tan(∠ACB / 2) * 12     r = 2.4

 Dec 31, 2020

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