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?
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
+128475 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 :
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?
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AB = 5 BC = 12
∠ACB = tan-1(5 / 12)
Radius r = tan(∠ACB / 2) * 12 r = 2.4
