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A circle is inscribed in a rt triangle. The radius of the circle is 6 cm, and the hypotenuse is 29 cm. Find the lengths of the 2 segments of the hypotenuse that are determined by the pt of tangency.

 

The solutions manual sets up the answer using

 

(35 - x)^2 + (6 + x)^2 = 29^2

 

I understand the 6+x, but I can't figure out how they got the 35.

 

Thanks

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I call the segments of the hypotenuse x and y.

We know x+y=29

 

Because the triangle is right, Pythagoras holds: a^2+b^2=29^2 with a and b the two sides.

Using the inscribed circle, we can express the two short sides as a=x+r and b=9+r (r=6, radius of circle)

(sides of triangle are tangent on circle, thus radius makes right angle with side, we see similar triangles...)

 

Now writing Pythagoras:

(x+6)^2+(y+6)^2=29^2

 

Using the first equation, x+y=29, we can express y=29-x

Now put in Pythagoras:

(x+6)^2+((29-x)+6)^=29^2

so

(x+6)^2+(35-x)^2=29^2

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I call the segments of the hypotenuse x and y.

We know x+y=29

 

Because the triangle is right, Pythagoras holds: a^2+b^2=29^2 with a and b the two sides.

Using the inscribed circle, we can express the two short sides as a=x+r and b=9+r (r=6, radius of circle)

(sides of triangle are tangent on circle, thus radius makes right angle with side, we see similar triangles...)

 

Now writing Pythagoras:

(x+6)^2+(y+6)^2=29^2

 

Using the first equation, x+y=29, we can express y=29-x

Now put in Pythagoras:

(x+6)^2+((29-x)+6)^=29^2

so

(x+6)^2+(35-x)^2=29^2

 

 

 

Thank you! I didn't think of setting it up as 29=x+y. That makes it perfectly clear. I appreciate the help.

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