SOLUTION: the tangency point of a circle,inscribed in a right angled triangle ,divides its hypotenuse by the segment 5cm and 12cm.find the triangles legs

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Question 1028878: the tangency point of a circle,inscribed in a right angled triangle ,divides its hypotenuse by the segment 5cm and 12cm.find the triangles legs
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16946) About Me  (Show Source):
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(x+6)^2+(x+12)^2=(12+6)^2

x^2+12x+36+x^2+24x+144=324

2x^2+36x+180=324
2x^2+36x-144=0
/2
x^2+18x-72=0
Find the roots of the equation by quadratic formula

a= 1 b= 18 c= -72

b^2-4ac= 324 - -288
b^2-4ac= 612 sqrt%28%09612%09%29= 24.74
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=( -18 + 24.74 )/ 2
x1= 3.37
x2=( -18 -24.74 ) / 2
x2= -21.37
Ignore negative value
x = 3.37
The legs are 9.37cm & 15.37cm

Answer by ikleyn(52794) About Me  (Show Source):
You can put this solution on YOUR website!
.
the tangency point of a circle,inscribed in a right angled triangle ,divides its hypotenuse by the segment 5cm and 12cm.find the triangles legs
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Answer. The legs are 8 cm and 15 cm.

Solution 

Let "r" be the radius of the inscribed circle.

Then one leg is (5+r) cm long, and the other leg is (12+r) cm.
(Make a sketch and use the fact that the tangent segments from the point outside the circle are congruent).

Then you have this quadratic equation to find "r" (Pythagorean, of course).

%28r%2B5%29%5E2+%2B+%28r%2B12%29%5E2 = %285+%2B+12%29%5E2.

Simplify and solve it. It is arithmetics.

The solution is r = 3.

Then the legs are 5+3 = 8 cm  and  12+3 = 15 cm.