Perimeter of a Right TriangleDate: 08/14/2001 at 21:47:06 From: Hanul Oh Subject: High School mathematics contest What is the perimeter of a right triangle with hypotenuse 65 that can be circumscribed about a circle with radius 12? Thank you. Date: 08/15/2001 at 12:54:13 From: Doctor Greenie Subject: Re: High School mathematics contest Hello, Hanul - Here is a rough picture of the problem (I have only drawn the radii of the circle - I didn't want to try to draw a circle using typed text). A * * * * * D * * * * * * * * * * * F ********* O * * * * * * * * * * ********************************************* C E B The right triangle is ABC; the circle has center O and is tangent to triangle sides AB, BC, and CA at points D, E, and F, respectively. AB has length 65. OD, OE, and OF are radii of the circle and so have length 12. Because the angle at C is a right angle, CE and CF also have length 12. We know, with circle O being inscribed in triangle ABC, that AD = AF and BD = BE. Let AF = x; then AD = x also. Because we are given that AB = 65, BD is (65-x); then BE is (65-x) also. You can solve the problem by finding the area of triangle ABC by two different methods and equating the expressions: (1) area = (1/2) base * height = (1/2)(AC)(BC) = (1/2)(x+12)(77-x) (2) area = (radius of inscribed circle) * (semiperimeter of triangle) = (12)(1/2)[(65)+(x+12)+(77-x)] So you have (1/2)(x+12)(77-x) = (12)(1/2)[(65)+(x+12)+(77-x)] Solve this equation for x (it's not as ugly as it looks); then you will have all the information you need to find the perimeter of the triangle. Write back if you have further questions on this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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