09302003, 06:59 PM  #1 
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This problem is killing my head. I can't figure it out, and I've been looking at it for like 20 minutes.
[i]Triangle PQR is a right angled at Q. PQ = 16cm and QR = 30cm. A circle is drawn inside the triangle, touching all sides. Find the radius of the cirlcle
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09302003, 07:16 PM  #2 
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funny you should bring this up, I took the mensa test yesterday on jwa site and had to refresh my memory on finding the hypotenuse of a right triangle...
find the square root of PQ squared + QR squared and find that the hypotenus, or PR is 34 there's a start... if I had some graph paper I could help you better...
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09302003, 07:50 PM  #3 
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Nevermind I solved it.
But here's a picture to help you solve it, if you still want to try. Here's the solution, kinda messy and hard to understand. Given: PQ = 16, QR = 30 PQ perp. QR, circle inside the triangle RTF: Radius of that circle ® Solution:  Call points of intersection between circle and triangle: alpha, beta and gamma, intersecting at the lines PQ, QR, PR respectively  Construct: Radius from O(origin/centre of circle) to alpha (forms 90 degree angle, tangent to center theorm) Radius from O to beta Radius from O to gamma Construct: Bisector of angle P which will fall on the line OP since O is equidistant from the two sides of the angle Bisector of angle Q falls on line OQ Bisector of angle R falls on line OR Proof: r^2 + (24z)^2 = a^2 (pyth. theorm) r^2 + (16 y)^2 = a^2 r^2 + z^2 = b^2 r^2 + y^2 = c^2 Since <Q = 90 and the angle formed at alpha = 90 and the angle formed at beta = 90 That section must be a rectangle (sum int angles = 360) Since the line segment from O to alpha and O to beta are equal AND opposite sides of a rectangle are equal we know that all the sides are equal Therefore: y = r So! If: r^2 + (24z)^2 = a^2 (pyth. theorm) r^2 + (16 y)^2 = a^2 AND: z + r = 30 z = 30  r Then: r^2 + (34 z)^2 = r^2 + (16y)^2 34^2 68z + z^2 = 16^2  32y + y^2 34^2  2040 + 68R + 30^2  60R + R^2 = 16^2  32R + R^2 16 + 8R = 16^2  32R 40R = 240 R = 6 Therefore the radius is 6! QED
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09302003, 07:53 PM  #4 
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nevermind...
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09302003, 07:59 PM  #5 
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ok now you have to tell me how you found the value of y and therefore c...
man it's been too long since I've done this stuff
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10042003, 09:26 PM  #6 
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Sorry I didn't help, I haven't been on the site for a long time. That special circle is called the "Incircle" and the radius is the "Inradius". The circle has some special properties, and you can study them here:
http://mathworld.wolfram.com/Incircle.html 
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10042003, 09:32 PM  #7 
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i know the answer! wait...flashback to the math stuff i posted about a year ago... ahh!

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