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September 30, 2014

September 30, 2014

Posted by **sarah** on Saturday, April 17, 2010 at 12:19am.

Then Write an expression for the inscribed radius r in terms of the variable w , then find the value of w, to the nearest hundredth, that gives the maximum value of r.

- geometry -
**Reiny**, Saturday, April 17, 2010 at 8:46amThe inscribed circle has its centre on the bisectors of the angles.

Because of the properties of isosceles triangles that angle bisector also becomes the right-bisector of the non-equal side, or our 2w base.

I will do the w=8 case.

draw a 10-10-16 triangle, 16 as the base

label the base angle 2Ø, thus each of the bisected base angles are Ø.

label the length of the radius on the right-bisector as r, (where the angle bisector meets the righ-bisector of 16)

cos 2Ø = 8/10 = 4/5

we know cos 2Ø = 2cos^ Ø - 1

4/5 = 2cos^2 Ø -

cos^2Ø = (4/5 + 1)/2 = 9/10

cosØ = 3/√10

then sinØ = 1/√10 , using Pythagoras

back to the triangle, in the smaller right-angled triangle

tanØ = r/8

sinØ/cosØ = r/8

(1/√10)/(3/√10) = r/8

1/3 = r/8

r = 8/3

Now you repeat the calculations for w = 6 and see if you can see the pattern.

then generalize, or go through the same process, for w.

- geometry Thanks Reiny -
**Sarah**, Saturday, April 17, 2010 at 12:38pmThanks for the help.

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