Complex Inscribed Quadrilateral and Circumscribed Triangle

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thousandthpapercrane

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Mar 22, 2006
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Re: Complex Inscribed Quadrilateral and Circumscribed Triang

Hello, thousandthpapercrane!

Here's the first one.
They gave us way too much information . . .
Code: 
                L
              * * *
           *    |6   *
        H*- - - + - - -*I
        * \    M| 18  / *
       *   R\   |   /R   *
              \ | /
                *
                Q
We have: \(\displaystyle \,QH\,=\,QL\,=\,QI\,=\,R\) (radius)

And: \(\displaystyle \,LM\,=\,6,\;HM\,=\,MI\,=\,18\)

Then \(\displaystyle MQ\:=\:R\,-\,6\)


In right triangle \(\displaystyle IMQ\), we have: \(\displaystyle \,R^2\;=\;(R\,-\,6)^2\,+\,18^2\)

\(\displaystyle \;\;\)and we get: \(\displaystyle \,120R\,=\,360\;\;\Rightarrow\;\;R\,=\,30\)


Therefore, the diameter is \(\displaystyle 60\) . . . answer (d)
 
Wow. It's always the simple stuff, right? I sat there for AGES trying to figure it out too. >.< Oh well, you win some, you lose some. Thank you SOOOOO much for your help!!!
 
Your 3rd one contains a goof: SY and UY are not equal to 5, but to 5.19~

Since ZT = ZU = 35 and XT = XS = 7, then XZ = 35 + 7 = 42.

Let x = SY = SU; then XY = x + 7 and ZY = x + 35. So:
(x + 7)^2 + (x + 35)^2 = 42^2
Solve for x: x = 5.19....

By the way, the length of SY and UY is NOT required in this problem.
 
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