Complex Inscribed Quadrilateral and Circumscribed Triangle

[thousandthpapercrane]

I have three problems I absolutely cannot solve without some assistance. I provided the steps I was able to take, along with the problems.

I would REALLY appreciate some help, but please don't just solve the problem for me. I need to know how to do this stuff for future tests. Please tell me what I'm doing wrong! T_T

http://www.maj.com/gallery/thousandthpa ... /math1.bmp
http://www.maj.com/gallery/thousandthpa ... /math2.bmp
http://www.maj.com/gallery/thousandthpa ... /math3.bmp

 

[soroban]

Re: Complex Inscribed Quadrilateral and Circumscribed Triang

Hello, thousandthpapercrane!

Here's the first one.
They gave us way too much information . . .

                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)

 

[thousandthpapercrane]

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!!!

 

[Denis]

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.

 

[thousandthpapercrane]

=X Thank you oodles! <3