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Anybody a Math Head?

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Ok anyone here a math genius?

 

or if you're just bored.. like I seem to be this Saturday.. LOL

 

My son just got home (age 14) from taking his first ever practice SAT exam. (Man, they start getting kids ready for college a lot earlier than in my days) anyway, he told me he didn’t have too much of a problem with it, but there was one question that he defiantly wasn’t sure he got right..

 

Being the ex-math head I was back in the day. I asked what it was and he was able to give me the details.. he was shocked when the *old man spit out the answer fairly fast.. LOL! old goat shouted out “Ma! I still got it Ma!†to which my wonderful wife exclaimed from the living rooms upstairs. “you sure do baby!†god I luv that woman she’s so good for my ego! ;)

 

Here’s the question.. I included a quick reproduction of how I imagined it appeared on his exam. Who’s got the math man skills to give the answer.. and no your own your honor not to do it by google’n around till you find it.. heck I’m not sure you even can.. since Sat exams are usually fairly good at giving questions that aren’t well known

 

circleincircle.jpg

 

 

The radius of each of the larger circles is 5 what is the radius of the smaller circle?

 

I’ll post my solution and how I got it later.. any one gets it right.. get’s an official “no Prize†along with some mad/luv from Da Goat.

 

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Guest XtremeDuty

Good job Puddy. This is an easy one. Square root of (x2(5) - x1(0))squared + (y2(5) - y1(0)). This will give you the square root of 50 which is the length of a line from the center of one circle to the center of the grid, 0,0. Now, SQRT (50) is about 7.07, so just subtract 5, because the inner circle is touching the outer circle, and you get 2.07.

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XtremeDuty ]

Good job Puddy.  This is an easy one.  Square root of (x2(5) - x1(0))squared + (y2(5) - y1(0)).  This will give you the square root of 50 which is the length of a line from the center of one circle to the center of the grid, 0,0.  Now, SQRT (50) is about 7.07, so just subtract 5, because the inner circle is touching the outer circle, and you get 2.07.

 

Damnit, you beat me to it, I couldnt of said it better myself.

 

Dub

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A^2 + B^2 = C^2

 

25 + 25 = C^2

 

50 = C^2 which = 7.07

 

7.07 - 5 = 2.07

 

pythagorean theory then assuming that circles are touching at tangent the distance from the center of the big circle to the edge is 5 so that is where you get the minus 5.

 

 

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Puddy!! mad/luv!

 

LOL

 

Xtreme gets minor credit for the calculations..

 

Answer: Like was mentioned

 

It’s simpler than it looks if you forgo thinking it involves division or multiplying with Pi..

 

the way I did it was this,,,

 

If you think in terms of trig and right angles than you can make an imaginary right triangle similar to this..

 

 

 

by forming a right triangle using three of the center points of any of the tangent larger circles.. you can easily see that two sides of the triangle would be 10.

 

solucir.jpg

 

Using a formula most everyone can remember.. i.e. you factor the Hypotenuse

 

Or the sq. root of Rsq + Rsq. The sq root of 10x10 + 10x10 or sq. root of 200 which is equal to (rounded off) 14.142 this is the Hypotenuse..

 

The diameter of the smaller circle is now easily found by simply subtracting the radius of the larger circles twice.. or 14.142 – 10 = 4.142 dia.

 

Since they asked for the radius.. you simply divide by two and whalla the radius of the small circle is 2.07 (rounded).

 

I asked if 2.07 was one of the answers he said it was.. but frowned since it’s not the one he went with..

 

I’m sure he did fine though, don’t tell him I said this.. but he’s probably smarter than the old man was at his age! LOL.

 

I surprised myself that I even remembered enough trig to get this so fast.. Fricking $35,000 college education.. for what? So I can impress my son???

 

Worth every penny!!! LOL!

 

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