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#### chingchongling

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##### Help me with maths questions :\
« on: June 03, 2014, 09:51:49 pm »
+1

Any help appreciated.

#### Phy124

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##### Re: Help me with maths questions :\
« Reply #1 on: June 03, 2014, 10:03:37 pm »
+6
$P = 2\times 25 + 2\times(10+12) = 50 + 44 = 94$ (The distance of the horizontal line at the bottom must be equal to the sum of the horizontal lines at the top of the figure, so as the bottom one is 25 then the top 3 must sum together to equal 25 as such the horizontal lines total to 2*25, same thing for left and right vertical lines 10+25 on the left, 10+25 on the right)

$L = \sqrt{a^2+b^2+c^2} = \sqrt{6^2 + 8^2 + 2^2} = \sqrt{104} = \sqrt{4\times 26} = 2\sqrt{26}$
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#### chingchongling

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##### Re: Help me with maths questions :\
« Reply #2 on: June 04, 2014, 08:18:22 pm »
0
Thanks that really helps with the first question. I still don't understand how you're calculating for the second question though..
« Last Edit: June 04, 2014, 08:26:23 pm by chingchongling »

#### Hannibal

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##### Re: Help me with maths questions :\
« Reply #3 on: June 04, 2014, 08:41:39 pm »
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Thanks that really helps with the first question. I still don't understand how you're calculating for the second question though..
I worked out EG before AG, because you need it first. EG is 10, because of the triad 6,8,10. Then you use 10 from EG and 2 from EA to find AG.
2^2+10^2=AG^2
4+100=AG^2
So root of 104 = AG
That is root of 4 x root of 26, which simplifies to 2x root of 26