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M16-22

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Re: M16-22  [#permalink]

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New post 21 Sep 2017, 20:44
Simba9 wrote:
Hi Bunuel

Since you say it's better to solve (when we get an equation like a^2 − 32a −32=0, as we cannot assume that one value will be positive and the other neg.) Would you mind posting a complete resolution to this problem? i.e. including your calculations to get to the positive and negative (invalid) solution.

I am asking this as I think you might be doing the calculus in a much more straightforward way than I am.

Thanks in advance!


Factoring Quadratics: http://www.purplemath.com/modules/factquad.htm
Solving Quadratic Equations: http://www.purplemath.com/modules/solvquad.htm
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Re: M16-22  [#permalink]

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New post 06 Nov 2017, 20:02
Can someone correct where am i going wrong?

let diameter = d
long side: d-2
short side: d /3

using Pythagorean theorem
d^2 = (d-2)^2 + (d/3)^2 or
d^2 = d^2 + 4 - 4d + d^2/9 subtracting d^2 on both sides
0 = 4 - 4d + d^2/9
multiplying both sides by 9
d^2 - 36d - 36 = 0

this is a different equation that derived in the original expert ans. What am I missing here?
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New post 06 Nov 2017, 20:49
sevenplusplus wrote:
Can someone correct where am i going wrong?

let diameter = d
long side: d-2
short side: d /3

using Pythagorean theorem
d^2 = (d-2)^2 + (d/3)^2 or
d^2 = d^2 + 4 - 4d + d^2/9 subtracting d^2 on both sides
0 = 4 - 4d + d^2/9
multiplying both sides by 9
d^2 - 36d - 36 = 0

this is a different equation that derived in the original expert ans. What am I missing here?


You have: d^2 = (d-2)^2 + (d/3)^2
Original solution: a^2 -32a-32=0

Do you see the difference? You have d, while the OE has a (longer side).
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M16-22  [#permalink]

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New post 11 Feb 2018, 11:52
We know the ratio of the length of the right triangle (1: \sqrt{3}: 2) and since the larger side is 2 meters less than the diagonal.
Can't we just assume that d is equal to 2x = x\sqrt{3} + 2 ?
And so we just need statement A to figure out the perimeter ?

Thank you
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New post 11 Feb 2018, 12:11
alvarezy wrote:
We know the ratio of the length of the right triangle (1: \sqrt{3}: 2) and since the larger side is 2 meters less than the diagonal.
Can't we just assume that d is equal to 2x = x\sqrt{3} + 2 ?
And so we just need statement A to figure out the perimeter ?

Thank you


Do we know that the diagonal will divide the rectangle into 30-60-90 triangles? No. Why it cannot be say 1-89-90? Or 7.5-82.5-90?
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M16-22  [#permalink]

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New post 11 Feb 2018, 15:06
Understood. Very clear. Thank you for the quick reply.


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Re: M16-22 &nbs [#permalink] 11 Feb 2018, 15:06

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