-2
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So the question is to find the areas A - G.

You are told that the vertical length is 4 and horizontal length is 28

enter image description here


I started making a load of simultaneous equations but found there was too many variables that left me not being able to solve what first appeared to be a straight forward question.

Does anyone know of a simpler route forward? Or is it just identifying which simultaneous equations need using when?

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closed as off-topic by Namaste, Shailesh, Cesareo, user10354138, Rebellos Dec 8 '18 at 8:03

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  • $\begingroup$ Are all lengths and areas integers? $\endgroup$ – Frpzzd Dec 7 '18 at 21:58
  • $\begingroup$ Instead of setting up the entire system, try to look for "corners" of rectangles where you know three values but not the fourth, For example, you can say $10 / 2 = 5/G$ because these pairs of rectangles have the same heights. $\endgroup$ – platty Dec 7 '18 at 22:03
  • $\begingroup$ @Mason "You are told that the vertical length is 4 and horizontal length is 28". Hence my question. Please read the question carefully, as I did. $\endgroup$ – Namaste Dec 7 '18 at 22:04
  • $\begingroup$ @Frpzzd. I think that they cannot be. $\endgroup$ – Mason Dec 7 '18 at 22:04
  • $\begingroup$ The rectangle below A has area 10, @Mason, we don't know yet what the area of rectangle A is. $\endgroup$ – Namaste Dec 7 '18 at 22:16
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Hint:

$10: 2$ as $5: G$. That type of thinking should get you there.

More hints:

[enter image description here] Applying this reasoning and you should arrive at something like the image above. And now we have to solve the following $$36c+4/c=72\implies 9c^2+1=18c$$ And this has two solutions $c= 1\pm\frac{2\sqrt 2}{3}$

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  • $\begingroup$ Did you come to a unique solution? $\endgroup$ – AlexanderJ93 Dec 7 '18 at 22:50
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    $\begingroup$ I am familiar with quadratics, I will see where I get to with this hint, thank you @Mason. $\endgroup$ – Ben Franks Dec 7 '18 at 23:13
  • $\begingroup$ @AlexanderJ93. Two solutions. $\endgroup$ – Mason Dec 7 '18 at 23:24

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