Find the area using simultaneous equations [closed]












-2












$begingroup$


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?






geometry systems-of-equations







share|cite|improve this question











$endgroup$



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


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Shailesh, Cesareo, user10354138, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    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.
    $endgroup$
    – amWhy
    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.
    $endgroup$
    – amWhy
    Dec 7 '18 at 22:16
















-2












$begingroup$


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?






geometry systems-of-equations







share|cite|improve this question











$endgroup$



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


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Shailesh, Cesareo, user10354138, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    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.
    $endgroup$
    – amWhy
    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.
    $endgroup$
    – amWhy
    Dec 7 '18 at 22:16














-2












-2








-2





$begingroup$


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?






geometry systems-of-equations







share|cite|improve this question











$endgroup$




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?






geometry systems-of-equations




geometry systems-of-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 12:18









Harry Peter

5,46911439




5,46911439










asked Dec 7 '18 at 21:55









Ben FranksBen Franks

263110




263110




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


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Shailesh, Cesareo, user10354138, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.







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


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Shailesh, Cesareo, user10354138, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    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.
    $endgroup$
    – amWhy
    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.
    $endgroup$
    – amWhy
    Dec 7 '18 at 22:16


















  • $begingroup$
    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.
    $endgroup$
    – amWhy
    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.
    $endgroup$
    – amWhy
    Dec 7 '18 at 22:16
















$begingroup$
Are all lengths and areas integers?
$endgroup$
– Frpzzd
Dec 7 '18 at 21:58




$begingroup$
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




$begingroup$
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.
$endgroup$
– amWhy
Dec 7 '18 at 22:04






$begingroup$
@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.
$endgroup$
– amWhy
Dec 7 '18 at 22:04














$begingroup$
@Frpzzd. I think that they cannot be.
$endgroup$
– Mason
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.
$endgroup$
– amWhy
Dec 7 '18 at 22:16




$begingroup$
The rectangle below A has area 10, @Mason, we don't know yet what the area of rectangle A is.
$endgroup$
– amWhy
Dec 7 '18 at 22:16










1 Answer
1






active

oldest

votes


















1












$begingroup$

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 $$36c+4/c=72implies 9c^2+1=18c$$ And this has two solutions $c= 1pmfrac{2sqrt 2}{3}$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Did you come to a unique solution?
    $endgroup$
    – AlexanderJ93
    Dec 7 '18 at 22:50






  • 1




    $begingroup$
    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


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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 $$36c+4/c=72implies 9c^2+1=18c$$ And this has two solutions $c= 1pmfrac{2sqrt 2}{3}$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Did you come to a unique solution?
    $endgroup$
    – AlexanderJ93
    Dec 7 '18 at 22:50






  • 1




    $begingroup$
    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
















1












$begingroup$

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 $$36c+4/c=72implies 9c^2+1=18c$$ And this has two solutions $c= 1pmfrac{2sqrt 2}{3}$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Did you come to a unique solution?
    $endgroup$
    – AlexanderJ93
    Dec 7 '18 at 22:50






  • 1




    $begingroup$
    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














1












1








1





$begingroup$

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 $$36c+4/c=72implies 9c^2+1=18c$$ And this has two solutions $c= 1pmfrac{2sqrt 2}{3}$







share|cite|improve this answer











$endgroup$



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 $$36c+4/c=72implies 9c^2+1=18c$$ And this has two solutions $c= 1pmfrac{2sqrt 2}{3}$








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 '18 at 0:11

























answered Dec 7 '18 at 22:42









MasonMason

1,9551530




1,9551530












  • $begingroup$
    Did you come to a unique solution?
    $endgroup$
    – AlexanderJ93
    Dec 7 '18 at 22:50






  • 1




    $begingroup$
    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


















  • $begingroup$
    Did you come to a unique solution?
    $endgroup$
    – AlexanderJ93
    Dec 7 '18 at 22:50






  • 1




    $begingroup$
    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
















$begingroup$
Did you come to a unique solution?
$endgroup$
– AlexanderJ93
Dec 7 '18 at 22:50




$begingroup$
Did you come to a unique solution?
$endgroup$
– AlexanderJ93
Dec 7 '18 at 22:50




1




1




$begingroup$
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




$begingroup$
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




$begingroup$
@AlexanderJ93. Two solutions.
$endgroup$
– Mason
Dec 7 '18 at 23:24



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