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



-

Popular posts from this blog

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
Read more

build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

Image
0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
Read more