Find the area using simultaneous equations [closed]

-1

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?

edited Dec 10 at 12:18

Harry Peter

5,44111439

asked Dec 7 at 21:55

Ben Franks

266110

closed as off-topic by amWhy, Shailesh, Cesareo, user10354138, Rebellos Dec 8 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.

Are all lengths and areas integers?
– Frpzzd
Dec 7 at 21:58

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.
– platty
Dec 7 at 22:03

@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.
– amWhy
Dec 7 at 22:04

@Frpzzd. I think that they cannot be.
– Mason
Dec 7 at 22:04

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

|
show 3 more comments

-1

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?

edited Dec 10 at 12:18

Harry Peter

5,44111439

asked Dec 7 at 21:55

Ben Franks

266110

closed as off-topic by amWhy, Shailesh, Cesareo, user10354138, Rebellos Dec 8 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.

Are all lengths and areas integers?
– Frpzzd
Dec 7 at 21:58

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.
– platty
Dec 7 at 22:03

@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.
– amWhy
Dec 7 at 22:04

@Frpzzd. I think that they cannot be.
– Mason
Dec 7 at 22:04

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

|
show 3 more comments

-1

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?

edited Dec 10 at 12:18

Harry Peter

5,44111439

asked Dec 7 at 21:55

Ben Franks

266110

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

edited Dec 10 at 12:18

Harry Peter

5,44111439

asked Dec 7 at 21:55

Ben Franks

266110

edited Dec 10 at 12:18

Harry Peter

5,44111439

asked Dec 7 at 21:55

Ben Franks

266110

edited Dec 10 at 12:18

Harry Peter

5,44111439

edited Dec 10 at 12:18

Harry Peter

5,44111439

edited Dec 10 at 12:18

Harry Peter

5,44111439

asked Dec 7 at 21:55

Ben Franks

266110

asked Dec 7 at 21:55

Ben Franks

266110

asked Dec 7 at 21:55

Ben Franks

266110

closed as off-topic by amWhy, Shailesh, Cesareo, user10354138, Rebellos Dec 8 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 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.

Are all lengths and areas integers?
– Frpzzd
Dec 7 at 21:58

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.
– platty
Dec 7 at 22:03

@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.
– amWhy
Dec 7 at 22:04

@Frpzzd. I think that they cannot be.
– Mason
Dec 7 at 22:04

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

|
show 3 more comments

Are all lengths and areas integers?
– Frpzzd
Dec 7 at 21:58

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.
– platty
Dec 7 at 22:03

@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.
– amWhy
Dec 7 at 22:04

@Frpzzd. I think that they cannot be.
– Mason
Dec 7 at 22:04

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

Are all lengths and areas integers?
– Frpzzd
Dec 7 at 21:58

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.
– platty
Dec 7 at 22:03

@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.
– amWhy
Dec 7 at 22:04

@Frpzzd. I think that they cannot be.
– Mason
Dec 7 at 22:04

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

|
show 3 more comments

1 Answer
1

active

oldest

votes

Hint:

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

More hints:

[] 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 $c= 1pmfrac{2sqrt 2}{3}$

edited Dec 8 at 0:11

answered Dec 7 at 22:42

Mason

1,8781529

Did you come to a unique solution?
– AlexanderJ93
Dec 7 at 22:50

1

I am familiar with quadratics, I will see where I get to with this hint, thank you @Mason.
– Ben Franks
Dec 7 at 23:13

@AlexanderJ93. Two solutions.
– Mason
Dec 7 at 23:24

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

Hint:

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

More hints:

[] 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 $c= 1pmfrac{2sqrt 2}{3}$

edited Dec 8 at 0:11

answered Dec 7 at 22:42

Mason

1,8781529

Did you come to a unique solution?
– AlexanderJ93
Dec 7 at 22:50

1

I am familiar with quadratics, I will see where I get to with this hint, thank you @Mason.
– Ben Franks
Dec 7 at 23:13

@AlexanderJ93. Two solutions.
– Mason
Dec 7 at 23:24

add a comment |

Hint:

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

More hints:

[] 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 $c= 1pmfrac{2sqrt 2}{3}$

edited Dec 8 at 0:11

answered Dec 7 at 22:42

Mason

1,8781529

Did you come to a unique solution?
– AlexanderJ93
Dec 7 at 22:50

1

I am familiar with quadratics, I will see where I get to with this hint, thank you @Mason.
– Ben Franks
Dec 7 at 23:13

@AlexanderJ93. Two solutions.
– Mason
Dec 7 at 23:24

add a comment |

Hint:

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

More hints:

[] 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 $c= 1pmfrac{2sqrt 2}{3}$

edited Dec 8 at 0:11

answered Dec 7 at 22:42

Mason

1,8781529

Hint:

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

More hints:

[] 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 $c= 1pmfrac{2sqrt 2}{3}$

edited Dec 8 at 0:11

answered Dec 7 at 22:42

Mason

1,8781529

edited Dec 8 at 0:11

answered Dec 7 at 22:42

Mason

1,8781529

answered Dec 7 at 22:42

Mason

1,8781529

answered Dec 7 at 22:42

Mason

1,8781529

Did you come to a unique solution?
– AlexanderJ93
Dec 7 at 22:50

1

I am familiar with quadratics, I will see where I get to with this hint, thank you @Mason.
– Ben Franks
Dec 7 at 23:13

@AlexanderJ93. Two solutions.
– Mason
Dec 7 at 23:24

add a comment |

Did you come to a unique solution?
– AlexanderJ93
Dec 7 at 22:50

1

I am familiar with quadratics, I will see where I get to with this hint, thank you @Mason.
– Ben Franks
Dec 7 at 23:13

@AlexanderJ93. Two solutions.
– Mason
Dec 7 at 23:24

Did you come to a unique solution?
– AlexanderJ93
Dec 7 at 22:50

I am familiar with quadratics, I will see where I get to with this hint, thank you @Mason.
– Ben Franks
Dec 7 at 23:13

@AlexanderJ93. Two solutions.
– Mason
Dec 7 at 23:24

add a comment |

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