Math Challenge #2

[midwestfisherman]

O.K. here's a new one for you all.

Surveyors are working in a rectangular lot. The longer sides of the rectangle are each 11 meters longer than the distance between them. The area of the lot is 80 square meters.

Find the length of the lot and the distance between the surveyors.

 

[MikeMike]

29 meters between them

40 meters long by 2 m wide??

MIKE

 

[TuxDave]

I refuse to dive into petty algebra problems. And....

29 meters between them

40 meters long by 2 m wide??

MIKE

No...

 

[sygyzy]

Are these challenges? Or your homework?

 

[MikeMike]

Originally posted by: TuxDave
I refuse to dive into petty algebra problems. And....

29 meters between them

40 meters long by 2 m wide??

MIKE

No...

why tf does it not work??

the longer sides of the rectangle are 11m longer than the distance between the surveyors. 40-29 = 11. 40x2 = 80m^2

MIKE

 

[TuxDave]

Originally posted by: sygyzy
Are these challenges? Or your homework?

I was starting to suspect that. Perhaps it's one of those 'extra credit' things that his math teacher gives and if you get it you get an extra point on your midterm or something.

 

[BigJ]

This is obviously his homework because this is nowhere near a difficult math problem to solve. Next time come outright and say it, and maybe we'll help you next time.

 

[minendo]

5x16

 

[shuan24]

try x^2 + 11X - 80 = 0

x = [-b +- sqrt(b^2 - 4ac)] /2

 

[TuxDave]

Originally posted by: nourdmrolNMT1

Originally posted by: TuxDave
I refuse to dive into petty algebra problems. And....

29 meters between them

40 meters long by 2 m wide??

MIKE

No...

why tf does it not work??

the longer sides of the rectangle are 11m longer than the distance between the surveyors. 40-29 = 11. 40x2 = 80m^2

MIKE

Sorry.. didn't mean to brush you off. Sorry again. The area enclosed is the length times the distance between them. So if you say the length is 40, the distance between them is 29, the area becomes 1160.

 

[bubbadu]

The longer sides are 16 and the shorter ones are 5.. 16X5 equals 80 and 16-5 equals 11.

 

[Kyteland]

Originally posted by: midwestfisherman
O.K. here's a new one for you all.

Surveyors are working in a rectangular lot. The longer sides of the rectangle are each 11 meters longer than the distance between them. The area of the lot is 80 square meters.

Find the length of the lot and the distance between the surveyors.

Not enough information. Your question never states where these mystical surveyor are within the rectangle.

The length is 14.5
The width is 25.5
The distance between corners is 29.334...

Edit: misread the question. I thought the perimeter was 80, not the area.

 

[MikeMike]

Originally posted by: TuxDave

Originally posted by: nourdmrolNMT1

Originally posted by: TuxDave
I refuse to dive into petty algebra problems. And....

29 meters between them

40 meters long by 2 m wide??

MIKE

No...

why tf does it not work??

the longer sides of the rectangle are 11m longer than the distance between the surveyors. 40-29 = 11. 40x2 = 80m^2

MIKE

Sorry.. didn't mean to brush you off. Sorry again. The area enclosed is the length times the distance between them. So if you say the length is 40, the distance between them is 29, the area becomes 1160.

i did not know that you were to assume that the surveyors were standing on the sides of the lot. this is not given to me, thus i did not assume it, and we all know assuming makes an ass outta u n me.

minendo has it, if they are standing on the sides then.

MIKE

 

[minendo]

Originally posted by: Kyteland

Originally posted by: midwestfisherman
O.K. here's a new one for you all.

Surveyors are working in a rectangular lot. The longer sides of the rectangle are each 11 meters longer than the distance between them. The area of the lot is 80 square meters.

Find the length of the lot and the distance between the surveyors.

Not enough information. Your question never states where these mystical surveyor are within the rectangle.

The length is 14.5
The width is 25.5
The distance between corners is 29.334...

The distance between them (the longer sides) would be the width.

 

[MikeMike]

i get a page cannot be displayed when i reply, so i dp'd

MIKE

 

[TuxDave]

Originally posted by: Kyteland

Originally posted by: midwestfisherman
O.K. here's a new one for you all.

Surveyors are working in a rectangular lot. The longer sides of the rectangle are each 11 meters longer than the distance between them. The area of the lot is 80 square meters.

Find the length of the lot and the distance between the surveyors.

Not enough information. Your question never states where these mystical surveyor are within the rectangle.

The length is 14.5
The width is 25.5
The distance between corners is 29.334...

lol... I'm beginning to see why people are getting confused.

them = surveyors
them = sides of the rectangle <---- got my vote.

 

[royaldank]

Not enough information. Your question never states where these mystical surveyor are within the rectangle.

That was my thought.

Or, given the problem as stated, the answer could be 16x5 since it just say the surveyors are inside the rectangle. In that case, the fact they are surveyors or even there is not needed.

 

[TuxDave]

Originally posted by: nourdmrolNMT1
i did not know that you were to assume that the surveyors were standing on the sides of the lot. this is not given to me, thus i did not assume it, and we all know assuming makes an ass outta u n me.

minendo has it, if they are standing on the sides then.

MIKE

Read above post for center of confusion. My only assumption if anything would be what "them" stands for. I don't think that makes me an ass.

 

[midwestfisherman]

Originally posted by: BigJ
This is obviously his homework because this is nowhere near a difficult math problem to solve. Next time come outright and say it, and maybe we'll help you next time.

Ummmm....no it is not my homework! Sheesh

 

[minendo]

Originally posted by: midwestfisherman

Originally posted by: BigJ
This is obviously his homework because this is nowhere near a difficult math problem to solve. Next time come outright and say it, and maybe we'll help you next time.

Ummmm....no it is not my homework! Sheesh

One of your kids homework?

 

[bubbadu]

Its 16X5. End of dicussion

 

[TuxDave]

*waits for Math Challenge #3*

 

[sygyzy]

I don't understand how there could be so much discussion about this. Post the next challenge.

 

[MrChad]

80 = (x + 11)(x) = x^2 + 11x

0 = x^2 + 11x - 80

0 = (x - 5)(x + 16)

x = 5

 

[midwestfisherman]

Originally posted by: minendo

Originally posted by: midwestfisherman

Originally posted by: BigJ
This is obviously his homework because this is nowhere near a difficult math problem to solve. Next time come outright and say it, and maybe we'll help you next time.

Ummmm....no it is not my homework! Sheesh

One of your kids homework?

This is no one's homework. We (a group of us at work) bring these in to do at work when we have some downtime. Some are more chellenging than others. I just thought I would post a few on here for fun. Some of you guys are so cynical.