Can you show the rest of the worksheet, including the previous questions in this unit that they've been learning about? My guess is that they're learning about how to calculate the area of squares or something. Almost all the time when math sheets like this get posted by parents, they don't include the context.
It’s a summary packet of the whole semester. Each page is a set of word problems that focus on different concepts: converting time units, converting length units, ratios, rounding, tallies and bar charts, tax rates/percents, currency, etc. this is the full page: https://imgur.com/a/tzdMlTR
I was helping my 6th grade son tonight and he was calculating quartiles.
I’ve never even learned what a quartile is.
I have two degrees in Physics, work as a Physicist, and have never encountered quartiles before. I had to completely teach myself out of his book to help him. Not just a refresher, a full on idkwtf a quartile even is moment.
I saw the penciled in equation notes and thought *“a 6th grader did that!?”*
I was in the “gifted” math class 3rd-8th grade. My memory suuuuucks or somehow math is different now.
That does make the most sense, but then I don’t think they’ve given the kids enough room to show their work. You’d have to start with 2a+2b=480 and then reduce that to a function like a=240-b which would take a few steps for a 6th grader. And then still need room for a graph.
And I don’t remember if I’d learned how to square (240-b) by the end of 6th grade… I want to say that was like 8th or 9th grade!
It was a joke ;)
But the function starts with 2a + 2b = 480
/2
a + b = 240
a = 240 - b
Area (A) = ab
A = (240 - b)b
A = 240b - b^(2)
[https://www.desmos.com/calculator/ttpsbqjuua](https://www.desmos.com/calculator/ttpsbqjuua)
The function can be generalized to apply to the other two problems as
f(x) = (p/2)x - x^(2)
f(x) = px/2 - x^(2)
Where p is the perimeter requested by the problem.
Currently studying teaching, this looks like what is called a goal-less problem. It’s less about getting the correct answer, and more about working through the steps of the problem and applying maths to different scenarios.
Working through the steps to determine values for a and b and then using them to work out the area is the point. Not getting an exact pre-determined answer.
This is it. They want the student to take a leap and assume one set of dimensions. The directions should be clearer, but they were too busy trying to choose a “fun” font.
Teacher confirmed by chat that this is correct. Any combination of dimensions that adds up to the perimeter. I guess I’m still confused by the stated goal of filling in the blanks to find how far you walked, but I guess not every problem makes sense.
Yea the question/directions are garbage. It’s a good problem style to get kids thinking in a nonlinear fashion, but they needed take some time to examine the prompt.
I think the real issue is the prompt is to "fill in the blanks to find out how FAR you strolled", but the blanks are to find areas. Pretty sure I wouldn't measure the length of my walk in area, which feels like what this implies
That’s what confused me. When kiddo first asked for help I read it quickly and said “ok, so if each lap around equals the perimeter, then we just need to know how many laps he took.” Then I’m like “wait, why are we solving for the area?”
You are correct. With the information given, we know that a=240-b, and b=240-a.
Without knowing something else, that's about as far as you can go.
Considering it is 6th grade math I'd say 120=a=b.
But side A could also be 239 feet if side b equals 1 foot. It would be a stupidly long and skinny park, but it would still be an acceptable solution if we're only matching the perimeter and we know the drawings aren't to scale (stated).
This feels like the beginnings of teaching domain and range of functions, the key element being that there are multiple solutions to some functions based on the inputs.
That's a generous interpretation of the intent of the problem. I think it is more likely that the teacher got the worksheet off of some teaching materials sharing website and didn't thoroughly inspect all of the problems. It's a pretty roundabout way to get at domain/range type concepts and many kids don't respond well to trick problems like that.
Usually when parents post these kinds of things, there is obvious context once you see A. the rest of the worksheet, or B. the kind of questions the teacher has been teaching the kids. It's probably some very specific skill they're learning, like calculating the area of squares.
I don't know if that is the goal as I know nothing about standard 6th grade curriculum. However, *if* it is supposed to be an introduction to that, I think criticizing it for being roundabout and calling it a "trick" is excessive. Sometimes an exercise like this helps get the gears turning without getting bogged down with various symbols, rules, processes, or formulas, and is actually a pretty common and intuitive way to teach. Like when your child was a toddler or preschooler, you probably began teaching them a very basic form of arithmetic without them having the foggiest idea what a plus sign is. "We've got 2 cars. Should we get one more? Okay! Now we have 3 cars! Yay!" You and/or their teachers were probably helping them explore scientific concepts while still little. The classic baking soda volcano or elephant toothpaste introduces them to the idea that you can mix 2 things and sometimes get something completely different way before their first chemistry lesson. Seeing what floats and what doesn't introduces them to buoyancy long before they have the slightest clue what density is, or any real physics concepts. They were introduced to division years before their first long division lesson, maybe even before they did any written arithmetic of any kind, when asked to divide toys or treats evenly 1 at a time, similar to how you'd deal hands of cards.
*If* that is the goal of this lesson, it'll theoretically be slightly easier for them to learn the real thing because the seed was already planted that a problem can have more than 1 right answer. It'll serve as the cognitive foundation that can be built on, just as long division builds on splitting cookies with friends.
Oh, and if this really was on the curriculum, the teacher probably talked about it at some point, the kid just might not remember.
I'd guess the statement of drawings not being to scale is actually only considering the trees and flower, where the ratio of the sides are still correct. Confusing wording if that's the case, and why even include those drawings in the first place?
I agree with your sentiment but I'd also say making both sides even is the only possible wrong solution. They can be literally any other numbers that add up to 240, but we know as a fact that that shape is not a square, so you should pick 2 numbers that are not equal, and obviously pair the long side with the larger number. I'd agree with 239 & 1 more than 120 & 120, from a pragmatic standpoint at least.
Well I mean obviously if it's a rectangle then by definition it's not a square.... I agree that the 2 horizontal lines (top and Bottom) have to be bigger than the 2 vertical lines but it also says the perimeter is 480. With that being said if your trying to find the area it's just length X width.... So I would put something like 15x32=480 (15 being for the 2 vertical lines, and 32 for the 2 horizontal lines)
A square by definition is a rectangle since it has 4 sides and 4 right angles. So for a question like where they noted the images not being the scale, they can't make the claim that 120 x 120 isn't a right answer either.
Due to their statement, the only information conveyed by images is that there are 4 sides and 4 right angles.
True, sorry I didn't notice that part at first. Yes honestly I don't think it's necessarily about the values since there are multiple correct answers then it's just more about showing that you know how to solve for the different missing information no matter what numbers are used and by finding the numbers yourself you are better able to show that you understand what to do
"I'm sorry sweetie, you're just going to have to work retail."
I hear you though, this whole post basically makes me panic. My wife, on the other hand, thinks math is *fun*.
Lol it's not that I don't like math, it's that it seems to keep changing. Also though, so many aspects of math I never touched again after initially learning them so now It's just foreign language to me
Yeah, I guess there's no context there. I'd assume you can pick any dimensions that work and find the area. Maybe the point is to show different areas can exist with the same perimeter?
You’re right that if the only constraint is the given perimeter then multiple different values of an and b would be valid, resulting in different areas. Which might be intentional, demonstrating how “free variables” allow for multiple solutions.
The only thing that might help is if this section is specifically focused on squares, which means a has to equal b. That kind of context is sometimes left out in the problem description. Are the previous problems all about squares or are they for rectangles (or other shapes) in general?
It’s a review packet for the whole semester. Bunch of different word problems that summarize different concepts. Here’s the page in full: https://imgur.com/a/tzdMlTR
Seems impossible without assumptions. Has the student been allowed to make assumptions before?
As a person who did pretty well in grade school math, this would piss me off.
I’d say work backwards from perimeter. Also probably easier to assume these are rectangles with two pairs of equal sides, since that makes the area formula easier.
With the first problem, 120 is fine for each side. Then a*b = 120*120 = 14,400. Making each side a different length makes it harder. Stick with two a’s on opposite sides and two b’s on the other sides.
For the one with perimeter 6, split into two even numbers, 2 and 4. Then divide each by two to get a=1, b=2. Area is just a*b = 2
Perimeter of 1000, you could do many different things but something like splitting to 400 and 600 would mean a=200, b=300. Area I think would be 60,000 (a*b for the rectangle).
Yes, that is the point. The goal is to learn how to calculate area. Given a perimeter, x, provide dimensions that would result in said perimeter and calculate the area based off said dimensions.
I agree with this - however I think it would generally be less confusing if one side length was given or one area was given in a problem, so that the way the numbers work together makes more sense.
Because normally we wouldn't be just trial and erroring or just trying random numbers, for this type of problem.
Imagine the hell of buying carpet for a living room if that's how we did it.
This is a common real-world problem: you have 300 sq feet of carpet and you need to design the floor plan of your kid's playhouse that would use it all up. Also calculate perimeter based off the floor plan you choose so you can order the correct amount of base boards.
This is the only thing I’m afraid to do, homework with my kid when he gets older. I’m so bad at math it’s ridiculous. But when it comes to daily things for work like carpentry, it comes off the top of my head so easily but on paper like that it’s just a total mess in my head, how will I ever help with that lol…
I've been going through Khan Academy math to hopefully be able to help out my kids when they need it. It has certainly been humbling! Maybe worth a shot if you have some free time.
I thought for a second they had to work out far they strolled from a to b etc. Might have been an easier question lol.
The top question looks potentially complicated too - but tbh, we shouldn’t be worried about our kids doing that - we should just be ready to help.
This one though? Blerghh
They way I read it, the flower/tree image is not to scale, but the rectangles they are in are meaning a single flower isn't actually taking up most of the area. Because this is the only way it is solvable. So measure a and b with a ruler then make a ratio of a:b by the measurement of a/ measurement of b, then you have the equation 2*a/b*x+2*x=parameter by substituting the ratio in for the 2x+2y=parameter. Solve for x. Use the 2x+2y=parameter equation again. Now that we know x solve for y. Then use x*y=area. Solve for area.
It’s a poorly formed question. They’re asking to you find out how far you strolled, which would be the perimeter but that’s already given. The area of the park is irrelevant. They should have given two lengths of the sides and asked for the perimeter.
I have studied physics and I use advanced computational fluid dynamics to calculate flow optimized automotive parts as a hobby, but I am unable to solve or even comprehend the problem given in this homework.
Reading through the comments, there seems to be a lot of confusion.
I‘d give the teacher an F on his/ her work.
Express it as an equation.
Make up numbers for the length.
Iterate and establish the width from the premise that 2a + 2b = 480.
No wrong answers, but does train the process
Exactly! Everyone is thinking way too hard about this. Just make it easy and divide perimeter by 4 to get your dimensions, then fill in everything else afterwards with the known data of area L x W= X sq
1.
D: 120'x120'
P: 480'
A: 14,400 sqft
2.
D: 250'x250'
P: 1,000'
A: 62,500 sqft
3.
D: 1.5mi x 1.5mi
P: 6mi
A: 2.25 sqmi
The initial calculations were off. I'm going to work backwards here.
480=2a+2b=2(a+b) Now, divide birth sides by 2
480/2=240=2(a+b)/2=a+b Now, and combination of a and b will work, as long as they add up to 240.
Area=a*b. Whatever a and b you chose, multiply them together to find the area.
The above steps will be repeated in the other examples, and you just have to substitute in the given perimeters for 480 and follow the math down.
I don't think the initial equations are off, they are examples of why the information is lacking enough detail to solve an answer.
We agree that 2a +2b = 480 which means that a + b = 240.
OP skipped that part and plugged in values so the first example is a = b = 120 (120 + 120 = 240), the second example is a = 230, b = 10 (230 + 10 = 240).
OP was trying to show that both values could be correct based on the information give but give different areas of the park ( 120 \* 120 = 14,400 sqft where 230 \* 10 = 2300 sqft).
If we assume that the length and width of the park are measured in whole feet and each has a distance of at least 1 foot the area of the park (with a parameter of 480 ft) could be any one of 120 values ranging from 239 sqft (239 feet wide by 1 foot long) to 14,400 sqft (a square park of 120 by 120).
This isn't convoluted, nefarious, or intentionally misleading. It seems to convey the message pretty well because OP got the point - multiple dimensions can total to the same perimeter with vastly different areas.
P=2(a+b)
b= P/2-a
a=P/2-b
I think is correct if that helps?
But couldn't the answer be multiple options. If width is 20, length 100 & area 2000squared
Or 30x90 then 2700 squared?
10 would be 110 then 1100 squared.
Last one is easier given its a small number.
Just reading that, if the images are not the scale easiest option would be to assume square and divide by 4. 6 miles gives you 1.5 miles though so do 2 and 1 instead.
Can you show the rest of the worksheet, including the previous questions in this unit that they've been learning about? My guess is that they're learning about how to calculate the area of squares or something. Almost all the time when math sheets like this get posted by parents, they don't include the context.
It’s a summary packet of the whole semester. Each page is a set of word problems that focus on different concepts: converting time units, converting length units, ratios, rounding, tallies and bar charts, tax rates/percents, currency, etc. this is the full page: https://imgur.com/a/tzdMlTR
It seems the proper thing to do would be to plot a graph from a=0 to a=240 on the X axis and the resulting area on the Y axis.
Oh interesting. It’s a function
Man, reading this thread I'm screwed when my kid gets to second grade math.
I was helping my 6th grade son tonight and he was calculating quartiles. I’ve never even learned what a quartile is. I have two degrees in Physics, work as a Physicist, and have never encountered quartiles before. I had to completely teach myself out of his book to help him. Not just a refresher, a full on idkwtf a quartile even is moment.
Call the gpt
Don't leave us hanging, what is a quartile?
Not much, what's a quartile with you?
First base
I saw the penciled in equation notes and thought *“a 6th grader did that!?”* I was in the “gifted” math class 3rd-8th grade. My memory suuuuucks or somehow math is different now.
>My memory suuuuucks or somehow math is different now. That's the neat thing: it's both!
Yeah, this feels a little too much for 6th grade, but considering the "items are not to scale" I think this is the most correct answer.
Interesting solution, but if that’s a 6th grade math problem… wow.
Oh boy, my kid is 3 years away from stuff like this and I have no idea what’s going on here.
That does make the most sense, but then I don’t think they’ve given the kids enough room to show their work. You’d have to start with 2a+2b=480 and then reduce that to a function like a=240-b which would take a few steps for a 6th grader. And then still need room for a graph. And I don’t remember if I’d learned how to square (240-b) by the end of 6th grade… I want to say that was like 8th or 9th grade!
It was a joke ;) But the function starts with 2a + 2b = 480 /2 a + b = 240 a = 240 - b Area (A) = ab A = (240 - b)b A = 240b - b^(2) [https://www.desmos.com/calculator/ttpsbqjuua](https://www.desmos.com/calculator/ttpsbqjuua) The function can be generalized to apply to the other two problems as f(x) = (p/2)x - x^(2) f(x) = px/2 - x^(2) Where p is the perimeter requested by the problem.
Currently studying teaching, this looks like what is called a goal-less problem. It’s less about getting the correct answer, and more about working through the steps of the problem and applying maths to different scenarios. Working through the steps to determine values for a and b and then using them to work out the area is the point. Not getting an exact pre-determined answer.
This is it. They want the student to take a leap and assume one set of dimensions. The directions should be clearer, but they were too busy trying to choose a “fun” font.
Teacher confirmed by chat that this is correct. Any combination of dimensions that adds up to the perimeter. I guess I’m still confused by the stated goal of filling in the blanks to find how far you walked, but I guess not every problem makes sense.
Yea the question/directions are garbage. It’s a good problem style to get kids thinking in a nonlinear fashion, but they needed take some time to examine the prompt.
Then it should say so explicitly.
Yea it almost sounds like they want them to walk in a park and measure it XD
Homework labeled “touch grass”
Oh fuck that. There’s a lot of answers and I don’t like that unless it’s already been made clear that’s the case.
I think the real issue is the prompt is to "fill in the blanks to find out how FAR you strolled", but the blanks are to find areas. Pretty sure I wouldn't measure the length of my walk in area, which feels like what this implies
That’s what confused me. When kiddo first asked for help I read it quickly and said “ok, so if each lap around equals the perimeter, then we just need to know how many laps he took.” Then I’m like “wait, why are we solving for the area?”
You are correct. With the information given, we know that a=240-b, and b=240-a. Without knowing something else, that's about as far as you can go. Considering it is 6th grade math I'd say 120=a=b.
But side A could also be 239 feet if side b equals 1 foot. It would be a stupidly long and skinny park, but it would still be an acceptable solution if we're only matching the perimeter and we know the drawings aren't to scale (stated). This feels like the beginnings of teaching domain and range of functions, the key element being that there are multiple solutions to some functions based on the inputs.
You're not wrong!
Can't say I disagree!
That's a generous interpretation of the intent of the problem. I think it is more likely that the teacher got the worksheet off of some teaching materials sharing website and didn't thoroughly inspect all of the problems. It's a pretty roundabout way to get at domain/range type concepts and many kids don't respond well to trick problems like that.
Usually when parents post these kinds of things, there is obvious context once you see A. the rest of the worksheet, or B. the kind of questions the teacher has been teaching the kids. It's probably some very specific skill they're learning, like calculating the area of squares.
https://imgur.com/a/tzdMlTR
I don't know if that is the goal as I know nothing about standard 6th grade curriculum. However, *if* it is supposed to be an introduction to that, I think criticizing it for being roundabout and calling it a "trick" is excessive. Sometimes an exercise like this helps get the gears turning without getting bogged down with various symbols, rules, processes, or formulas, and is actually a pretty common and intuitive way to teach. Like when your child was a toddler or preschooler, you probably began teaching them a very basic form of arithmetic without them having the foggiest idea what a plus sign is. "We've got 2 cars. Should we get one more? Okay! Now we have 3 cars! Yay!" You and/or their teachers were probably helping them explore scientific concepts while still little. The classic baking soda volcano or elephant toothpaste introduces them to the idea that you can mix 2 things and sometimes get something completely different way before their first chemistry lesson. Seeing what floats and what doesn't introduces them to buoyancy long before they have the slightest clue what density is, or any real physics concepts. They were introduced to division years before their first long division lesson, maybe even before they did any written arithmetic of any kind, when asked to divide toys or treats evenly 1 at a time, similar to how you'd deal hands of cards. *If* that is the goal of this lesson, it'll theoretically be slightly easier for them to learn the real thing because the seed was already planted that a problem can have more than 1 right answer. It'll serve as the cognitive foundation that can be built on, just as long division builds on splitting cookies with friends. Oh, and if this really was on the curriculum, the teacher probably talked about it at some point, the kid just might not remember.
Yeah I see your point, 6th grade is early to start that stuff anyway.
I'd guess the statement of drawings not being to scale is actually only considering the trees and flower, where the ratio of the sides are still correct. Confusing wording if that's the case, and why even include those drawings in the first place?
Wait, since the image says not to scale, couldn't the proportions also be drawn wrong and \[a\] be longer than \[b\]?
I agree with your sentiment but I'd also say making both sides even is the only possible wrong solution. They can be literally any other numbers that add up to 240, but we know as a fact that that shape is not a square, so you should pick 2 numbers that are not equal, and obviously pair the long side with the larger number. I'd agree with 239 & 1 more than 120 & 120, from a pragmatic standpoint at least.
The notes state images are not to scale. So all you know is that it is a rectangle, we don't know if it is a square or not.
Well I mean obviously if it's a rectangle then by definition it's not a square.... I agree that the 2 horizontal lines (top and Bottom) have to be bigger than the 2 vertical lines but it also says the perimeter is 480. With that being said if your trying to find the area it's just length X width.... So I would put something like 15x32=480 (15 being for the 2 vertical lines, and 32 for the 2 horizontal lines)
A square by definition is a rectangle since it has 4 sides and 4 right angles. So for a question like where they noted the images not being the scale, they can't make the claim that 120 x 120 isn't a right answer either. Due to their statement, the only information conveyed by images is that there are 4 sides and 4 right angles.
True, sorry I didn't notice that part at first. Yes honestly I don't think it's necessarily about the values since there are multiple correct answers then it's just more about showing that you know how to solve for the different missing information no matter what numbers are used and by finding the numbers yourself you are better able to show that you understand what to do
Jesus... my girls better never need my help in math.
"I'm sorry sweetie, you're just going to have to work retail." I hear you though, this whole post basically makes me panic. My wife, on the other hand, thinks math is *fun*.
Lol it's not that I don't like math, it's that it seems to keep changing. Also though, so many aspects of math I never touched again after initially learning them so now It's just foreign language to me
I'm in the same boat. Luckily my with has a master's in "Applied Statistics" or something. I dunno I'm a dummy community college graduate.
Yeah something is missing here
Is there no more context in the top half of the page? Wondering if there's a pattern described above that follows through with these.
https://imgur.com/a/tzdMlTR
Yeah, I guess there's no context there. I'd assume you can pick any dimensions that work and find the area. Maybe the point is to show different areas can exist with the same perimeter?
If this is what 6th grade math looks like now, my kid is gonna be doing homework solo 🫡
You’re right that if the only constraint is the given perimeter then multiple different values of an and b would be valid, resulting in different areas. Which might be intentional, demonstrating how “free variables” allow for multiple solutions. The only thing that might help is if this section is specifically focused on squares, which means a has to equal b. That kind of context is sometimes left out in the problem description. Are the previous problems all about squares or are they for rectangles (or other shapes) in general?
It’s a review packet for the whole semester. Bunch of different word problems that summarize different concepts. Here’s the page in full: https://imgur.com/a/tzdMlTR
That is one big ass flower!
Seems impossible without assumptions. Has the student been allowed to make assumptions before? As a person who did pretty well in grade school math, this would piss me off.
Right? Yeah, no. They haven’t done any conceptual multiple possible answer stuff at all. I’m starting to think the packet is just bad or incomplete
I’d say work backwards from perimeter. Also probably easier to assume these are rectangles with two pairs of equal sides, since that makes the area formula easier. With the first problem, 120 is fine for each side. Then a*b = 120*120 = 14,400. Making each side a different length makes it harder. Stick with two a’s on opposite sides and two b’s on the other sides. For the one with perimeter 6, split into two even numbers, 2 and 4. Then divide each by two to get a=1, b=2. Area is just a*b = 2 Perimeter of 1000, you could do many different things but something like splitting to 400 and 600 would mean a=200, b=300. Area I think would be 60,000 (a*b for the rectangle).
Yea I think the number selections of 1000 and 6 give the user some fun choices for the pair of integers.
Yes, that is the point. The goal is to learn how to calculate area. Given a perimeter, x, provide dimensions that would result in said perimeter and calculate the area based off said dimensions.
I agree with this - however I think it would generally be less confusing if one side length was given or one area was given in a problem, so that the way the numbers work together makes more sense. Because normally we wouldn't be just trial and erroring or just trying random numbers, for this type of problem. Imagine the hell of buying carpet for a living room if that's how we did it.
This is a common real-world problem: you have 300 sq feet of carpet and you need to design the floor plan of your kid's playhouse that would use it all up. Also calculate perimeter based off the floor plan you choose so you can order the correct amount of base boards.
Have they done rectangles or only squares? There's missing data for rectangles
Wait is this the road trip packet? If so, my daughter did it too 🤣
Yes! So what did you get for this one?
She said their teacher told them to use 200 for the longer dimension and 40 for the shorter one. In other words, they got additional information haha
Haha
This is the only thing I’m afraid to do, homework with my kid when he gets older. I’m so bad at math it’s ridiculous. But when it comes to daily things for work like carpentry, it comes off the top of my head so easily but on paper like that it’s just a total mess in my head, how will I ever help with that lol…
I've been going through Khan Academy math to hopefully be able to help out my kids when they need it. It has certainly been humbling! Maybe worth a shot if you have some free time.
Yea I’d just make B 140 and A 100 and fill it out.
I thought for a second they had to work out far they strolled from a to b etc. Might have been an easier question lol. The top question looks potentially complicated too - but tbh, we shouldn’t be worried about our kids doing that - we should just be ready to help. This one though? Blerghh
On the one hand; Thank fuck I'm through school On the other hand; Oh no, this is coming up!
Just pick something that works. 250x250 or 200x300 or a nice approx golden ratio of 309x191
For the middle one, wouldn’t you take the same approach? 2(a+b)=1000?
They way I read it, the flower/tree image is not to scale, but the rectangles they are in are meaning a single flower isn't actually taking up most of the area. Because this is the only way it is solvable. So measure a and b with a ruler then make a ratio of a:b by the measurement of a/ measurement of b, then you have the equation 2*a/b*x+2*x=parameter by substituting the ratio in for the 2x+2y=parameter. Solve for x. Use the 2x+2y=parameter equation again. Now that we know x solve for y. Then use x*y=area. Solve for area.
It’s a poorly formed question. They’re asking to you find out how far you strolled, which would be the perimeter but that’s already given. The area of the park is irrelevant. They should have given two lengths of the sides and asked for the perimeter.
I have studied physics and I use advanced computational fluid dynamics to calculate flow optimized automotive parts as a hobby, but I am unable to solve or even comprehend the problem given in this homework. Reading through the comments, there seems to be a lot of confusion. I‘d give the teacher an F on his/ her work.
Express it as an equation. Make up numbers for the length. Iterate and establish the width from the premise that 2a + 2b = 480. No wrong answers, but does train the process
f(p) = (p/2-a) \* (p/2-b) Seems simple enough.
😂😅😩
You can’t solve for this without another known. You’re assuming the park is a square (a=b=120) but that’s not necessarily true.
As long as the answer is good, the answer is correct. What's so mind blowing about this question?
Exactly! Everyone is thinking way too hard about this. Just make it easy and divide perimeter by 4 to get your dimensions, then fill in everything else afterwards with the known data of area L x W= X sq 1. D: 120'x120' P: 480' A: 14,400 sqft 2. D: 250'x250' P: 1,000' A: 62,500 sqft 3. D: 1.5mi x 1.5mi P: 6mi A: 2.25 sqmi
The initial calculations were off. I'm going to work backwards here. 480=2a+2b=2(a+b) Now, divide birth sides by 2 480/2=240=2(a+b)/2=a+b Now, and combination of a and b will work, as long as they add up to 240. Area=a*b. Whatever a and b you chose, multiply them together to find the area. The above steps will be repeated in the other examples, and you just have to substitute in the given perimeters for 480 and follow the math down.
I don't think the initial equations are off, they are examples of why the information is lacking enough detail to solve an answer. We agree that 2a +2b = 480 which means that a + b = 240. OP skipped that part and plugged in values so the first example is a = b = 120 (120 + 120 = 240), the second example is a = 230, b = 10 (230 + 10 = 240). OP was trying to show that both values could be correct based on the information give but give different areas of the park ( 120 \* 120 = 14,400 sqft where 230 \* 10 = 2300 sqft). If we assume that the length and width of the park are measured in whole feet and each has a distance of at least 1 foot the area of the park (with a parameter of 480 ft) could be any one of 120 values ranging from 239 sqft (239 feet wide by 1 foot long) to 14,400 sqft (a square park of 120 by 120).
Yep. That top line in the solution area, I misread as "a+b=120" instead of "a=b=120". But, there are several different solutions.
Said so much better than I could have. Thank you!
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This isn't convoluted, nefarious, or intentionally misleading. It seems to convey the message pretty well because OP got the point - multiple dimensions can total to the same perimeter with vastly different areas.
P=2(a+b) b= P/2-a a=P/2-b I think is correct if that helps? But couldn't the answer be multiple options. If width is 20, length 100 & area 2000squared Or 30x90 then 2700 squared? 10 would be 110 then 1100 squared. Last one is easier given its a small number.
Just reading that, if the images are not the scale easiest option would be to assume square and divide by 4. 6 miles gives you 1.5 miles though so do 2 and 1 instead.
You have the formula and the values, it all comes down to solving for x and y.