This blog is the online extension of our intermediate classrooms. Our goal is to enhance and document our learning experience throughout the school year, and share this journey with teachers, parents and students. We welcome your constructive feedback, and we look forward to learning with you!
So we are back into the swing of POTW. Please check your solutions from #16 below and the latest POTW #17 is below as well. Good luck on Caribou #3 today!
The first thing that I realized is that Figure AGFE is congruent to FIgure AFEB. But, we do not know for sure if FD is eel to BE and if AF is equal to EC. WE first need to proof (yay, this is fun. I’ve done a lot of proofing outside of school). FG is also equal to FD and GA is equal to DC. We will label the length of BE x and FD y. This means that EC = BC - BE. EC = 40 - x AF = AD - FD. AF = 40 - y
Since we know that ABE is right angled, we can use the Pythagoroan Theorem to find the lengths of x and y. AB^2 + BE^2 = AE^2 x^2 + 30^2 = (40-x)^2 x^2 +900 = 1600 - 80x + x^2 80x = 700 x = 8.75
Already, I see that this is the same as the equation above, so I know that:
x=y=8.75
Now, we can do what I have always wanted to do… eat Lindor. Nah, jk (not really, I want Lindor). Anyways, now we can use the Pythagorean Theorem to find the length of EF which is the crease that we are trying to find. If we pretend to draw a line straight down from F, and label this Z, we see that EFZ is a triangle.
We know the length of EZ (ha, too… EZ!!! IM SO FUNNNNYYYY), which is 40-x-y, and we know the length of FZ, which is 30 because it is the same side as AB. Now we can splunk them into an equation FZ^2 + EZ^2 = EF^2 30^2 + (22.5)^2 = EF^2 900 + 506.25 = EF^2 1406.25 = EF^2 37.5 = EF^2
To maximize the area, first, we need to maximize the two side lengths. And, to maximize the side lengths, we need to stretch these rectangles out as much as possible. However, how do we know all this? Let's take a step back.
Let's say you were trying to maximize the area of a single rectangle, with perimeter 20. To make it easier, divide 20 by 2, making it so we're dealing with only 2 side lengths. Now, let's convert this into algebra.
10x-x^2
We simply need to find the maximum value of this parabola to find the maximum side length, and thus the area of the given rectangle. The side lengths are both 5, making the maximum size of the rectangle P/4^2 every time, or in this case, 25u^2.
So, we can apply this to two rectangles, right?
Not quite.
It is easy to prove that this is not the case.
Because if we were to do this, we would end up with a square shaped like this: (sq. root of 6 + sq. root of 15)^2 = 40u^2. And we can beat the area of that square with a simple (5+2)(3+3) gained by picking random factors of 6 and 15, and we would get 90 units squared. The solution for this is quite the opposite. If you were to visualize stretching out both rectangles as far as possible, with a minimum side length at any particular point being one, you are actually stretching the overall rectangle more. In stretching the rectangles as much as possible, we get that the 6cm^2 rectangle is 6 by 1, and the 15cm^2 15 by 1. Now, we just add 15 to one, six to one, and multiply! 16*7=112cm^2
Therefore, the maximum area is 112cm^2.
(unless you include decimal numbers; then, it stretches to infinity.)
Grade 7 POTW This POTW is very straight forward. In order to find out the greatest possible area for the rectangle ACEG, I will have to stretch the side lengths of ABJH and JDEF. ABJH would be 1 by 6 cm and JDEF would be 1 by 15 cm. The width of ACEG (CE) is 16 cm as 1 cm + 15 cm = 16 cm. The length of ACEG (AC) is 7 cm as 1 cm + 6 cm = 7 cm. Since these are the greatest possible side lengths for ACEG, I will have to multiply them to get the area of the rectangle as the formula is LxW. 16x7=112 cm squared.
The largest possible area for ACEG is 112 cm squared.
To increase the area of the total rectangle, we need to make both the rectangles as long as possible so they stretch out the perimeter more. Since it cannot be decimal numbers, then the rectangles must have an area of 1x6 and 1x15. Then, I drew it on paper and found out the area of the rectangle would be 112, or 7x16. If we used decimals, then the number could easily be infinity. That was short. GOODNIGHT wait its like 4 pm
To make the area of ACEG maximum size you have to stretch out the side lengths of JDEF and ABJH. After some trial and error on paper i found the greatest side lengths would be 1cm by 6cm for ABJH and 1cm by 15cm for JDEF. Then i simply calculated the area on paper and found that the max area of rectangle ACEG is 112cm2
Grade 7 POTW (because I am definitely in Grade 7, anyways…):
To maximize the area, we need to maximize its dimensions, which means that we would have to take the large factor of each area. This would mean that we have larger dimensions, and therefore, a larger area.
Anyways, We know that the area of the green rectangle (ABJH) is 6 cm^2, which means that we have to use the largest factor or 6, which is….. 6….. ANYWAYSSS, Then we have to determine the extra line segment that is called British Columbia (BC). To do this, we need to find the dimensions of the pink rectangle (JDFE). Since the area is 15 cm^2, we know that 15*1=15.
Now, we need to use guess and check to find the largest area with the dimension 6,1, 15 and 1. There are only 2 possible choices.
First, we would the possible side lengths of each rectangle. The green rectangle can either be 1*6, or 2*3. The pink rectangle can be either 3*5 or 1* 15. (2 + 3) * (3 + 5) = 40 cm squared (2 + 3) * (1 + 15) = 80 cm squared (1 + 6) * (3 + 5) = 56 cm squared (1 + 6) * (1 + 15) = 112 cm squared Of these, 112 cm squared is the largest number, therefore the largest possible area of this rectangle is 112 cm squared.
Grade 8 POTW I am going to go straight into it. What is the length of FE? First let’s list the lengths we know. We know AB = CD = 30cm and AD = BC = 40cm. Also, figure AEFG = CEFD in terms of area. We also know ABE is a right triangle. GF = FD (we don’t know for sure if BE also equals this). AE = EC. To find out FE, we can use most of the information listed here (I didn’t list everything as that would take too long). Okay, so let’s start at the end. To figure out the length of FE, what can we do. The first thing that comes in mind is forming a right triangle. This can be done by creating a line segment parallel to AB and DC that is made vertically above E (we’ll call the point at the end of the line segment H). This would be 30cm like the other lines. Then, we can use the pythagorean theorem to figure out FE. It would be 30^2 + (FH)^2 = FE^2. To do this, we need to find the length of FH. Using what we know, FH = BC - BE - FD = 40 - BE - FD. Now we need to find the values of BE and FD. You can notice that ABE is a right triangle. This means the length of BE can be solved using the pythagorean theorem again. 30^2 + (BE)^2 = (AE)^2. The length of AE needs to be found. It can be seen that AE and EC are equal in length. EC = 40 - BE. This means that AE = 40 - BE. Back to the equation, 30^2 + (BE)^2 = (40 - BE)^2. 30^2 + (BE)^2 = (40 - BE)^2 900 + (BE)^2 = 1600 - 80(BE) + (BE)^2 900 + (BE)^2 - 900 - (BE)^2 + 80(BE) = 1600 - 80(BE) + (BE)^2 - 900 - (BE)^2 + 80(BE) 80(BE) = 700 80(BE)/80 = 700/80 BE = 8.75cm To find out FD, we could also form a right triangle by adding a new line segment, but then we have no equal values to go off of other than 30cm. However, we know FD = GF. We also know AG = CD = 30cm as well as the fact that AGF is a right angle. Now, we can form another right triangle by creating line segment AF. Okay, so AF = 40 - FD = 40 - GF. 30^2 + (GF)^2 = 40 - (GF) 900 + (GF)^2 = (40 - GF)^2 … Note that the equation above is the same as before, except BE is replaced with GF. Due to this, BE = GF = FD = 8.75cm. Now we can finally move back to the original question, the length of FE. We now have the values we need: 30^2 + (40 - 8.75 - 8.75)^2 = (FE)^2 900 + 22.5^2 = (FE)^2 900 + 506.25 = (FE)^2 1406.25 = (FE)^2 √(1406.25) = √((FE)^2) 37.5cm = FE Therefore, line segment FE is equal to 37.5cm.
Grade 7 POTW I will go right into it. ABJH = 6cm^2. JDEF = 15cm^2. We must use integer side lengths to make this work. An integer is a positive/negative whole number (or 0). To gain the greatest area, we must have the largest side lengths. In other words, we just have to think of combinations of side lengths that when 1 pair is multiplied, the product is 6cm^2, the other pair results in 15cm^2, and when a number from each pair are added, and then multiplied by the sum of the other remaining numbers, results in the largest area. The sum of the numbers from each pair has to both be as large as possible. We could try different possibilities, but it is easy to realize that to get 6, we just need to do 6 x 1 and 15 is 15 x 1. You add the highest number in one pair to the lowest number in the other (same with the other 2). 6 + 1 = 7 and 15 + 1 = 16. 7 x 16 = 112 cm^2. To double check , we can just try to use other numbers. For example, 2 x 3 and 3 x 5. No matter how you pair up the numbers from both pairs (hard to explain), the area would be less than 112cm^2 when they are multiplied. Therefore, the largest possible area of the entire rectangle ACEG is 112cm^2.
To figure out the area of the rectangle ACEG, we need to determine the greatest possible dimension of the green and pink rectangles: JDEF: 1 x 15 (cm) ABJH: 1 x 6 (cm) The width of rectangle JDEF is 15 cm. We need to add the width of the rectangle ABJH, 1 cm, so 15 + 1 = 16, so the width of rectangle ACEG is 16 cm. The length of rectangle ABJH is 6 cm. We need to add the length of the rectangle JDEF, 1 cm, so 6 + 1 = 7, so the length of the rectangle ACEG is 7 cm.
Now to calculate the area: 16 cm x 7 cm = 112 cm2
The greatest possible area of the rectangle ACEG is 112 cm2.
We only had 8 POTW posters last week. Come on people, just give it a try!
ReplyDeleteI tried...
DeleteGrade 8 POTW:
ReplyDeleteThe first thing that I realized is that Figure AGFE is congruent to FIgure AFEB. But, we do not know for sure if FD is eel to BE and if AF is equal to EC. WE first need to proof (yay, this is fun. I’ve done a lot of proofing outside of school). FG is also equal to FD and GA is equal to DC. We will label the length of BE x and FD y.
This means that EC = BC - BE.
EC = 40 - x
AF = AD - FD.
AF = 40 - y
Since we know that ABE is right angled, we can use the Pythagoroan Theorem to find the lengths of x and y.
AB^2 + BE^2 = AE^2
x^2 + 30^2 = (40-x)^2
x^2 +900 = 1600 - 80x + x^2
80x = 700
x = 8.75
FG^2+AG^2 = AF^2
y^2 + 30^2 = (40-y^2)
y^2 + 900 = 1600 - 80y + y^2.
Already, I see that this is the same as the equation above, so I know that:
x=y=8.75
Now, we can do what I have always wanted to do… eat Lindor. Nah, jk (not really, I want Lindor). Anyways, now we can use the Pythagorean Theorem to find the length of EF which is the crease that we are trying to find. If we pretend to draw a line straight down from F, and label this Z, we see that EFZ is a triangle.
We know the length of EZ (ha, too… EZ!!! IM SO FUNNNNYYYY), which is 40-x-y, and we know the length of FZ, which is 30 because it is the same side as AB. Now we can splunk them into an equation
FZ^2 + EZ^2 = EF^2
30^2 + (22.5)^2 = EF^2
900 + 506.25 = EF^2
1406.25 = EF^2
37.5 = EF^2
Yay, the crease is 37.5 cm.
We need more people to do POTW :D
Whoops, forgot to do POTW and RATL last week
ReplyDeleteTo maximize the area, first, we need to maximize the two side lengths. And, to maximize the side lengths, we need to stretch these rectangles out as much as possible. However, how do we know all this? Let's take a step back.
Let's say you were trying to maximize the area of a single rectangle, with perimeter 20. To make it easier, divide 20 by 2, making it so we're dealing with only 2 side lengths. Now, let's convert this into algebra.
10x-x^2
We simply need to find the maximum value of this parabola to find the maximum side length, and thus the area of the given rectangle. The side lengths are both 5, making the maximum size of the rectangle P/4^2 every time, or in this case, 25u^2.
So, we can apply this to two rectangles, right?
Not quite.
It is easy to prove that this is not the case.
Because if we were to do this, we would end up with a square shaped like this:
(sq. root of 6 + sq. root of 15)^2 = 40u^2.
And we can beat the area of that square with a simple (5+2)(3+3) gained by picking random factors of 6 and 15, and we would get 90 units squared.
The solution for this is quite the opposite. If you were to visualize stretching out both rectangles as far as possible, with a minimum side length at any particular point being one, you are actually stretching the overall rectangle more. In stretching the rectangles as much as possible, we get that the 6cm^2 rectangle is 6 by 1, and the 15cm^2 15 by 1.
Now, we just add 15 to one, six to one, and multiply!
16*7=112cm^2
Therefore, the maximum area is 112cm^2.
(unless you include decimal numbers; then, it stretches to infinity.)
Grade 7 POTW
ReplyDeleteThis POTW is very straight forward. In order to find out the greatest possible area for the rectangle ACEG, I will have to stretch the side lengths of ABJH and JDEF. ABJH would be 1 by 6 cm and JDEF would be 1 by 15 cm. The width of ACEG (CE) is 16 cm as 1 cm + 15 cm = 16 cm. The length of ACEG (AC) is 7 cm as 1 cm + 6 cm = 7 cm. Since these are the greatest possible side lengths for ACEG, I will have to multiply them to get the area of the rectangle as the formula is LxW. 16x7=112 cm squared.
The largest possible area for ACEG is 112 cm squared.
For the first question:
ReplyDeleteTo increase the area of the total rectangle, we need to make both the rectangles as long as possible so they stretch out the perimeter more. Since it cannot be decimal numbers, then the rectangles must have an area of 1x6 and 1x15. Then, I drew it on paper and found out the area of the rectangle would be 112, or 7x16.
If we used decimals, then the number could easily be infinity.
That was short.
GOODNIGHT
wait its like 4 pm
To make the area of ACEG maximum size you have to stretch out the side lengths of JDEF and ABJH. After some trial and error on paper i found the greatest side lengths would be 1cm by 6cm for ABJH and 1cm by 15cm for JDEF. Then i simply calculated the area on paper and found that the max area of rectangle ACEG is 112cm2
ReplyDeleteGrade 7 POTW (because I am definitely in Grade 7, anyways…):
ReplyDeleteTo maximize the area, we need to maximize its dimensions, which means that we would have to take the large factor of each area. This would mean that we have larger dimensions, and therefore, a larger area.
Anyways, We know that the area of the green rectangle (ABJH) is 6 cm^2, which means that we have to use the largest factor or 6, which is….. 6….. ANYWAYSSS, Then we have to determine the extra line segment that is called British Columbia (BC). To do this, we need to find the dimensions of the pink rectangle (JDFE). Since the area is 15 cm^2, we know that 15*1=15.
Now, we need to use guess and check to find the largest area with the dimension 6,1, 15 and 1. There are only 2 possible choices.
Choice 1: AB= 6, AH = 1, JD = 15, JF=1
: 21 x 2
: 42 cm^2
Choice 2: AB=6, AH= 1, JD= 1, JF = 15
: 7 x 16
: 112 cm^2
Choice 2 is must larger than Choice 1, so the maximum area of rectangle ACEG is 112 cm^2.
I did on paper and found out that
ReplyDeleteA-H=1
H-J=6
J-D=1
J-F=15
=122 units^2
First, we would the possible side lengths of each rectangle.
ReplyDeleteThe green rectangle can either be 1*6, or 2*3.
The pink rectangle can be either 3*5 or 1* 15.
(2 + 3) * (3 + 5) = 40 cm squared
(2 + 3) * (1 + 15) = 80 cm squared
(1 + 6) * (3 + 5) = 56 cm squared
(1 + 6) * (1 + 15) = 112 cm squared
Of these, 112 cm squared is the largest number, therefore the largest possible area of this rectangle is 112 cm squared.
Grade 8 POTW
ReplyDeleteI am going to go straight into it. What is the length of FE? First let’s list the lengths we know. We know AB = CD = 30cm and AD = BC = 40cm. Also, figure AEFG = CEFD in terms of area. We also know ABE is a right triangle. GF = FD (we don’t know for sure if BE also equals this). AE = EC. To find out FE, we can use most of the information listed here (I didn’t list everything as that would take too long). Okay, so let’s start at the end. To figure out the length of FE, what can we do. The first thing that comes in mind is forming a right triangle. This can be done by creating a line segment parallel to AB and DC that is made vertically above E (we’ll call the point at the end of the line segment H). This would be 30cm like the other lines. Then, we can use the pythagorean theorem to figure out FE. It would be 30^2 + (FH)^2 = FE^2. To do this, we need to find the length of FH. Using what we know, FH = BC - BE - FD = 40 - BE - FD. Now we need to find the values of BE and FD. You can notice that ABE is a right triangle. This means the length of BE can be solved using the pythagorean theorem again. 30^2 + (BE)^2 = (AE)^2. The length of AE needs to be found. It can be seen that AE and EC are equal in length. EC = 40 - BE. This means that AE = 40 - BE. Back to the equation, 30^2 + (BE)^2 = (40 - BE)^2.
30^2 + (BE)^2 = (40 - BE)^2
900 + (BE)^2 = 1600 - 80(BE) + (BE)^2
900 + (BE)^2 - 900 - (BE)^2 + 80(BE) = 1600 - 80(BE) + (BE)^2 - 900 - (BE)^2 + 80(BE)
80(BE) = 700
80(BE)/80 = 700/80
BE = 8.75cm
To find out FD, we could also form a right triangle by adding a new line segment, but then we have no equal values to go off of other than 30cm. However, we know FD = GF. We also know AG = CD = 30cm as well as the fact that AGF is a right angle. Now, we can form another right triangle by creating line segment AF. Okay, so AF = 40 - FD = 40 - GF. 30^2 + (GF)^2 = 40 - (GF)
900 + (GF)^2 = (40 - GF)^2
…
Note that the equation above is the same as before, except BE is replaced with GF. Due to this, BE = GF = FD = 8.75cm. Now we can finally move back to the original question, the length of FE. We now have the values we need:
30^2 + (40 - 8.75 - 8.75)^2 = (FE)^2
900 + 22.5^2 = (FE)^2
900 + 506.25 = (FE)^2
1406.25 = (FE)^2
√(1406.25) = √((FE)^2)
37.5cm = FE
Therefore, line segment FE is equal to 37.5cm.
Grade 7 POTW
ReplyDeleteI will go right into it. ABJH = 6cm^2. JDEF = 15cm^2. We must use integer side lengths to make this work. An integer is a positive/negative whole number (or 0). To gain the greatest area, we must have the largest side lengths. In other words, we just have to think of combinations of side lengths that when 1 pair is multiplied, the product is 6cm^2, the other pair results in 15cm^2, and when a number from each pair are added, and then multiplied by the sum of the other remaining numbers, results in the largest area. The sum of the numbers from each pair has to both be as large as possible. We could try different possibilities, but it is easy to realize that to get 6, we just need to do 6 x 1 and 15 is 15 x 1. You add the highest number in one pair to the lowest number in the other (same with the other 2). 6 + 1 = 7 and 15 + 1 = 16. 7 x 16 = 112 cm^2. To double check , we can just try to use other numbers. For example, 2 x 3 and 3 x 5. No matter how you pair up the numbers from both pairs (hard to explain), the area would be less than 112cm^2 when they are multiplied.
Therefore, the largest possible area of the entire rectangle ACEG is 112cm^2.
POTW:
ReplyDeleteTo figure out the area of the rectangle ACEG, we need to determine the greatest possible dimension of the green and pink rectangles:
JDEF: 1 x 15 (cm)
ABJH: 1 x 6 (cm)
The width of rectangle JDEF is 15 cm. We need to add the width of the rectangle ABJH, 1 cm, so 15 + 1 = 16, so the width of rectangle ACEG is 16 cm.
The length of rectangle ABJH is 6 cm. We need to add the length of the rectangle JDEF, 1 cm, so 6 + 1 = 7, so the length of the rectangle ACEG is 7 cm.
Now to calculate the area:
16 cm x 7 cm = 112 cm2
The greatest possible area of the rectangle ACEG is 112 cm2.