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Here is how I went about solving the Grade 8 POTW:
There are two areas, 6 cm^2 and 15 cm^2. Since the rectangles don't have to be to scale, you can put the lengths as "6 x 1" and "15 x 1". Now in order to get the largest possible area, you would have to put: FE = 1 cm GF = 6 cm HG = 15 cm AH = 1 cm
6 + 1 = 7 cm
15 + 1 = 16 cm
And as we all know, the formula for calculating the area of a rectangle is "length x width". So we would do "7 x 16" to get 112. Therefore, the largest possible area for rectangle ACEG is 112 cm^2.
Because the laundry room is a square, the square root of 4 is 2. 2 x 2 = 4 The dining room is also a square, and the square root of 25 is 5. 5 x 5 = 25. By using the lengths of the laundry room and the dining room, I know the width of the entire room. It is 9. 4 + 5 = 9 I know that the living room has the same width of the dining room. Now I just have to find the length by dividing 30 by 5. That equals 6. 30 / 5 = 6 Now I know that the length of the room. The length is 11 6 + 5 = 11 Now, to find the the kitchen area, I just find the area of the entire room subtracting all the rooms except the kitchen. (11 x 9) - 30 - 25 - 4 = 99 - 30 - 25 - 4 = 40 The area of the kitchen is 40 m2.
The area of the kitchen is 18m squared. I found this by finding the numbers used to get the area of the other rooms. The dining room is 25m squared, which means it is 5m times 5m. So the outer part is 5m. The living room is next to the dining room, and is 30 m squared. This can then be 6m times 5m. So the top has 6m and 5m. Together, the top line of the house is 11m. The other side is 5m. The laundry room is 4m squared. So the sides are both 2m. The other side is then added together, 7m. Since it is a rectangle, the other two sides are the same as the other. The right side is 7m, and since we know the dining room side is 5m, the rest is 2m. On the bottom side, the laundry room is 2m, and since the top side is 11m, the rest must be 9m. Multiplying 9 and 2 would be 18m squared. Therefore the kitchen is 18m squared.
so lets begin... first I need to realize the area for each room. then I divided each area per room so that it would make sense based on it's shape so laundry room is 2x2, dinning is 5x5 and living is 5x5. so to figure out the side, I add 2 to 5 and I get 7. because dinning and living have the same height, I know that the side that has 6 for living is on the top.then to figure out the length of the top, I just add 6 to 5 to get 11. the Landry room has a width of 2 so I subtract that to 11 to get 7. so now I know the length of the kitchen is 7 and it's on the same level of laundry so it's height is 2. 2x7 gives me an area of 14 so the answer is that the kitchen has an area of 14 cm^2
First, I need to find the lengths of each side of every other room that I have information about. First, I know that the laundry room is 4 m squared. From basic knowledge of multiplication, I know that 2 x 2 = 4 and since the room is a square, it must have equal sides of 2 m . The dining room is also a square and I also know that 5 x 5 = 25 so each side must be 5 m. For the living room, I know that the 2 shorter sides are 5 m long since it is up against the dining room. I know that 5 + 5 = 10 so I subtract 30 - 10 = 20. There are two sides left, so I know that 20/2 = 10. So know, I know that: Laundry room = 2 m each side Dining room = 5 m each side Living room = 5 m on 2 sides and 10 m on the other 2 Now I need to find out the long living room length minus the laundry room length to find out how long the kitchen is without the dining room part (where it is up against it). 10 - 2 = 8 (living room - laundry room = 8 m). I know that the rest of the kitchen length is 5, since the dining room is 5. 8 + 5 = 13. Since the other side is against the laundry room, I know that the other two sides are 2 m. To then find the area, I did 2 x 13 = 26 Therefore, the area of the kitchen is 26 m squared.
I don't think that the living room has a length of 10 meters because the question states that the area of the living room is 30 meters squared, not the perimeter.
Grade 8 POTW: First of all, I need the maximum lengths for the green and pink rectangle to create the largest possible area for the required rectangle. Since the areas are 15 and 6, the side lengths for these rectangles can be 15*1 and 6*1, creating side lengths of 15 and 6. Multiplying these together, this would give me 90 cm squared.
Grade 7 POTW: Since I was given those two areas, it would automatically give me the side length of the laundry room, which is 2, and the side length of the dining room, which is 5 cm. Because the living room shares a side with the dining room, one of its side lengths is 5 cm. Since its area is 30 cm squared, its other side is 6 cm long. To figure out the side length of the kitchen, I add up the length of the living room and the dining room, subtracting the side length of the laundry room, which gives me 5+6-2 = 9 cm. The other side is 2 cm, since it shares a side with the laundry room. The area would then be 9*2 = 18 cm squared. -Alan
Grade 8 POTW: I messed up, I wrote the max for area of rectangle HJGF, not the large rectangle. The largest area is actually(16+1)(5+1) = 17*6 = 112 cm squared. -Alan
Grade 7 POTW First let's find out what we already know. We know the areas of the 3 rooms except the kitchen. Laundry Room = 4 m squared Dining Room = 25 m squared Living Room = 30 m squared Now we can start the work. We need to know the square roots of the laundry room and the dining room to find out the side lengths. Laundry Room Side = 2 m Dining Room Side = 5 m Now by dividing the area of the living room (30m squared) by 1 side length (5m) you get the other side length. 30 divided by 5 is 6. The other side length is six When you subtract the side length of the 6 m side from 2 m of the laundry room to get part of 1 side length of the kitchen, you get 4 (6-2=4). Add the side length of 5 m to get the full length of 1 side. 5 + 4 is 9. The other side is 2 due the laundry room having an equal side length to that length. The area of a rectangle is length x width and 9 x 2 is 18. Therefore, the final area of the kitchen is 18 m squared.
Grade 8 POTW I won’t explain the repeated information too much because the question explains most of it already. We need to find the largest area for rectangle ACEG. ABJH = 6 cm squared JDEF = 15 cm squared Here are more things we know. AB/HJ/GF + BC/JD/FE = AC/GE. Same thing goes for AH + JF = AG. Here are all possibilities: AH and AB can be, in cm, 1, 2, 3, or 6. JD and JF can be 1, 3, 5, or 15. I won’t explain which side lengths are which, that’s not too important. Now for the work. 1,1=2 6,15=21 2 x 21 = 42 1,3=4 6,5=11 4 x 11 = 44 As you can see, getting a bit closer will raise the area. 1,5=6 6,3=9 6 x 9 = 54 1,15=16 6,1=7 7 x 16 = 112 Now we did all 1 and 6s for ABJH. 2,1=3 3,15=18 3 x 18 = 54 2,3=5 3,5=8 8 x 5 = 40 2,5=10 3,3=6 10 x 6 = 60 2,15=17 3,1=4 17 x 4 = 68 Those are all the combinations. Clearly 112 is the largest area. 1,15=16 6,1=7 7 x 16 = 112. AH can be 1. JF would be 15. AB will be 6. JD would be 1. AG will be 16. AC will be 7. Okay so the largest possible area is 112 cm squared.
I accidentally read Grade 8 POTW wrong therefore solving it wrong.
Grade 8 POTW:
ABJH=6cm2 JDEF=15cm2
What I did to find the largest possible area for ACEG. To find the whole length, you'd have to add up the length and width of the different rectangles.
For example, if ABJH's length was 6 and the width for JEDF is 1, then the width of ACEG would be 7 and the width of ABJH would be 1 and JEDF's length would be 15, the length of ACEG is 16.
In this case 16*7= 112cm2, and if you check all the possible outcomes, this will be the largest possible area.
So if the area of ABJH is 6cm squared, while JDEF is 15cm squared, and we have to find the area of ACEG, then we simply make sure that the longest sides of ABJH, and JDEF are as long as possible. This allows us to have 15x1 for JDEF, and 6x1, on ABJH. Now you add the 1 to 15, and 1 to 6. Therefore the largest area is 112cm squared.
This problem is asking us to find the largest possible area for the rectangle ACEG.
Now regularly, you would multiply the length and width of a rectangle to get the area. But you'll notice that there are sections divided in the rectangle. That means that in order to get the area, we first have to add up all the areas of the rectangles within the bigger rectangle and then multiply them together to get the area.
Since it says that the image given is not to scale, we could say that the widths of the rectangles are 1 cm. We also know that the lengths of two of the rectangles are 6 cm and 15 cm, so lets solve the equation!
These are the lengths/widths of the sides:
AB- 6cm HJ- 6 cm AH- 1 cm BJ- 1 cm FE- 1 cm JD- 1 cm JF- 15 cm DE- 15 cm
For the grade 8 potw I decided to try and find every possible option. For the first square the possible combinations are 1 by 6 or a 2 by 3. Then for the other block it can be 3 by 5 or 1 by 15. Since we know that they are at the 2 corners of the square we can find the dimensions of the square. Going through the combinations (eg. vertical 1 by 6 with a vertical 1 by 15 gets you a 2 by 21 square or 42cm2). Then a horizontal 1 by 6 and a vertical 1 by 15 to get a 7 by 16 or 112cm2 etc. Going through all the combinations the largest number is the 112cm2 rectangle.
Since we want to find the largest possible area, we want to ensure that we are making it so that the lengths are as large as possible. For rectangle ABJH, it is 6cm2 so in order to make the length as large as possible, we would make it 6x1. For rectangle ACEG, it is 15cm2 so we would make it 15x1. Now, we want to add 6+1 to get 7 and 15+1 to get 16. We then multiply 7 by 16 to get 112.
I need the lengths and widths of the 3 other rooms to find the area of the kitchen. Lengths and Widths: - Laundry Room: 2 x 2 (Area is 4 square meters and it is square) - Dining Room:5 x 5 (Area is 25 square meters and it is square) - Living Room:6 x 5 (Area is 30 square meters and it is rectangular. The shorter side is next to the Dining Room, so it means that the shorter side is 5 m long. The only option for the length is 6 m. - Kitchen: ?
To find the kitchen's width, you can take the Laundry Room's width because they are equal in width. Kitchen: 2 x ? To find the kitchen's length,you need to add the length of the dining room and the length of the (living room - laundry room). Equation: Dining room length + (living room- laundry room) = 5 + (6-2) = 5 + (4) = 9 The length of the kitchen is 9.
Since the length is 9 and the width is 2, the area has to be 18 square meters.
Grade 7 POTW lets start with the laundry room. we know that in order for it to be 4 m squared it has to be 2x2. this means that the short side of the kitchen is also 2m. if the kitchen is 30m squared then the other side is 15m: 2x15=30, so we know the the total length of the bottom is 17 m. Since the dining room is 25m Squared it has to be 5x5. this means that the living room is also 5m tall,if we subtract 5 from 17 we get 12. so if we do 12x5 we get 60, which is the area of the living room.
I got 2x9. I got this by first finding the dimensions of the rooms. The dining room was 5x5 and the living room was 5x6. Then I added the bottoms of the living and dining rooms together and subtracted 2 from the laundry room. I also made the height 2 because the laundry room is 2x2.
Dining room side = 5 m2 Living room sides = 5m2 and 6m2 Side = 7 m2 Top = 11 m2
The area of the kitchen is 18m2
First I had to find the length of each wall. So I know that if the laundry room’s area is 4 m2 then that means that both sides are 2 m2 because it is a square. Also since the kitchen’s side length is the same is the laundry room it is also 2 m2. So next, it said the areas of the living room and the dining room. The dining room area is 25 m2 and it is a square. This means that each side of the dining room is the same length. Since 5x5 is 25 that means that both sides of the dining room is 5m. The living room area is 30 m2. So since the top side of the living room is longer that means the top side of the living room is 6m and the side length of the living room is 5m. I know this because The dining room and living room are the same height in 5m. Since I know the length of each top side of both rooms I now know the length of the whole top side which is 11m. Now to find the area of the kitchen I have to find the length of each side. I already know that the shorter side kitchen is 2m. So to find the length of the longer side I subtract 2m from 11m. 11 m is the length of the top side and the 2m is the length of the laundry room. So to get the longer side of the kitchens length I subtract 2 from 11 to get 9 which is the length of the longer side of the kitchen. 9 x 2 = 18 which is the area of the kitchen.
I decided to make ABJH and JDEF as long as possible and to make them into a rectangle, with JDEF being the length side and ABJH being the width. The length would be 16cm (15 from JDEF+1 from ABJH) and the width would be 7cm (6 from ABJH and one from JDEF). Multiplied together, it would make the largest possible area of ACEG, or 112 cm2
We know taht ABJH is 6cm^2, and DEFJ is 15cm^2. The important thing is that the diagram is not to scale. That means that we can say rectangle ABJH is 6 x 1 and rectangle DEFJ is 15 x 1. Using these numbers we can determine that the length is 7 and the width is 16. (6+1=7) length (15+1=16) width Therefore when we multiply these two we should get the largest area possible, which is 112cm^2.
in case one we say the purple one is 1 and 15 and the green one is 1 and 6 this gives us c e to be 16 and e g to be 7 giving us a total area of 7 times 16 being 112
in case 2 we would say that the purple one is 3 and 5 and the green one is 1 and 6 giving us c e to be 6 and e g to be 9 giving the total to be 6 times 9 which is 54
in case 3 we would say that the purple one is 1 and 15 and the green one is 2 and 3 giving us c e to be 17 and e g to be 4 giving the total area to be 4 times 17 which is 68
in case 4 we would say that the purple one is 5 and 3 and the green one is 2 and 3 making c e 7 and e g to be 6 which makes the total area to be 6 times 7 which is 42
Laundry Room = 4 m squared 2m x 2m dimensions, as a square room
Dining Room = 25 m squared 5m x 5m, as another square room
Living room = Rectangular 30 m squared Shares a side with dining room which is 5m one side. 5m x 6m = 30m squared
Kitchen: It shares a side with everything. It's width is the same as the Laundry room's, making it 2m. It also shares a side with the dining and living room. THe lengths combined of those two rooms are 11m (6m + 5m). However, some of that length is taken up by the laundry room, which is 2m. So, the kitchen takes up 9m length, and 2m width. Finding the area of that.
To find out the kitchen's area, we have to first figure out the areas of rooms around it. We know that 4 and 25 are perfect squares. 4=2x2 25=5x5 Since the living room's area isn't a perfect square, we have to use the dining room to determine what the length and width of the living room is. The dining room is a square since it's a perfect square. So we already know the length of the living room because of the dining room (5x5) So the length of the living room has to be 5 then. The overall area is 30. 30/5 = 6 The living room's area is 6x5 Next, Since the laundry room is 2x2 and it and the kitchen have the same length so we know that the length of the kitchen is 2. Then, To find out the width, we have to use the living room and the laundry room. The living room's width is 6 and the laundry room's width is 2. 2 can go into 6 3 times. 6/2 = 3. The kitchen is already wider than the 5x5 dining room, therefore, the kitchen is more then 5 m. The kitchen and the laundry room both equal the same width as the dining room and the living room. The laundry room, if it was a perfect rectangle, would be a third of the living room since 6/2=3. So, if you add the width of the living room and dining room, you get 11. If you take away the laundry room, you take away a third of a living room. 6 - 2 = 4 The dining room is 5 m wide. 4 + 5 = 9 Multiply 9 by the length. 9 x 2 = 18 The area of the kitchen is 18m2
The question already tells us the areas of two of the rooms; the dining room and the laundry room. The square root of 4 is 2, so now we know that the dining room is 2 by 2. The square root of 25 is 5, so now we know that the laundry room is 5 by 5. Because the laundry room is the same width as the kitchen, it means that the width of the kitchen is 2 m. The dining room is the same width as the living room, so that means the width of the living room is 5 m. The question already tells us the area of the living room, which is 30 cm. 30 / 5 = 6. This means that the length of the living room must be 6. The dining room's length is 5. 6 + 5 = 11. This means that the length of the entire floor must be 11 m. The "bottom" half of the floor only has 2 rooms: the kitchen and the laundry room. Because we already know the length of the entire floor and the length of the laundry room, I just need to subtract them to find the length of the other side of the kitchen. 11 - 2 = 9. We already know the length of the other side of the kitchen (2 m), so now we just need to multiply the two lengths to get the area. 9 x 2 = 18. The area of the kitchen is 18 m squared. :)
The largest possible side lengths would be 7 cm by 16 cm. This is because we can say that side AC would be 6+1cm and EG would be 15+1cm This ultimately would add up to 7cm and 16cm, giving us the biggest possibly area of rectangle ACEG, 112 cm^2
The areas of rectangle ABJH and JDEF are already given to us. In order to find the largest possible area of rectangle ACEG, we can utilize the fact that the image is not to scale. Which means, the measurements of rectangle ABJH and JDEF can be made up. To get the largest area, the largest possible length and width must be found. In this case, for rectangle ABJH, the dimensions can be 6cm by 1 cm. Rectangle JDEF's dimensions can be 15cm by 1 cm.
Now to find the area of rectangle ACEG. Length: 7cm 6cm + 1cm = 7cm Width: 16cm 15cm + 1cm = 16 cm Area = Length x Width Area = 7cm x 16 cm Area = 112 cm^2
Therefore, the largest possible area for rectangle ACEG is 112 cm^2
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ReplyDeleteHere is how I went about solving the Grade 8 POTW:
ReplyDeleteThere are two areas, 6 cm^2 and 15 cm^2. Since the rectangles don't have to be to scale, you can put the lengths as "6 x 1" and "15 x 1". Now in order to get the largest possible area, you would have to put:
FE = 1 cm
GF = 6 cm
HG = 15 cm
AH = 1 cm
6 + 1 = 7 cm
15 + 1 = 16 cm
And as we all know, the formula for calculating the area of a rectangle is "length x width". So we would do "7 x 16" to get 112. Therefore, the largest possible area for rectangle ACEG is 112 cm^2.
Because the laundry room is a square, the square root of 4 is 2.
ReplyDelete2 x 2 = 4
The dining room is also a square, and the square root of 25 is 5.
5 x 5 = 25.
By using the lengths of the laundry room and the dining room, I know the width of the entire room. It is 9.
4 + 5 = 9
I know that the living room has the same width of the dining room. Now I just have to find the length by dividing 30 by 5. That equals 6.
30 / 5 = 6
Now I know that the length of the room. The length is 11
6 + 5 = 11
Now, to find the the kitchen area, I just find the area of the entire room subtracting all the rooms except the kitchen.
(11 x 9) - 30 - 25 - 4 =
99 - 30 - 25 - 4 =
40
The area of the kitchen is 40 m2.
The area of the kitchen is 18m squared. I found this by finding the numbers used to get the area of the other rooms. The dining room is 25m squared, which means it is 5m times 5m. So the outer part is 5m. The living room is next to the dining room, and is 30 m squared. This can then be 6m times 5m. So the top has 6m and 5m. Together, the top line of the house is 11m. The other side is 5m. The laundry room is 4m squared. So the sides are both 2m. The other side is then added together, 7m. Since it is a rectangle, the other two sides are the same as the other. The right side is 7m, and since we know the dining room side is 5m, the rest is 2m. On the bottom side, the laundry room is 2m, and since the top side is 11m, the rest must be 9m. Multiplying 9 and 2 would be 18m squared. Therefore the kitchen is 18m squared.
ReplyDeleteso the area of the kitchen is 14
ReplyDeleteso lets begin...
first I need to realize the area for each room.
then I divided each area per room so that it would make sense based on it's shape so laundry room is 2x2, dinning is 5x5 and living is 5x5.
so to figure out the side, I add 2 to 5 and I get 7.
because dinning and living have the same height, I know that the side that has 6 for living is on the top.then to figure out the length of the top, I just add 6 to 5 to get 11.
the Landry room has a width of 2 so I subtract that to 11 to get 7. so now I know the length of the kitchen is 7 and it's on the same level of laundry so it's height is 2. 2x7 gives me an area of 14 so the answer is that the kitchen has an area of 14 cm^2
my bad, 11 - 2 is actually 9 not 7 so the answer would be 18 cm^2
DeleteGrade 7 POTW #3
ReplyDeleteFirst, I need to find the lengths of each side of every other room that I have information about. First, I know that the laundry room is 4 m squared. From basic knowledge of multiplication, I know that 2 x 2 = 4 and since the room is a square, it must have equal sides of 2 m . The dining room is also a square and I also know that 5 x 5 = 25 so each side must be 5 m. For the living room, I know that the 2 shorter sides are 5 m long since it is up against the dining room. I know that 5 + 5 = 10 so I subtract 30 - 10 = 20. There are two sides left, so I know that 20/2 = 10. So know, I know that:
Laundry room = 2 m each side
Dining room = 5 m each side
Living room = 5 m on 2 sides and 10 m on the other 2
Now I need to find out the long living room length minus the laundry room length to find out how long the kitchen is without the dining room part (where it is up against it). 10 - 2 = 8 (living room - laundry room = 8 m).
I know that the rest of the kitchen length is 5, since the dining room is 5. 8 + 5 = 13. Since the other side is against the laundry room, I know that the other two sides are 2 m.
To then find the area, I did 2 x 13 = 26 Therefore, the area of the kitchen is 26 m squared.
I don't think that the living room has a length of 10 meters because the question states that the area of the living room is 30 meters squared, not the perimeter.
DeleteGrade 8 POTW:
ReplyDeleteFirst of all, I need the maximum lengths for the green and pink rectangle to create the largest possible area for the required rectangle. Since the areas are 15 and 6, the side lengths for these rectangles can be 15*1 and 6*1, creating side lengths of 15 and 6. Multiplying these together, this would give me 90 cm squared.
Grade 7 POTW:
ReplyDeleteSince I was given those two areas, it would automatically give me the side length of the laundry room, which is 2, and the side length of the dining room, which is 5 cm.
Because the living room shares a side with the dining room, one of its side lengths is 5 cm. Since its area is 30 cm squared, its other side is 6 cm long.
To figure out the side length of the kitchen, I add up the length of the living room and the dining room, subtracting the side length of the laundry room, which gives me 5+6-2 = 9 cm.
The other side is 2 cm, since it shares a side with the laundry room.
The area would then be 9*2 = 18 cm squared.
-Alan
Grade 8 POTW:
ReplyDeleteI messed up, I wrote the max for area of rectangle HJGF, not the large rectangle. The largest area is actually(16+1)(5+1) = 17*6 = 112 cm squared.
-Alan
Grade 7 POTW
ReplyDeleteFirst let's find out what we already know. We know the areas of the 3 rooms except the kitchen.
Laundry Room = 4 m squared
Dining Room = 25 m squared
Living Room = 30 m squared
Now we can start the work. We need to know the square roots of the laundry room and the dining room to find out the side lengths.
Laundry Room Side = 2 m
Dining Room Side = 5 m
Now by dividing the area of the living room (30m squared) by 1 side length (5m) you get the other side length. 30 divided by 5 is 6. The other side length is six
When you subtract the side length of the 6 m side from 2 m of the laundry room to get part of 1 side length of the kitchen, you get 4 (6-2=4). Add the side length of 5 m to get the full length of 1 side. 5 + 4 is 9. The other side is 2 due the laundry room having an equal side length to that length.
The area of a rectangle is length x width and 9 x 2 is 18. Therefore, the final area of the kitchen is 18 m squared.
Grade 8 POTW
ReplyDeleteI won’t explain the repeated information too much because the question explains most of it already. We need to find the largest area for rectangle ACEG.
ABJH = 6 cm squared
JDEF = 15 cm squared
Here are more things we know. AB/HJ/GF + BC/JD/FE = AC/GE. Same thing goes for AH + JF = AG.
Here are all possibilities:
AH and AB can be, in cm, 1, 2, 3, or 6.
JD and JF can be 1, 3, 5, or 15.
I won’t explain which side lengths are which, that’s not too important. Now for the work.
1,1=2 6,15=21 2 x 21 = 42
1,3=4 6,5=11 4 x 11 = 44
As you can see, getting a bit closer will raise the area.
1,5=6 6,3=9 6 x 9 = 54
1,15=16 6,1=7 7 x 16 = 112
Now we did all 1 and 6s for ABJH.
2,1=3 3,15=18 3 x 18 = 54
2,3=5 3,5=8 8 x 5 = 40
2,5=10 3,3=6 10 x 6 = 60
2,15=17 3,1=4 17 x 4 = 68
Those are all the combinations. Clearly 112 is the largest area. 1,15=16 6,1=7 7 x 16 = 112. AH can be 1. JF would be 15. AB will be 6. JD would be 1. AG will be 16. AC will be 7.
Okay so the largest possible area is 112 cm squared.
I accidentally read Grade 8 POTW wrong therefore solving it wrong.
ReplyDeleteGrade 8 POTW:
ABJH=6cm2
JDEF=15cm2
What I did to find the largest possible area for ACEG. To find the whole length, you'd have to add up the length and width of the different rectangles.
For example, if ABJH's length was 6 and the width for JEDF is 1, then the width of ACEG would be 7 and the width of ABJH would be 1 and JEDF's length would be 15, the length of ACEG is 16.
In this case 16*7= 112cm2, and if you check all the possible outcomes, this will be the largest possible area.
The largest possible area of ACEG is 112cm2
Grade 7 POTW:
Laundry room: 4m2= 2*2
Dining room: 25m2= 5*5
Living room: 30cm2= 6*5
Kitchen: ???
25+30+4=59m2
59m2 is the area of all the rooms we know of.
Area of house:
5(Living room side) + 2 (laundry room) = 7
6(Other living room side) + 5 (Dining room) = 11
7*11= 77
77-59=18m2
There area of the kitchen is 18m2.
Grade 8 POTW:
ReplyDeleteTo get the largest possible area, I have to maximize the lengths of the rectangles.
To do this, rectangle JDEF would 15x1 and rectangle ABJH would be 6x1.
That means side lengths
FE = 1
DE = 15
AH = 1
BA = 6
15 + 1 = 16
6 + 1 = 7
Area of a rectangle = L*W
16 * 7 = 112
The largest possible area for rectangle ACEG is 122 cm squared.
TYPO sorry. I meant to write 112 as my final answer
DeleteNot too hard, for POTW Grade 8.
ReplyDeleteSo if the area of ABJH is 6cm squared, while JDEF is 15cm squared, and we have to find the area of ACEG, then we simply make sure that the longest sides of ABJH, and JDEF are as long as possible. This allows us to have 15x1 for JDEF, and 6x1, on ABJH. Now you add the 1 to 15, and 1 to 6. Therefore the largest area is 112cm squared.
Grade 8 POTW:
ReplyDeleteThis problem is asking us to find the largest possible area for the rectangle ACEG.
Now regularly, you would multiply the length and width of a rectangle to get the area. But you'll notice that there are sections divided in the rectangle. That means that in order to get the area, we first have to add up all the areas of the rectangles within the bigger rectangle and then multiply them together to get the area.
Since it says that the image given is not to scale, we could say that the widths of the rectangles are 1 cm. We also know that the lengths of two of the rectangles are 6 cm and 15 cm, so lets solve the equation!
These are the lengths/widths of the sides:
AB- 6cm
HJ- 6 cm
AH- 1 cm
BJ- 1 cm
FE- 1 cm
JD- 1 cm
JF- 15 cm
DE- 15 cm
6+1= 7
15+1= 16
7x16= 112
Therefore the largest possible area is 112 cm2
For the grade 8 potw I decided to try and find every possible option. For the first square the possible combinations are 1 by 6 or a 2 by 3. Then for the other block it can be 3 by 5 or 1 by 15. Since we know that they are at the 2 corners of the square we can find the dimensions of the square. Going through the combinations (eg. vertical 1 by 6 with a vertical 1 by 15 gets you a 2 by 21 square or 42cm2). Then a horizontal 1 by 6 and a vertical 1 by 15 to get a 7 by 16 or 112cm2 etc. Going through all the combinations the largest number is the 112cm2 rectangle.
ReplyDeleteArea of green rectangle: 6cm^2 Side option: 1x6 or 2x3
ReplyDeleteArea of pink rectangle: 15cm^2 Side options: 1x15 or 3x5.
To have the biggest areas, I will choose 1x6 and 1x15.
So, side AG will be 16cm and side AC will be 7cm. 7cmx16cm=112cm^2
Grade 8 POTW:
ReplyDeleteSince we want to find the largest possible area, we want to ensure that we are making it so that the lengths are as large as possible. For rectangle ABJH, it is 6cm2 so in order to make the length as large as possible, we would make it 6x1. For rectangle ACEG, it is 15cm2 so we would make it 15x1. Now, we want to add 6+1 to get 7 and 15+1 to get 16. We then multiply 7 by 16 to get 112.
The largest possible area is 112cm2
Grade 7 POTW
ReplyDeleteI need the lengths and widths of the 3 other rooms to find the area of the kitchen.
Lengths and Widths:
- Laundry Room: 2 x 2 (Area is 4 square meters and it is square)
- Dining Room:5 x 5 (Area is 25 square meters and it is square)
- Living Room:6 x 5 (Area is 30 square meters and it is rectangular. The shorter side is next to the Dining Room, so it means that the shorter side is 5 m long. The only option for the length is 6 m.
- Kitchen: ?
To find the kitchen's width, you can take the Laundry Room's width because they are equal in width.
Kitchen: 2 x ?
To find the kitchen's length,you need to add the length of the dining room and the length of the (living room - laundry room).
Equation:
Dining room length + (living room- laundry room)
= 5 + (6-2)
= 5 + (4)
= 9
The length of the kitchen is 9.
Since the length is 9 and the width is 2, the area has to be 18 square meters.
Grade 7 POTW
ReplyDeletelets start with the laundry room. we know that in order for it to be 4 m squared it has to be 2x2. this means that the short side of the kitchen is also 2m. if the kitchen is 30m squared then the other side is 15m: 2x15=30, so we know the the total length of the bottom is 17 m.
Since the dining room is 25m Squared it has to be 5x5. this means that the living room is also 5m tall,if we subtract 5 from 17 we get 12. so if we do 12x5 we get 60, which is the area of the living room.
Grade 7 POTW
ReplyDeleteI got 2x9. I got this by first finding the dimensions of the rooms. The dining room was 5x5 and the living room was 5x6. Then I added the bottoms of the living and dining rooms together and subtracted 2 from the laundry room. I also made the height 2 because the laundry room is 2x2.
Grade 7 POTW
ReplyDeleteLaundry room = 4 m2
Dining room = 25 m2
Living room = 30 m2
Dining room side = 5 m2
Living room sides = 5m2 and 6m2
Side = 7 m2
Top = 11 m2
The area of the kitchen is 18m2
First I had to find the length of each wall. So I know that if the laundry room’s area is 4 m2 then that means that both sides are 2 m2 because it is a square. Also since the kitchen’s side length is the same is the laundry room it is also 2 m2. So next, it said the areas of the living room and the dining room. The dining room area is 25 m2 and it is a square. This means that each side of the dining room is the same length. Since 5x5 is 25 that means that both sides of the dining room is 5m. The living room area is 30 m2. So since the top side of the living room is longer that means the top side of the living room is 6m and the side length of the living room is 5m. I know this because The dining room and living room are the same height in 5m. Since I know the length of each top side of both rooms I now know the length of the whole top side which is 11m. Now to find the area of the kitchen I have to find the length of each side. I already know that the shorter side kitchen is 2m. So to find the length of the longer side I subtract 2m from 11m. 11 m is the length of the top side and the 2m is the length of the laundry room. So to get the longer side of the kitchens length I subtract 2 from 11 to get 9 which is the length of the longer side of the kitchen. 9 x 2 = 18 which is the area of the kitchen.
Assuming BCDJ is 1cm^2, I could create the greatest perimeter, thus creating the greatest number of 112 cm2.
ReplyDeleteGrade 8 POTW
ReplyDeleteI decided to make ABJH and JDEF as long as possible and to make them into a rectangle, with JDEF being the length side and ABJH being the width. The length would be 16cm (15 from JDEF+1 from ABJH) and the width would be 7cm (6 from ABJH and one from JDEF). Multiplied together, it would make the largest possible area of ACEG, or 112 cm2
We know taht ABJH is 6cm^2, and DEFJ is 15cm^2. The important thing is that the diagram is not to scale. That means that we can say rectangle ABJH is 6 x 1 and rectangle DEFJ is 15 x 1. Using these numbers we can determine that the length is 7 and the width is 16.
ReplyDelete(6+1=7) length
(15+1=16) width
Therefore when we multiply these two we should get the largest area possible, which is 112cm^2.
in case one we say the purple one is 1 and 15 and the green one is 1 and 6 this gives us c e to be 16 and e g to be 7 giving us a total area of 7 times 16 being 112
ReplyDeletein case 2 we would say that the purple one is 3 and 5 and the green one is 1 and 6 giving us c e to be 6 and e g to be 9 giving the total to be 6 times 9 which is 54
in case 3 we would say that the purple one is 1 and 15 and the green one is 2 and 3 giving us c e to be 17 and e g to be 4 giving the total area to be 4 times 17 which is 68
in case 4 we would say that the purple one is 5 and 3 and the green one is 2 and 3 making c e 7 and e g to be 6 which makes the total area to be 6 times 7 which is 42
112 cm2 hardcopy.
ReplyDeleteAn Grade 7 POTW
ReplyDeleteLaundry Room = 4 m squared
2m x 2m dimensions, as a square room
Dining Room = 25 m squared
5m x 5m, as another square room
Living room = Rectangular 30 m squared
Shares a side with dining room which is 5m one side. 5m x 6m = 30m squared
Kitchen: It shares a side with everything. It's width is the same as the Laundry room's, making it 2m. It also shares a side with the dining and living room. THe lengths combined of those two rooms are 11m (6m + 5m). However, some of that length is taken up by the laundry room, which is 2m. So, the kitchen takes up 9m length, and 2m width. Finding the area of that.
9m x 2m = 18m squared
THe kitchen is 18m squared.
To find out the kitchen's area, we have to first figure out the areas of rooms around it. We know that 4 and 25 are perfect squares.
ReplyDelete4=2x2
25=5x5
Since the living room's area isn't a perfect square, we have to use the dining room to determine what the length and width of the living room is. The dining room is a square since it's a perfect square. So we already know the length of the living room because of the dining room (5x5) So the length of the living room has to be 5 then. The overall area is 30.
30/5 = 6
The living room's area is 6x5
Next, Since the laundry room is 2x2 and it and the kitchen have the same length so we know that the length of the kitchen is 2.
Then, To find out the width, we have to use the living room and the laundry room. The living room's width is 6 and the laundry room's width is 2. 2 can go into 6 3 times.
6/2 = 3.
The kitchen is already wider than the 5x5 dining room, therefore, the kitchen is more then 5 m. The kitchen and the laundry room both equal the same width as the dining room and the living room. The laundry room, if it was a perfect rectangle, would be a third of the living room since 6/2=3.
So, if you add the width of the living room and dining room, you get 11. If you take away the laundry room, you take away a third of a living room.
6 - 2 = 4
The dining room is 5 m wide.
4 + 5 = 9
Multiply 9 by the length.
9 x 2 = 18
The area of the kitchen is 18m2
Grade 7 POTW #3:
ReplyDeleteThe question already tells us the areas of two of the rooms; the dining room and the laundry room.
The square root of 4 is 2, so now we know that the dining room is 2 by 2.
The square root of 25 is 5, so now we know that the laundry room is 5 by 5.
Because the laundry room is the same width as the kitchen, it means that the width of the kitchen is 2 m.
The dining room is the same width as the living room, so that means the width of the living room is 5 m. The question already tells us the area of the living room, which is 30 cm. 30 / 5 = 6. This means that the length of the living room must be 6. The dining room's length is 5. 6 + 5 = 11. This means that the length of the entire floor must be 11 m.
The "bottom" half of the floor only has 2 rooms: the kitchen and the laundry room. Because we already know the length of the entire floor and the length of the laundry room, I just need to subtract them to find the length of the other side of the kitchen.
11 - 2 = 9. We already know the length of the other side of the kitchen (2 m), so now we just need to multiply the two lengths to get the area.
9 x 2 = 18. The area of the kitchen is 18 m squared.
:)
Grade 8 POTW:
ReplyDeleteABJH can be assumed to have 2 sides being 6cm and the other 2 being 1cm
DEFJ can be assumed to have 2 sides being 15cm and the other 2 being 1cm
6cm+1cm=7cm
15cm+1cm=16cm
7cm*16cm=122cm^2
The largest possible area for ACEG is 122cm^2
Grade 8 POTW:
ReplyDeleteThe largest possible side lengths would be 7 cm by 16 cm. This is because we can say that side AC would be 6+1cm and EG would be 15+1cm
This ultimately would add up to 7cm and 16cm, giving us the biggest possibly area of rectangle ACEG, 112 cm^2
Grade 8 POTW
ReplyDeleteThe areas of rectangle ABJH and JDEF are already given to us. In order to find the largest possible area of rectangle ACEG, we can utilize the fact that the image is not to scale. Which means, the measurements of rectangle ABJH and JDEF can be made up. To get the largest area, the largest possible length and width must be found. In this case, for rectangle ABJH, the dimensions can be 6cm by 1 cm. Rectangle JDEF's dimensions can be 15cm by 1 cm.
Now to find the area of rectangle ACEG.
Length: 7cm
6cm + 1cm = 7cm
Width: 16cm
15cm + 1cm = 16 cm
Area = Length x Width
Area = 7cm x 16 cm
Area = 112 cm^2
Therefore, the largest possible area for rectangle ACEG is 112 cm^2