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Grade 7 POTW: First, I needed to find the side lengths of square ABCD. I just need to square root the area of the square. This means that one side length is equaled to 8 cm. Next, I need to find the area of each triangle, so I can figure out the area of the smaller square. For each triangle, the base is 6 cm and the height of the triangle is 2 cm. To find the area of a triangle, you have to do B*H/2. So you would do 6*2, which is 12, then divide it by 2, which is 6. Also, since there are 4 triangles, we would multiply 6, the area of the triangle, by 4, to give us 24. Then, we would subtract the area of square ABCD by the area of all 4 triangles. So we would do 64-24, leaving us with 40. Therefore, the area of the square EFGH is 40 cm squared
Here how Aleks and Ameya went about solving the Grade 7 POTW: We know that the shape was a square, and the area was 64cm^2. This means all the side lengths were 8cm long. Then we found out that the spaces in between the corners of the larger square and the vertices of the smaller square, was 2cm long. That means the the space after that was 6cm long because the side length was originally 8, but then we subtracted 2. That mean that we can now find out the area of a triangle, that repeats itself 4 times in the square. The area of the triangle was 6 (6*2/2). And then we multiplied that by 4 (6*4=24cm^2). Then we subtracted that from the total (which was 64cm^2) and came out with a difference of 40cm^2. That means that the area of the inside square was 40cm^2.
To get the area of the inside square, I need to subtract the triangles from the outside square.
I know the that area of the outside square is 64 square cm. That means that each side is 8 cm (square root of 64 = 8). I know that AE, BF, HD and CG are 2 cm so the rest of the side is 6 cm. That means the base of the triangle is 2 cm and the height is 6 cm. Area of a triangle = (B*H)/2 A = (2*6)/2 A = 12/2 A= 6 square cm
There are 4 squares 6*4 = 24 square cm
I need to subtract this from the outside square. 64 - 24 = 40 square cm
How I figured this out was to figure out the side length, then calculate each small triangle. After using the area of the large square subtracted by the 4 triangles, I would get the area of the small square.
Since ABCD is a square, the side length would of course be 8 cm. However, we also know that the small length is 2, so I can subtract 8-2=6. The length of the triangle is 6.
The dimensions for the triangle would then be 6 and 2. 6*2/2*4=24 cm squared, since there are 4 triangles and I am calculating area for a triangle.
Therefore, I need to subtract 24 from the area of the larger square, which is 64-20=40 cm squared.
Conclusion: The area of square EFGH is 40 cm squared
Grade 8 POTW: How I figured out was probably complicated, but hopefully it works. All you need to know is algebra. If you think its complicated, just bear with me ok?
Here we go!
Let x be the length of the rectangle Let y be the width of the rectangle
The illogical way of solving the grade 7 POTW: (with a check) This isn't the best strategy but I used the finger strategy again because AE, CG, DH, and BF looked like 1/4 of the whole length, so I multiplied 2m by 4, getting 8 the side length of the outer square. I now know that the area of the big square is 64m^2
To find the area of the triangles, I simply subtracted 8 by 2, and I now know that the base and height of the triangle is 6*2.
(6*2)/2=6 6*4= 24m^2 I multiplied it by four because I saw four triangles. The area of the triangles is 24m^2. I subtract 64 by 24 giving me 40m^2. The area of the inner square is 40m^2. Now, I checked to see if 8 was actually correct, by finding the hypotenuse, which is a squared + b squared= c squared. I got something around 6.32, and I multiplied that by it self and got 39.9424 which is relatively close to 40, that is how I checked my illogical way of finding the side length of the big triangle. I am not sure if my way of checking even works so please tell me if there was a logical way of solving it.
Here is how I solved the POTW Grade 8: So first I labeled the parts of the sides as variables: QA = AB = BR = x PC = CQ = y PS = QR = 3x SR = PQ = 2y Then I made the formula: A = PQ * QR = PQ * QR = (AR * SR)/2 + (CQ * QA)/2 + ( PC * PS)/2 + 10 Next, I solved it: = PQ * QR = (AR * SR)/2 + (CQ * QA)/2 + ( PC * PS)/2 + 10 = 2y * 3x = (2x * 2y)/2 + (y * x)/2 + (y * 3x)/2 + 10 = 6xy = 2xy + (xy)/2 + (3xy)2 + 10 I then multiplied both sides by two to get rid of the /2: = 12xy = 4xy + xy + 3xy + 20 = 4xy = 20 = xy = 5 = 6(5) = 30
Each side of the square is 8cm long, and there is a 2cm gap in between each corner of the square, so the inside squares' side lengths must be each 6cm long. Using the formula to find the area of a square (LxW), we can find that the area of EFGH is 36cm squared.
POTW #17 Grade 8 Solution: 
Square EFGH is 40m squared.
ReplyDeleteI did the work in my math notebook.
The area of square EFGH is 40 cm2. I did my work in my math notebook.
ReplyDeleteGrade 7 POTW:
ReplyDeleteFirst, I needed to find the side lengths of square ABCD. I just need to square root the area of the square. This means that one side length is equaled to 8 cm.
Next, I need to find the area of each triangle, so I can figure out the area of the smaller square. For each triangle, the base is 6 cm and the height of the triangle is 2 cm. To find the area of a triangle, you have to do B*H/2. So you would do 6*2, which is 12, then divide it by 2, which is 6. Also, since there are 4 triangles, we would multiply 6, the area of the triangle, by 4, to give us 24.
Then, we would subtract the area of square ABCD by the area of all 4 triangles. So we would do 64-24, leaving us with 40.
Therefore, the area of the square EFGH is 40 cm squared
Here how Aleks and Ameya went about solving the Grade 7 POTW:
ReplyDeleteWe know that the shape was a square, and the area was 64cm^2. This means all the side lengths were 8cm long. Then we found out that the spaces in between the corners of the larger square and the vertices of the smaller square, was 2cm long. That means the the space after that was 6cm long because the side length was originally 8, but then we subtracted 2. That mean that we can now find out the area of a triangle, that repeats itself 4 times in the square. The area of the triangle was 6 (6*2/2). And then we multiplied that by 4 (6*4=24cm^2). Then we subtracted that from the total (which was 64cm^2) and came out with a difference of 40cm^2. That means that the area of the inside square was 40cm^2.
The outside square has side lengths of 8cm. There is a 2cm gap between each. 6cm*6cm=36cm squared. Or- 4 6cm squared triangles. 60-6= 36 cm squared.
ReplyDeleteGrade 7 POTW
ReplyDeleteTo get the area of the inside square, I need to subtract the triangles from the outside square.
I know the that area of the outside square is 64 square cm. That means that each side is 8 cm (square root of 64 = 8). I know that AE, BF, HD and CG are 2 cm so the rest of the side is 6 cm.
That means the base of the triangle is 2 cm and the height is 6 cm.
Area of a triangle = (B*H)/2
A = (2*6)/2
A = 12/2
A= 6 square cm
There are 4 squares
6*4 = 24 square cm
I need to subtract this from the outside square.
64 - 24 = 40 square cm
The area of square EFGH is 40 square cm.
Grade 7 POTW:
ReplyDeleteHow I figured this out was to figure out the side length, then calculate each small triangle. After using the area of the large square subtracted by the 4 triangles, I would get the area of the small square.
Since ABCD is a square, the side length would of course be 8 cm. However, we also know that the small length is 2, so I can subtract 8-2=6. The length of the triangle is 6.
The dimensions for the triangle would then be 6 and 2. 6*2/2*4=24 cm squared, since there are 4 triangles and I am calculating area for a triangle.
Therefore, I need to subtract 24 from the area of the larger square, which is 64-20=40 cm squared.
Conclusion: The area of square EFGH is 40 cm squared
--Alan
Grade 8 POTW:
ReplyDeleteHow I figured out was probably complicated, but hopefully it works. All you need to know is algebra. If you think its complicated, just bear with me ok?
Here we go!
Let x be the length of the rectangle
Let y be the width of the rectangle
xy-(y/2 * x/3 /2 +y * 2/3 x /2 + x * y/2 / 2)=10
xy-(xy/6 / 2 + 2/3xy / 2 + xy/2 / 2) = 10
xy-(xy/12 + 2xy/6+xy/4) = 6
xy-(xy/12 + 4xy/12 + 3xy/12) = 10
xy-8xy/12=10
xy-2/3xy=10
1/3xy=10
xy=30
Therefore, the area of the rectangle is 30 cm squared, from my calculations in the algebra equations.
--Alan
Grade 7 POTW:
ReplyDeleteThe illogical way of solving the grade 7 POTW: (with a check)
This isn't the best strategy but I used the finger strategy again because AE, CG, DH, and BF looked like 1/4 of the whole length, so I multiplied 2m by 4, getting 8 the side length of the outer square. I now know that the area of the big square is 64m^2
To find the area of the triangles, I simply subtracted 8 by 2, and I now know that the base and height of the triangle is 6*2.
(6*2)/2=6
6*4= 24m^2
I multiplied it by four because I saw four triangles.
The area of the triangles is 24m^2. I subtract 64 by 24 giving me 40m^2. The area of the inner square is 40m^2. Now, I checked to see if 8 was actually correct, by finding the hypotenuse, which is a squared + b squared= c squared. I got something around 6.32, and I multiplied that by it self and got 39.9424 which is relatively close to 40, that is how I checked my illogical way of finding the side length of the big triangle. I am not sure if my way of checking even works so please tell me if there was a logical way of solving it.
Whoops, I didn't see that the area of the square was 64cm^2.
DeleteI obtained an answer of 30 cm squared.
ReplyDelete40 cm2, but I did my work hardcopy
ReplyDeleteI"M LATEEE O~O
ReplyDeletebut I'll still do it!
Here is how I solved the POTW Grade 8:
So first I labeled the parts of the sides as variables:
QA = AB = BR = x
PC = CQ = y
PS = QR = 3x
SR = PQ = 2y
Then I made the formula:
A = PQ * QR
= PQ * QR = (AR * SR)/2 + (CQ * QA)/2 + ( PC * PS)/2 + 10
Next, I solved it:
= PQ * QR = (AR * SR)/2 + (CQ * QA)/2 + ( PC * PS)/2 + 10
= 2y * 3x = (2x * 2y)/2 + (y * x)/2 + (y * 3x)/2 + 10
= 6xy = 2xy + (xy)/2 + (3xy)2 + 10
I then multiplied both sides by two to get rid of the /2:
= 12xy = 4xy + xy + 3xy + 20
= 4xy = 20
= xy = 5
= 6(5) = 30
The area of the whole rectangle is 30 cm^2.
Grade 7 POTW
ReplyDeleteEach side of the square is 8cm long, and there is a 2cm gap in between each corner of the square, so the inside squares' side lengths must be each 6cm long. Using the formula to find the area of a square (LxW), we can find that the area of EFGH is 36cm squared.
The area of EFGH is 40cm^2. Sorry for the really late answer I was working on all the projects. The work was done in my math notebook.
ReplyDelete