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3 possible side lengths include 1. Triangle: 40 cm, total 120 cm 1. Square: 20 cm, total 80 cm 2. Triangle: 50 cm, total 150 cm 2. Square: 12.5 cm, total 50 cm 3. Triangle: 20 cm, total 60 cm 3. Square: 35 cm, total 140 cm
Grade 7 POTW There are many possibilities for the side lengths of the triangles and squares. 1. Triangle side length=20 cm Square side length=5 cm. In order to check that I am correct, I will have to review the question in reverse. One triangle side length=20 cm and there are 3 equilateral triangles. 3x3x20=180. One square side length = 5cm. 5x4=20. 180+20=200 cm of rope. 2. The triangle side length can be 8 cm and the square side length can be 32 cm. 8x3x3= 72. 32x4=128. 128+72=200 cm of rope. 3. The triangle side length can be 12 cm and the square side length can be 23 cm. 12x3x3=108. 23x4=92. 108+92=200 cm of rope. 4. The triangle side length can be 4 cm and the square side length can be 41 cm. 4x3x3=36. 41x4=164. 164+26=200 cm of rope. 5. The triangle side length can be 16 cm and the square side length can be 14 cm. 16x3x3= 144. 14x4=56. 144+56=200 cm of rope.
There are 5 possibilities for the side lengths of the squares and triangles. Starting at 200 cm, there was a pattern. Going down every 36 cm, there was a possible perimeter for a square and the difference of that perimeter and the total amount of rope was a correct amount for the perimeter of 3 equilateral triangles. I started by going down by 9 cm because there are 3 triangles with 3 sides each meaning that the total perimeter of the three triangles had to be a multiple of 9 cm. I then realized that only when 4 9's passed, would the square and triangle's perimeter's would be 200 cm. Starting from 0 and going up by 36, there are 5 multiples before the 6th multiple goes over 200 cm of rope.
Grade 8 POTW Okay, so we have 100cm of rope and we get to make one cut so that: there are 2 pieces, one piece forms a rectangle with a side length of 8cm, the second piece forms a square, and the area of both quadrilaterals are equal. The perimeters of both shapes should add up to 100cm. We know that the square would have all 4 sides equal. In other words square perimeter = 4s. The area would be s x s = s^2. The rectangle’s perimeter would be the remainder of that 100cm or 100 - 4s. We already know 2 sides (8cm), so the length of the other 2 sides would be 100 - 4s - 8 - 8 = 84 - 4s. The length of one side would be (84 - 4s)/2 = 42 - 2s. The area of the rectangle would then be l x w or 8 x (42 - 2s) = 336 - 16s. Since the areas of both the rectangle and square are equal, we can find the value of s by using quadratics: s^2 = 336 - 16s s^2 - 336 + 16s = 336 - 16s - 336 + 16s s^2 - 336 + 16s = 0 s^2 + 16s - 336 = 0 s^2 - 12s + 28s - 336 = 0 (you could go straight to the answer from here, but I’ll show steps) s(s - 12) + 28(s - 12) = 0 (s - 12)(s + 28) = 0 If s = 12, the equation would be true OR if s = -28, the equation would be true. Therefore, s can be equal to 12 or -28. However, s is a side length, so it can’t be negative meaning it must be 12cm. s^2 = 12^2 = 12 x 12 = 144cm^2 42 - 2s = 42 - 2(12) = 42 - 24 = 18cm 8 x 18 = 144cm^2 Both areas are equal. To find out the where to make the cut: 4s = 4(12) = 48cm 2(8 + 18) = 2(26) = 52cm 100 = 48 + 52 Therefore, the cut should be made at 48cm so that the 48cm is for the square (second piece) while the 52cm is for the rectangle (first piece).
Grade 7 POTW We have 200cm cut into four pieces. If 3 pieces are made to form identical equilateral triangles with integer side lengths, it means that 3 pieces are the same length. The fourth piece is used to form a square with integer side lengths. In integer is a whole number (not decimal). To determine all the side length values, we just need to start from the bottom. Let t = triangle side length and s = square side length. 3t x 3 = 9t 4s 9t + 4s = 200cm To save time, instead of trying every possibility, we can just skip it, as t must be divisible by 4 to allow s to be an integer. This is because if t isn’t for example, t = 3, then 200 - 9(3) = 173 which isn’t divisible by 4. This is how it works because if you subtract 200 (divisible by 4) by a number not divisible by 4, the result won’t be divisible by 4. Now, we can just list each number multiplied by 4 for the t value until right before we reach 200: 200 - 9(4) = 200 - 36 = 164 = 4s 200 - 9(8) = 200 - 72 = 128 = 4s 200 - 9(12) = 200 - 108 = 92 = 4s 200 - 9(16) = 200 - 144 = 56 = 4s 200 - 9(20) = 200 - 180 = 20 = 4s 200 - 9(24) = 200 - 216 (too high) 164/4 = 41 = s 128/4 = 32 = s 92/4 = 23 = s 56/4 = 14 = s 20/4 = 5 = s Therefore, the 5 possible side lengths for each triangle and square is: t = 4cm, s = 41cm t = 8cm, s = 32cm t = 12cm, s = 23cm t = 16cm, s = 14cm t = 20cm, s = 5cm
Grade 8 POTW Okay, so we have 100cm of rope and we get to make one cut so that: there are 2 pieces, one piece forms a rectangle with a side length of 8cm, the second piece forms a square, and the area of both quadrilaterals are equal. The perimeters of both shapes should add up to 100cm. We know that the square would have all 4 sides equal. In other words square perimeter = 4s. The area would be s x s = s^2. The rectangle’s perimeter would be the remainder of that 100cm or 100 - 4s. We already know 2 sides (8cm), so the length of the other 2 sides would be 100 - 4s - 8 - 8 = 84 - 4s. The length of one side would be (84 - 4s)/2 = 42 - 2s. The area of the rectangle would then be l x w or 8 x (42 - 2s) = 336 - 16s. Since the areas of both the rectangle and square are equal, we can find the value of s by using quadratics: s^2 = 336 - 16s s^2 - 336 + 16s = 336 - 16s - 336 + 16s s^2 - 336 + 16s = 0 s^2 + 16s - 336 = 0 s^2 - 12s + 28s - 336 = 0 (you could go straight to the answer from here, but I’ll show steps) s(s - 12) + 28(s - 12) = 0 (s - 12)(s + 28) = 0 If s = 12, the equation would be true OR if s = -28, the equation would be true. Therefore, s can be equal to 12 or -28. However, s is a side length, so it can’t be negative meaning it must be 12cm. s^2 = 12^2 = 12 x 12 = 144cm^2 42 - 2s = 42 - 2(12) = 42 - 24 = 18cm 8 x 18 = 144cm^2 Both areas are equal. To find out the where to make the cut: 4s = 4(12) = 48cm 2(8 + 18) = 2(26) = 52cm 100 = 48 + 52 Therefore, the cut should be made at 48cm so that the 48cm is for the square (second piece) while the 52cm is for the rectangle (first piece).
Since we need to make a square and a rectangle, we know that the perimeters must add up to 100. The square would have a perimeter of 4x cm if we use x as the length of one of the side lengths of the square. The area would then be x^2 cm.
And since we don’t know much of the specifics of the rectangle, we can form this equation: (100-4x) cm for the perimeter.
Since we already know that one of the sides of the rectangle is 8 cm. We can form this equation for the other side lengths. 100-4x-8-8 = (84-4x). Now we need to divide this by 2 to find one of the side lengths of the length (does that make sense?). (84-4x)/2 = (42-2x) cm. Now, we know that the length is (42-2x) and the width is 8. Since the two areas are equal…
(much wow. look its quadratics. now we have to find two numbers that multiply to -336 and add up (or subtract) to 16. would you look at that, -12 and 28 work)
This gives x the possibility of being 12 or -28 (reverse the negatives).
Since -28 cannot be the answer, we know that x= 12. The area of the square is 144^cm and the area of the rectangle is 8 x (42-2x) = 8 x (42=24) = 8 x 18 = 144.
The side length of the square is 4(12) = 48 leaving 52 cm for the rectangle.
The rope should be cut, leaving a 48cm long piece and a 52cm long piece.
POTW:
ReplyDelete3 possible side lengths include
1. Triangle: 40 cm, total 120 cm
1. Square: 20 cm, total 80 cm
2. Triangle: 50 cm, total 150 cm
2. Square: 12.5 cm, total 50 cm
3. Triangle: 20 cm, total 60 cm
3. Square: 35 cm, total 140 cm
Grade 7 POTW
ReplyDeleteThere are many possibilities for the side lengths of the triangles and squares.
1. Triangle side length=20 cm Square side length=5 cm. In order to check that I am correct, I will have to review the question in reverse. One triangle side length=20 cm and there are 3 equilateral triangles. 3x3x20=180. One square side length = 5cm. 5x4=20. 180+20=200 cm of rope.
2. The triangle side length can be 8 cm and the square side length can be 32 cm. 8x3x3= 72. 32x4=128. 128+72=200 cm of rope.
3. The triangle side length can be 12 cm and the square side length can be 23 cm. 12x3x3=108. 23x4=92. 108+92=200 cm of rope.
4. The triangle side length can be 4 cm and the square side length can be 41 cm. 4x3x3=36. 41x4=164. 164+26=200 cm of rope.
5. The triangle side length can be 16 cm and the square side length can be 14 cm. 16x3x3= 144. 14x4=56. 144+56=200 cm of rope.
There are 5 possibilities for the side lengths of the squares and triangles. Starting at 200 cm, there was a pattern. Going down every 36 cm, there was a possible perimeter for a square and the difference of that perimeter and the total amount of rope was a correct amount for the perimeter of 3 equilateral triangles. I started by going down by 9 cm because there are 3 triangles with 3 sides each meaning that the total perimeter of the three triangles had to be a multiple of 9 cm. I then realized that only when 4 9's passed, would the square and triangle's perimeter's would be 200 cm. Starting from 0 and going up by 36, there are 5 multiples before the 6th multiple goes over 200 cm of rope.
Grade 8 POTW
ReplyDeleteOkay, so we have 100cm of rope and we get to make one cut so that: there are 2 pieces, one piece forms a rectangle with a side length of 8cm, the second piece forms a square, and the area of both quadrilaterals are equal. The perimeters of both shapes should add up to 100cm. We know that the square would have all 4 sides equal. In other words square perimeter = 4s. The area would be s x s = s^2. The rectangle’s perimeter would be the remainder of that 100cm or 100 - 4s. We already know 2 sides (8cm), so the length of the other 2 sides would be 100 - 4s - 8 - 8 = 84 - 4s. The length of one side would be (84 - 4s)/2 = 42 - 2s. The area of the rectangle would then be l x w or 8 x (42 - 2s) = 336 - 16s. Since the areas of both the rectangle and square are equal, we can find the value of s by using quadratics:
s^2 = 336 - 16s
s^2 - 336 + 16s = 336 - 16s - 336 + 16s
s^2 - 336 + 16s = 0
s^2 + 16s - 336 = 0
s^2 - 12s + 28s - 336 = 0 (you could go straight to the answer from here, but I’ll show steps)
s(s - 12) + 28(s - 12) = 0
(s - 12)(s + 28) = 0
If s = 12, the equation would be true OR if s = -28, the equation would be true.
Therefore, s can be equal to 12 or -28. However, s is a side length, so it can’t be negative meaning it must be 12cm.
s^2 = 12^2 = 12 x 12 = 144cm^2
42 - 2s = 42 - 2(12) = 42 - 24 = 18cm
8 x 18 = 144cm^2
Both areas are equal. To find out the where to make the cut:
4s = 4(12) = 48cm
2(8 + 18) = 2(26) = 52cm
100 = 48 + 52
Therefore, the cut should be made at 48cm so that the 48cm is for the square (second piece) while the 52cm is for the rectangle (first piece).
Grade 7 POTW
ReplyDeleteWe have 200cm cut into four pieces. If 3 pieces are made to form identical equilateral triangles with integer side lengths, it means that 3 pieces are the same length. The fourth piece is used to form a square with integer side lengths. In integer is a whole number (not decimal). To determine all the side length values, we just need to start from the bottom. Let t = triangle side length and s = square side length.
3t x 3 = 9t
4s
9t + 4s = 200cm
To save time, instead of trying every possibility, we can just skip it, as t must be divisible by 4 to allow s to be an integer. This is because if t isn’t for example, t = 3, then 200 - 9(3) = 173 which isn’t divisible by 4. This is how it works because if you subtract 200 (divisible by 4) by a number not divisible by 4, the result won’t be divisible by 4.
Now, we can just list each number multiplied by 4 for the t value until right before we reach 200:
200 - 9(4) = 200 - 36 = 164 = 4s
200 - 9(8) = 200 - 72 = 128 = 4s
200 - 9(12) = 200 - 108 = 92 = 4s
200 - 9(16) = 200 - 144 = 56 = 4s
200 - 9(20) = 200 - 180 = 20 = 4s
200 - 9(24) = 200 - 216 (too high)
164/4 = 41 = s
128/4 = 32 = s
92/4 = 23 = s
56/4 = 14 = s
20/4 = 5 = s
Therefore, the 5 possible side lengths for each triangle and square is:
t = 4cm, s = 41cm
t = 8cm, s = 32cm
t = 12cm, s = 23cm
t = 16cm, s = 14cm
t = 20cm, s = 5cm
Grade 8 POTW
ReplyDeleteOkay, so we have 100cm of rope and we get to make one cut so that: there are 2 pieces, one piece forms a rectangle with a side length of 8cm, the second piece forms a square, and the area of both quadrilaterals are equal. The perimeters of both shapes should add up to 100cm. We know that the square would have all 4 sides equal. In other words square perimeter = 4s. The area would be s x s = s^2. The rectangle’s perimeter would be the remainder of that 100cm or 100 - 4s. We already know 2 sides (8cm), so the length of the other 2 sides would be 100 - 4s - 8 - 8 = 84 - 4s. The length of one side would be (84 - 4s)/2 = 42 - 2s. The area of the rectangle would then be l x w or 8 x (42 - 2s) = 336 - 16s. Since the areas of both the rectangle and square are equal, we can find the value of s by using quadratics:
s^2 = 336 - 16s
s^2 - 336 + 16s = 336 - 16s - 336 + 16s
s^2 - 336 + 16s = 0
s^2 + 16s - 336 = 0
s^2 - 12s + 28s - 336 = 0 (you could go straight to the answer from here, but I’ll show steps)
s(s - 12) + 28(s - 12) = 0
(s - 12)(s + 28) = 0
If s = 12, the equation would be true OR if s = -28, the equation would be true.
Therefore, s can be equal to 12 or -28. However, s is a side length, so it can’t be negative meaning it must be 12cm.
s^2 = 12^2 = 12 x 12 = 144cm^2
42 - 2s = 42 - 2(12) = 42 - 24 = 18cm
8 x 18 = 144cm^2
Both areas are equal. To find out the where to make the cut:
4s = 4(12) = 48cm
2(8 + 18) = 2(26) = 52cm
100 = 48 + 52
Therefore, the cut should be made at 48cm so that the 48cm is for the square (second piece) while the 52cm is for the rectangle (first piece).
Grade 8 POTW:
ReplyDeleteSince we need to make a square and a rectangle, we know that the perimeters must add up to 100. The square would have a perimeter of 4x cm if we use x as the length of one of the side lengths of the square. The area would then be x^2 cm.
And since we don’t know much of the specifics of the rectangle, we can form this equation: (100-4x) cm for the perimeter.
Since we already know that one of the sides of the rectangle is 8 cm. We can form this equation for the other side lengths. 100-4x-8-8 = (84-4x). Now we need to divide this by 2 to find one of the side lengths of the length (does that make sense?).
(84-4x)/2 = (42-2x) cm. Now, we know that the length is (42-2x) and the width is 8. Since the two areas are equal…
(much wow. look its quadratics. now we have to find two numbers that multiply to -336 and add up (or subtract) to 16. would you look at that, -12 and 28 work)
x^2= (336-16x)
x^2 + 16x - 336 = 0
(x-12)(x+28) = 0
This gives x the possibility of being 12 or -28 (reverse the negatives).
Since -28 cannot be the answer, we know that x= 12. The area of the square is 144^cm and the area of the rectangle is 8 x (42-2x) = 8 x (42=24) = 8 x 18 = 144.
The side length of the square is 4(12) = 48 leaving 52 cm for the rectangle.
The rope should be cut, leaving a 48cm long piece and a 52cm long piece.