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Info: - (I'm going to use "top right", "top left" and "middle" instead because the names are a little confusing) - Top right= 36 cm^2. - Top right-middle= middle-top left Find= - Area of middle and top left
To answer this, I would guess that we need to use algebra in some way. So, I'll give each rectangle a variable. Middle= x Top left= y 36-x=x-y (I'm just going to play around with this until I get something close to an answer) 36-x=x-y 36=2x-y 2x=36+y 36+y=2x (this tells us that y must be an even number and a multiple of 6 because it shares a side with the top right rectangle). Since the area isn't bringing us very far, we can use the side lengths now. Variables: Middle side length EF and DC= a Middle side length ED and FC= b 2(a x b)= 36+y 2a x 2b= 36 +y a x b=18+y/2
Ah this is all I can get for now. I'll get back to this when I find more information on the relationships between the different rectangles :)
Wait. J'ai une reponse. And it still uses algebra. We can use the side length of one of the sides of ABCD as x (this time, ignore what was originally said). Then using some quick math we know that if the top right rectangle is 36, that means that one side lenght is 6m meaning that side length AH (AHHHHHHHHHHHHHH) is equal to x-6 and so is side length ED (hi my name is ed) is also x-6. Then we can make an equation and solve l'equation. top right-middle=middle-top left 36- x(x-6)= x(x-6) - 6(x-6) 36 -x^2+6x=x^2 -6x -6x +36 36-x^2+18x=x^2+36 (add 12 to both sides) -x^2+18x=x^2 (subtract 36) 18x=2(x^2) (add x^2 This can only be 9 because x cannot be 0. The 2 would help the 9 get to 18 and from there on it is only x=x. So, 9 is the side legnth Middle= x(x-6) Middle= 9(9-6) Middle= 27cm^2 Top Left=6(x-6) Top Left=6(9-6) Top Left=18cm^2 Yay.
Maruti So first, you find the area of the square which is 36cm squared and label all the side lengths. Then, I cut the small square in triangles which fit into the rectangle on the left. The triangle side length is half the square's side length, so the bottom of the left triangle is 3cm and the left rectangle bottom side is 3cm, so the area is 18cm squared (6x3). Then you have to find the bottom rectangle area. They said it was a square so the side lengths of the rectangle are 3cm and 9cm and the area is 27cm squared. The question also tells us that the square minus the bottom rectangle area is equal to the bottom rectangle minus the top rectangle area. 36-27=27-18 is true so it satisfies the rule. The area of CDEF is 27cm squared and the area of AHGE is 18cm squared.
POTW BFGH - CDEF = CDEF - AHGE BFGH = 36 cm squared which means every side is 6cm as 6 x 6=36 From here on I used guess and check to solve this problem. I guessed and checked the side length of the large square to match the rectangles and squares in the equation. In the end, I got to a conclusion of the side length of the large square as 9cm. 9cm side length- 6cm BFGH side length= 3 cm width of AHGE= 3x6=18cm squared for AHGE 9 cm length x 3 cm width of CDEF=27cm squared. 36-27=9cm squared 27-18=9cm squared
This was how I got to the conclusion of AHGE being 18cm squared and CDEF being 27cm squared.
POTW Since the area of square BFGH = 36 cm squared, the sidelength would be 6 cm. Now, we have a measurement for two sides of rectangle AHGE. But what about the other two? Well, what I did was use guess and check a number of times, until I got the answer. Now, if the sidelength of square ABCD = 9, then everything comes together. 9 - 6 (the sidelength of the larger square minus the sidelength of the smaller square) is 3, which is how long the other two sides of rectangle AHGE are. Now, we have an area: 6 x 3 for rectangle AHGE, which is 18 cm squared. Also, now that we know that the sidelength of square ABCD is 9, then rectangle CDEF's area can be found as well. One side is 9 cm; the other is 9 - 6, which is 3. So, the area is 9 x 3, which is 27 cm squared. Now to check. The area of square BFGH - the area of rectangle CDEF (36 - 27) = 9. The area of rectangle CDEF - the area of rectangle AHGE (27 - 18) = 9. They are the same.
So, the area of rectangle AHGE was 18 cm squared, and the area of rectangle CDEF was 27 cm squared.
Grade 7/8 POTW Okay, so BFGH is a square equal to 36 cm. A square's area is l^2 meaning the square root of 36 is the side length of the square or 6 cm. Using this, you can translate the equation BFGH - CDEF = CDEF - AHGE into an algebraic equation using the value 6 already known. It becomes: 6*6 - x(x - 6) = x(x - 6) - 6(x - 6) Now to simplify this until we find the value of x. (*Note: x is equal to the side length of the largest square) 6*6 - x(x - 6) = x(x - 6) - 6(x - 6) First you solve all the numerical values you can (6 x 6) and use distributive property to remove the brackets. 36 - x^2 + 6x = x^2 - 6x - 6x + 36 You can then simplify - 6x - 6x to - 12x. 36 - x^2 + 6x = x^2 - 12x + 36 You add 12x to both sides. 36 - x^2 + 18x = x^2 + 36 You subtract 36 from both sides. -x^2 + 18x = x^2 You add x^2 (x squared) to both sides. 18x = 2x^2 You divide both sides by 2. 9x = x^2 You divide both sides by x. 9 = x Okay, using this and the known value of x (9), we can easily solve the areas of the rectangles. CDEF = x(x - 6) = 9(9 - 6) = 9(3) = 27 AHGE = 6(x - 6) = 6(9 - 6) = 6(3) = 18 Therefore, the area of rectangle CDEF is 27cm^2 or squared and the area of rectangle AHGE is 18cm^2 or squared. (We can also find the area of ABCD now too: ABCD = x * x = x^2 = 9 * 9 = 9^2 = 81 cm^2 or squared).
First I will find the dimensions of BFGH. 6x6=36cm2. This means that AHGE's length is 6cm. If you divide 6cm by 2 you get 3cm. Now we know the full dimensions of AHGE which is 18cm2. If we add 6cm and 3cm we get 9cm which is the length of CDEF and the overall square. 9cm x 9cm = 81cm2. Now that we know the full area we can subtract the areas of BHGF and AHGE. 81cm2-36cm2-18cm2= 27cm2(this is the area of CDEF).
In conclusion, the area of the overall square is 81cm2, the area of BGHE is 36cm2, the area of AHGE is 18cm2 and the area of CDEF is 27cm2.
From the statement "The area of BFGH is 36cm^2, we immediately know that each side is 6 centimetres long. Also, because the whole thing is a square, AH=CF=EG=ED. Therefore, we know that 36cm^2-CDEF is going to be equivalent to CDEF-AHGE.
So since BFGH is a square it’s side lengths must be 6. You can tell that one of ahge sides are 6. 36 - C = C - 6*x and using the fact that cdef is 3/4 of B so cdef = 27 you can find Ahge = 18
First, let’s say BFGH is now x, CDEF is now y, and AHGE is now z. We know that: x - y = y - z We can change this equation to y - z + y = 36 or 2y = 36 + z By this last equation, we know that the area of (x + z) is twice the area of y. Now, since they both have the same width, it means the height of y is half of (x + z). The height of (x + z) is 6 cm because the area of x is 36. Therefore, the height of y is 3 cm. Since ABCD is a square, the base is equal to the height, which is 6 + 3 = 9. Therefore, CDEF, or y is 9 * 3 = 27 cm squared, and AHGE, or z is equal to 3 * 6 = 18 cm squared.
To find the length and width of BFGH I found the square root of 36cm sq. The answer was 6cm. Let us assume AH is y The area of AHGE is 6*y=6y The area of CDEF is y*(6+y)= 6y+ysq So BFGH-CDEF=CDEF-AHGE becomes 36-6y+y sq.= 6y+y sq.- 6y After simplifying the equation this is what I got 36=2y sq.+6y 18=y sq.+3y 18=yx(y+3) Then I did guess and check and figured out y=3 AHGE=3x6=18cm sq. CDEF=3x(6+3)=3x9=27cm sq.
We know that BFGH has an area of 36 cm squared, making each side of the square 6 cm. We can see that AHGE has a side length of 6 cm squared. It's width is half of it's length; ( 6 / 2 = 3 cm). That makes the area of AHGE 6 x 3 = 18 cm squared. Now, we understand that the width of CDEF is 9 cm, as 6 + 3 = 9 ( 6 from width of BFGH, and 3 from width of AHGE). The length is 3 cm. 9 x 3 = 27 cm squared.
Scince we know that squares haave equal sides 36/4 = 6. This means that each side of the square is 6cm.
Next CDEF. The two longer sides of CDEF are 6 because they are same length as the ones of the square. next using visuals i found that we take 36cm2 - 27cm2 - 9. SCINCE THE AREA OF CDEF IS 27 WE CAN now say that because CDEF - AHGE is the same as ABCD - CDEF, 27 - X = 9 .... 27 - 18 = 9. (I used guess and check for this)
Therefor the area of CDEF is 27cm2 and the are AHGE is 18cm2.
Info:
ReplyDelete- (I'm going to use "top right", "top left" and "middle" instead because the names are a little confusing)
- Top right= 36 cm^2.
- Top right-middle= middle-top left
Find=
- Area of middle and top left
To answer this, I would guess that we need to use algebra in some way. So, I'll give each rectangle a variable.
Middle= x
Top left= y
36-x=x-y
(I'm just going to play around with this until I get something close to an answer)
36-x=x-y
36=2x-y
2x=36+y
36+y=2x (this tells us that y must be an even number and a multiple of 6 because it shares a side with the top right rectangle).
Since the area isn't bringing us very far, we can use the side lengths now.
Variables:
Middle side length EF and DC= a
Middle side length ED and FC= b
2(a x b)= 36+y
2a x 2b= 36 +y
a x b=18+y/2
Ah this is all I can get for now. I'll get back to this when I find more information on the relationships between the different rectangles :)
Wait. J'ai une reponse. And it still uses algebra.
DeleteWe can use the side length of one of the sides of ABCD as x (this time, ignore what was originally said). Then using some quick math we know that if the top right rectangle is 36, that means that one side lenght is 6m meaning that side length AH (AHHHHHHHHHHHHHH) is equal to x-6 and so is side length ED (hi my name is ed) is also x-6. Then we can make an equation and solve l'equation.
top right-middle=middle-top left
36- x(x-6)= x(x-6) - 6(x-6)
36 -x^2+6x=x^2 -6x -6x +36
36-x^2+18x=x^2+36 (add 12 to both sides)
-x^2+18x=x^2 (subtract 36)
18x=2(x^2) (add x^2
This can only be 9 because x cannot be 0. The 2 would help the 9 get to 18 and from there on it is only x=x. So, 9 is the side legnth
Middle= x(x-6)
Middle= 9(9-6)
Middle= 27cm^2
Top Left=6(x-6)
Top Left=6(9-6)
Top Left=18cm^2
Yay.
Maruti
ReplyDeleteSo first, you find the area of the square which is 36cm squared and label all the side lengths. Then, I cut the small square in triangles which fit into the rectangle on the left. The triangle side length is half the square's side length, so the bottom of the left triangle is 3cm and the left rectangle bottom side is 3cm, so the area is 18cm squared (6x3). Then you have to find the bottom rectangle area. They said it was a square so the side lengths of the rectangle are 3cm and 9cm and the area is 27cm squared. The question also tells us that the square minus the bottom rectangle area is equal to the bottom rectangle minus the top rectangle area. 36-27=27-18 is true so it satisfies the rule.
The area of CDEF is 27cm squared and the area of AHGE is 18cm squared.
CDEF is 27cm squared and AHGE is 18cm squared. I did my work in my head. Then wrote only the answers. And forgot how I did it. Heh.
ReplyDeleteNice
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DeletePOTW
ReplyDeleteBFGH - CDEF = CDEF - AHGE
BFGH = 36 cm squared which means every side is 6cm as 6 x 6=36
From here on I used guess and check to solve this problem. I guessed and checked the side length of the large square to match the rectangles and squares in the equation. In the end, I got to a conclusion of the side length of the large square as 9cm.
9cm side length- 6cm BFGH side length= 3 cm width of AHGE= 3x6=18cm squared for AHGE
9 cm length x 3 cm width of CDEF=27cm squared.
36-27=9cm squared
27-18=9cm squared
This was how I got to the conclusion of AHGE being 18cm squared and CDEF being 27cm squared.
This is mine. I am Edward. Idk why it calls me unknown
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DeletePOTW
ReplyDeleteSince the area of square BFGH = 36 cm squared, the sidelength would be 6 cm. Now, we have a measurement for two sides of rectangle AHGE. But what about the other two? Well, what I did was use guess and check a number of times, until I got the answer. Now, if the sidelength of square ABCD = 9, then everything comes together.
9 - 6 (the sidelength of the larger square minus the sidelength of the smaller square) is 3, which is how long the other two sides of rectangle AHGE are.
Now, we have an area: 6 x 3 for rectangle AHGE, which is 18 cm squared. Also, now that we know that the sidelength of square ABCD is 9, then rectangle CDEF's area can be found as well.
One side is 9 cm; the other is 9 - 6, which is 3. So, the area is 9 x 3, which is 27 cm squared.
Now to check. The area of square BFGH - the area of rectangle CDEF (36 - 27) = 9. The area of rectangle CDEF - the area of rectangle AHGE (27 - 18) = 9. They are the same.
So, the area of rectangle AHGE was 18 cm squared, and the area of rectangle CDEF was 27 cm squared.
Done on paper
ReplyDeleteCDEF= 27cm^2
AHGE= 18cm^2
Grade 7/8 POTW
ReplyDeleteOkay, so BFGH is a square equal to 36 cm. A square's area is l^2 meaning the square root of 36 is the side length of the square or 6 cm. Using this, you can translate the equation BFGH - CDEF = CDEF - AHGE into an algebraic equation using the value 6 already known. It becomes:
6*6 - x(x - 6) = x(x - 6) - 6(x - 6)
Now to simplify this until we find the value of x. (*Note: x is equal to the side length of the largest square)
6*6 - x(x - 6) = x(x - 6) - 6(x - 6)
First you solve all the numerical values you can (6 x 6) and use distributive property to remove the brackets.
36 - x^2 + 6x = x^2 - 6x - 6x + 36
You can then simplify - 6x - 6x to - 12x.
36 - x^2 + 6x = x^2 - 12x + 36
You add 12x to both sides.
36 - x^2 + 18x = x^2 + 36
You subtract 36 from both sides.
-x^2 + 18x = x^2
You add x^2 (x squared) to both sides.
18x = 2x^2
You divide both sides by 2.
9x = x^2
You divide both sides by x.
9 = x
Okay, using this and the known value of x (9), we can easily solve the areas of the rectangles.
CDEF = x(x - 6) = 9(9 - 6) = 9(3) = 27
AHGE = 6(x - 6) = 6(9 - 6) = 6(3) = 18
Therefore, the area of rectangle CDEF is 27cm^2 or squared and the area of rectangle AHGE is 18cm^2 or squared.
(We can also find the area of ABCD now too: ABCD = x * x = x^2 = 9 * 9 = 9^2 = 81 cm^2 or squared).
Aliyah,
ReplyDeleteFirst I will find the dimensions of BFGH. 6x6=36cm2. This means that AHGE's length is 6cm. If you divide 6cm by 2 you get 3cm. Now we know the full dimensions of AHGE which is 18cm2. If we add 6cm and 3cm we get 9cm which is the length of CDEF and the overall square. 9cm x 9cm = 81cm2. Now that we know the full area we can subtract the areas of BHGF and AHGE. 81cm2-36cm2-18cm2= 27cm2(this is the area of CDEF).
In conclusion, the area of the overall square is 81cm2, the area of BGHE is 36cm2, the area of AHGE is 18cm2 and the area of CDEF is 27cm2.
From the statement "The area of BFGH is 36cm^2, we immediately know that each side is 6 centimetres long. Also, because the whole thing is a square, AH=CF=EG=ED. Therefore, we know that 36cm^2-CDEF is going to be equivalent to CDEF-AHGE.
ReplyDeleteFor the sake of simplicity, x=AH/CF...
6*6-(6+x*x)=(6+x*x)-(6*x)
6*6-6x+x^2=12x+x^2
36-6x=6x
36-12x=0
x=3
Therefore, (6+x*x)=(9*3)=27cm^2
CDEF=27cm^2
AHGE=18cm^2
1 side of BFGH = 9 (36/4)
ReplyDeleteCDEF = ¾ of BFGH
AHGE = ½ pf BFGH
CDEF = (36/4)*3 = 27 cm2
AHGF = (36/2) = 18cm2
So since BFGH is a square it’s side lengths must be 6. You can tell that one of ahge sides are 6.
ReplyDelete36 - C = C - 6*x and using the fact that cdef is 3/4 of B so cdef = 27 you can find Ahge = 18
First, let’s say BFGH is now x, CDEF is now y, and AHGE is now z.
ReplyDeleteWe know that:
x - y = y - z
We can change this equation to
y - z + y = 36
or
2y = 36 + z
By this last equation, we know that the area of (x + z) is twice the area of y.
Now, since they both have the same width, it means the height of y is half of (x + z).
The height of (x + z) is 6 cm because the area of x is 36.
Therefore, the height of y is 3 cm.
Since ABCD is a square, the base is equal to the height, which is 6 + 3 = 9.
Therefore, CDEF, or y is 9 * 3 = 27 cm squared, and AHGE, or z is equal to 3 * 6 = 18 cm squared.
To find the length and width of BFGH I found the square root of 36cm sq. The answer was 6cm.
ReplyDeleteLet us assume AH is y
The area of AHGE is 6*y=6y
The area of CDEF is y*(6+y)= 6y+ysq
So BFGH-CDEF=CDEF-AHGE becomes
36-6y+y sq.= 6y+y sq.- 6y
After simplifying the equation this is what I got
36=2y sq.+6y
18=y sq.+3y
18=yx(y+3)
Then I did guess and check and figured out y=3
AHGE=3x6=18cm sq.
CDEF=3x(6+3)=3x9=27cm sq.
POTW:
ReplyDeleteWe know that BFGH has an area of 36 cm squared, making each side of the square 6 cm. We can see that AHGE has a side length of 6 cm squared. It's width is half of it's length; ( 6 / 2 = 3 cm). That makes the area of AHGE 6 x 3 = 18 cm squared.
Now, we understand that the width of CDEF is 9 cm, as 6 + 3 = 9 ( 6 from width of BFGH, and 3 from width of AHGE). The length is 3 cm. 9 x 3 = 27 cm squared.
AHGE: 18 cm squared
CDEF: 27 cm squared.
Scince we know that squares haave equal sides 36/4 = 6. This means that each side of the square is 6cm.
ReplyDeleteNext CDEF. The two longer sides of CDEF are 6 because they are same length as the ones of the square.
next using visuals i found that we take 36cm2 - 27cm2 - 9.
SCINCE THE AREA OF CDEF IS 27 WE CAN now say that because CDEF - AHGE is the same as ABCD - CDEF, 27 - X = 9 .... 27 - 18 = 9. (I used guess and check for this)
Therefor the area of CDEF is 27cm2 and the are AHGE is 18cm2.