I think you can write this in many different ways. I think the most interesting point is to eliminate one of the variables using rta + rtc =rtb Using this I come to Side length of red = 1/2(rta-rtc), so Area is the square of that. Much easier to work this in side lengths. I used the Greek letter equivalents.
Yes this is the only useful answer, as it uses the fact that red is a square to eliminate a variable. Same as what I got. Once you have assigned 2 variables the size of red box is fixed.
We first complete the rectangle Consider the two rectangles with length and width √b & √a -√b and √a -√b & √c Now area of the red square = above mentioned first rectangle - area of above mentioned second rectangle = (√a -√b) √b -(√a -√b) √c =(√a-√b) (√b -√c) [ from this form we may write √(ab ) - √(ac) -b +√(bc ) =√b(√a +√c) -b -√(ac) =√b*2√b - b -√(ac) =b -√(ac) This is also derived by two persons ]
It is also possible to write immediately that x = (sqrt(a) - sqrt(b))^2 = a + b -2.sqrt(a.b) or that x = (sqrt(b) - sqrt(c))^2 = b + c -2.sqrt(b.c). It is also possible to write that x = b - sqrt(a.c) (I let the proof to you). Naturally your formula is the best as it contains no radical.
Without radical same derivation in another way also done by me. May see. And with radical also a formula has been derived that is easy to memorise (√a -√b) (√b -√c)
@@PreMath The problem statement should have required that the solution not contain any radicals. Otherwise, I don't see why x = (√a - √b)² (from equation 1) and x = (√b - √c)² (from equation 2) are not valid solutions.
May see this solution side of Red square (x) =√a -√b = √b -√c Hence 2√b =√a +√c -- (1) Area of red square (x ^2) =(√a-√b) ^2=a+b - 2√(ab) --(2) Area of red square (x ^2) =(√b -√c) ^2 =b +c - 2√(bc ) -- (3) Adding 2& 3 2(Area of red square ) =a +2b +c -2√b(√a+√c) = a+ 2b +c -2√b*2√b (form 1) =a-2b +c Area of red square =(a - 2b +c) /2
Another sol We may complete the rectangle Then Area of red square =area of big rectangle -a-b-c - area of small rectangle =√a(√a+√b) -a-b-c -√c(√a-√b -√c) =√(ab) - b -√(ca) +√(bc) =(√a-√b) (√b -√c)
Let's find the area: . .. ... .... ..... Let A (B,C,X) be the side length of the green (blue,yellow,red) square. From the diagram we can conclude: A = √a B = √b C = √c B = C + X ∧ A = B + X ⇒ A = (C + X) + X = C + 2X ⇒ X = (A − C)/2 Therefore the area of the red square turns out to be: A(yellow) = X² = [(A − C)/2]² = (A² − 2AC + C²)/4 = (a − 2√(ac) + c)/4 This expression can be further simplified: AC = (B + X)(B − X) = B² − X² ⇒ X² = (A² − 2AC + C²)/4 = [A² − 2(B² − X²) + C²]/4 = A²/4 − B²/2 + X²/2 + C²/4 ⇒ X²/2 = A²/4 − B²/2 + C²/4 = a/4 − b/2 + c/4 ⇒ X² = a/2 − b + c/2 = (a − 2b + c)/2 Therefore the area of the yellow square turns out to be: A(yellow) = X² = (a − 2√(ac) + c)/4 = (a − 2b + c)/2 Best regards from Germany
Solution: d = Red Square Area √a = √b + √d ... ¹ √d = √b - √c √c = √b - √d ... ² Equation 1 whole square (√a)² = (√b + √d)² a = b + 2√bd + d ... ³ Equation 2 whole square (√c)² = (√b - √d)² c = b - 2√bd + d ... ⁴ Adding 3 and 4 a = b + 2√bd + d c = b - 2√bd + d ---------------------------- a + c = 2b + 2d ... ⁵ Solving for "d" in equation ⁵ a + c = 2b + 2d 2d = a + c - 2b Therefore: d = (a - 2b + c)/2 d = Red Square Area = RSA RSA = (a - 2b + c)/2 ✅
Bom dia Mestre
Obrigado pela aula
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I think you can write this in many different ways. I think the most interesting point is to eliminate one of the variables using rta + rtc =rtb
Using this I come to
Side length of red = 1/2(rta-rtc), so Area is the square of that.
Much easier to work this in side lengths. I used the Greek letter equivalents.
This is the way I did it as well. It is a nice easy solution. [Side length of red = (1/2)(rta-rtc)] Thanks for posting!
(√a - √b)^2 = a + b - 2√(ab)
or
(√b - √c)^2 = b + c - 2√(bc)
Thank you!
Using a, b, c and x as the sides of A, B, C and X, it is easy to see that X=x^2=(a-c)^2/4.
It was said to derive the area of red square in terms of a, b, c
But u did not do it
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Yes this is the only useful answer, as it uses the fact that red is a square to eliminate a variable. Same as what I got. Once you have assigned 2 variables the size of red box is fixed.
Except using your notation x = X^2, etc
Ans b-sqrt(ac)
We first complete the rectangle
Consider the two rectangles with length and width √b & √a -√b and √a -√b & √c
Now area of the red square
= above mentioned first rectangle - area of above mentioned second rectangle
= (√a -√b) √b -(√a -√b) √c
=(√a-√b) (√b -√c)
[ from this form we may write
√(ab ) - √(ac) -b +√(bc )
=√b(√a +√c) -b -√(ac)
=√b*2√b - b -√(ac)
=b -√(ac)
This is also derived by two persons ]
It is also possible to write immediately that x = (sqrt(a) - sqrt(b))^2 = a + b -2.sqrt(a.b) or that x = (sqrt(b) - sqrt(c))^2 = b + c -2.sqrt(b.c).
It is also possible to write that x = b - sqrt(a.c) (I let the proof to you). Naturally your formula is the best as it contains no radical.
I got the same answer (simple). If they need answer x in a,b,c form. Yes, the answer from admin would be great.
Without radical same derivation in another way also done by me.
May see.
And with radical also a formula has been derived that is easy to memorise
(√a -√b) (√b -√c)
@@PrithwirajSen-nj6qq There a several possible formulas, and this one is fine.
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@@PreMath The problem statement should have required that the solution not contain any radicals. Otherwise, I don't see why x = (√a - √b)² (from equation 1) and x = (√b - √c)² (from equation 2) are not valid solutions.
May see this solution
side of Red square (x)
=√a -√b
= √b -√c
Hence 2√b =√a +√c -- (1)
Area of red square (x ^2)
=(√a-√b) ^2=a+b - 2√(ab) --(2)
Area of red square (x ^2)
=(√b -√c) ^2
=b +c - 2√(bc ) -- (3)
Adding 2& 3
2(Area of red square )
=a +2b +c -2√b(√a+√c)
= a+ 2b +c -2√b*2√b (form 1)
=a-2b +c
Area of red square
=(a - 2b +c) /2
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b=c+d, a=b+d......troublesome 😮All I can do 😢. b-c=a-b, 2b=a+c......
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Thanks Sir
Good work
That’s very nice
Good luck with respects
❤❤❤❤
Another sol
We may complete the rectangle
Then Area of red square
=area of big rectangle -a-b-c - area of small rectangle
=√a(√a+√b) -a-b-c -√c(√a-√b -√c)
=√(ab) - b -√(ca) +√(bc)
=(√a-√b) (√b -√c)
Thanks for sharing ❤️
Let's find the area:
.
..
...
....
.....
Let A (B,C,X) be the side length of the green (blue,yellow,red) square. From the diagram we can conclude:
A = √a
B = √b
C = √c
B = C + X ∧ A = B + X ⇒ A = (C + X) + X = C + 2X ⇒ X = (A − C)/2
Therefore the area of the red square turns out to be:
A(yellow) = X² = [(A − C)/2]² = (A² − 2AC + C²)/4 = (a − 2√(ac) + c)/4
This expression can be further simplified:
AC = (B + X)(B − X) = B² − X²
⇒ X² = (A² − 2AC + C²)/4 = [A² − 2(B² − X²) + C²]/4 = A²/4 − B²/2 + X²/2 + C²/4
⇒ X²/2 = A²/4 − B²/2 + C²/4 = a/4 − b/2 + c/4
⇒ X² = a/2 − b + c/2 = (a − 2b + c)/2
Therefore the area of the yellow square turns out to be:
A(yellow) = X² = (a − 2√(ac) + c)/4 = (a − 2b + c)/2
Best regards from Germany
Fine. I like it. Thanks.
Excellent!
Thanks for sharing ❤️
Solution:
d = Red Square Area
√a = √b + √d ... ¹
√d = √b - √c
√c = √b - √d ... ²
Equation 1 whole square
(√a)² = (√b + √d)²
a = b + 2√bd + d ... ³
Equation 2 whole square
(√c)² = (√b - √d)²
c = b - 2√bd + d ... ⁴
Adding 3 and 4
a = b + 2√bd + d
c = b - 2√bd + d
----------------------------
a + c = 2b + 2d ... ⁵
Solving for "d" in equation ⁵
a + c = 2b + 2d
2d = a + c - 2b
Therefore:
d = (a - 2b + c)/2
d = Red Square Area = RSA
RSA = (a - 2b + c)/2 ✅
I love it. Thanks
Excellent!
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(√a - √b)^2 = area of 🟥.
What is the other question , I don't know? I didn't watch .
Veja:
√a ×√c = (√b + √x) × (√b - √x)
Use: (A+B) × (A-B)=A² - B²
√(ac) = b - x → *x = b - √(ac).*
This has been done by one. May see in comments. But he /she did not show the derivation.
Thanks for sharing ❤️
@@PrithwirajSen-nj6qq Tudo bem, isso já mostra que mais de uma pessoa viu essa parte da questão, pois nem sempre vejo os comentários.
I'm still trying to figure out what the square root of a tree is equal too.... 😊
😀
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My answer is b-(ac)^(0.5)
It is true. There are several answers.
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