I used the same method. The diagonal CE forms a right triangle with CE and ED. Just calculate CE using the Pythagorean Theorem, and that is the same as AB.
I think I have a simpler way to solve in only 2 steps. First, connect CE. The angle between the side of the square and the diagonal is always 45 degrees, so angles CEB and DEB are both 45 degrees. In triangle CED, angle CED is the CEB+DEB which is 45+45=90 degrees. So CED is a right-angled triangle. Next, we know 2 sides of triangle CED and we can calculate CE with the Pythagoras Theorem. CE²+(2√ 23)²=10² CE²+92=100 CE²=8 CE=2√ 2 We know CE and AB are the diagonals of square ACBE, so CE=AB=2√2, that's the answer.
Join CE to get 🔺 CDE Now we find angle CED is 90 degrees with Angie CED = 90 degrees [ arguments Yellow square diagonal divides the square into two isosceles rt 🔺 s Hence the other angles of the triangles is 45 degrees Likewise the diagonal of the green square divides the square into two isosceles rt 🔺 s. Hence the other angles of the triangles are 45 degrees each Angle CED =45+45=90 degrees ] Hence CE^2 =10^2 - (2√23)^2 = 100 - 4*23= 8 CE= √8=2√2 The length of AB = CE =2√2 units
The side length of the yellow square is (2.sqrt(23))/sqrt(2) = sqrt(46).We note c the side length of the green square. Now we use an orthonormal center A and first axis (AF). We have C(0; c), D(c + sqrt(46)); sqrt(46)), VectorCD(c + sqrt(46); sqrt(46) - c) So: CD^2 = c^2 + 2.c.sqrt(46) + 46 + c^2 -2.c.sqrt(46) + 46 = 2.(c^2) + 92. As CD^2 = 10^2 = 100, we have that 2.(c^2) + 92 = 100 That gives that c^2 = 4 and c= 2. Finally AB = c.sqrt(2) = 2.sqrt(2). (Was easy)
side of green square : x side of yellow square : 2√23/√2=√46 (x+√46)²+(√46-x)²=10² x²+46+x²+46=100 2x²-8=0 x²-4=0 (x+2)(x-2)=0 x>0 , x=2 AB²=2²+2²=8 AB=2√2
I saw the construction of adding CE ( = AB) and then calculating CE directly using right △CED. I was amazed to spot the simple solution since I'm usually so slow and feeble-minded with math. As for Mr. Premath's solution, I would not have been smart enough to extend CB like that.
Сторона жёлтого квадрата √(4*23/2)=√2*23=√46. Верхний отрезок даёт прямоугольный треугольник, основанием длиннее на сторону зелёного квадрата, а высотой на неё же короче. Обозначим за х, тогда (√46+х)²+(√46-х)²=10²=100. Двухчлен сокращается остаётся только два раза по квадрату числа и переменной, т. е. 2(46+х²)=100⇔46+х²=50⇔х²=4. Отсюда сторона квадрата 2, а, следовательно, диагональ 2√2.
Let the side length of green square ACBE be x. As AB is a diagonal of ACBE, then AB = √2x. As the diagonal of the yellow square is 2√23, then the side length of the yellow square is 2√23/√2 = √46. Extend CB to T on FD. As CT is parallel to AF, then it is perpendicular to FD. Triangle ∆CTD: CT² + TD² = DC² (x+√46)² + (√46-x)² = 10² x² + 2√46x + 46 + 46 - 2√46x + x² = 100 2x² = 100 - 92 = 8 x² = 8/2 = 4 x = √4 = 2 AB = √2x [ AB = 2√2 ]
Sir, as green box and yellow box are square, angle CED will be 90 degree and by applying Pythagoras theorem on triangle ACD , we can calculate CE. CE and AB are equal
Label the side of the yellow and green square as A and a. Extend CB intersecting DF at point B’. We have CB’= A+a and DB’= A-a --> sq (A+a) +sq (A-a)=sq CD -> 2 (sqA+sqa) =100 --> sqA+sqa=50 Because A=sqrt46 -> sqa=50-46=4 -> a =2 -> AB= 2sqrt2😅😅😅
Let's find the length: . .. ... .... ..... Let y be the side length of the yellow square and let g be the side length of the green square. Now we add point G on DF such that CDG is a right triangle. Then we apply the Pythagorean theorem step by step to the right triangles DEF, CDG and ABE: DF² + EF² = DE² y² + y² = (2√23)² 2y² = 92 CG² + DG² = CD² (BC + BG)² + (DF − FG)² = CD² (g + y)² + (y − g)² = CD² g² + 2gy + y² + y² − 2yg + g² = CD² 2g² + 2y² = CD² ⇒ 2g² = CD² − 2y² = 10² − 92 = 100 − 92 = 8 AB² = AE² + BE² = g² + g² = 2g² = 8 ⇒ AB = √8 = 2√2 Best regards from Germany
Solution: The two boxes are square, so we'll calculate the yellow square side EF = 2√23/√2 EF = 2√46/2 EF = √46 Let's label the green square side as "a" and extending a line from "C" to "P", such that "P" is on the line "DF", we have: CP = a + √46 DP = √46 - a CD = 10 Applying the Pythagorean Theorem (a + √46)² + (√46 - a) = 10² a² + 2√46a + 46 + 46 - 2√46a + a² = 100 a² + 46 + 46 + a² = 100 2a² + 92 = 100 2a² = 8 a² = 4 a = 2 AB = a√2 AB = 2√2 Square Units ✅ AB ≈ 2,8284 Square Units ✅
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
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Right triangle CDE. Pytagorean theorem :
d² = 10² - (2√23)²
d = 2√2 cm ( Solved √ )
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Вот этот прямой угол я как-то упустил из виду. %)
I used the same method. The diagonal CE forms a right triangle with CE and ED. Just calculate CE using the Pythagorean Theorem, and that is the same as AB.
I think I have a simpler way to solve in only 2 steps.
First, connect CE. The angle between the side of the square and the diagonal is always 45 degrees, so angles CEB and DEB are both 45 degrees. In triangle CED, angle CED is the CEB+DEB which is 45+45=90 degrees. So CED is a right-angled triangle.
Next, we know 2 sides of triangle CED and we can calculate CE with the Pythagoras Theorem. CE²+(2√ 23)²=10²
CE²+92=100
CE²=8
CE=2√ 2
We know CE and AB are the diagonals of square ACBE, so CE=AB=2√2, that's the answer.
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Join CE to get 🔺 CDE
Now we find angle CED is 90 degrees with Angie CED = 90 degrees
[ arguments
Yellow square diagonal divides the square into two isosceles rt 🔺 s
Hence the other angles of the triangles is 45 degrees
Likewise the diagonal of the green square divides the square into two isosceles rt 🔺 s.
Hence the other angles of the triangles are 45 degrees each
Angle CED =45+45=90 degrees ]
Hence
CE^2 =10^2 - (2√23)^2
= 100 - 4*23= 8
CE= √8=2√2
The length of AB = CE =2√2 units
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This is what I immediately did.....solved in 5 lines with detailed working.
@alkemis waiting for ur sol to the latest problem offered by Premath
AB = CE in case that's any help.
As they are both squares,
Good approach
@MrPaulc222:
Very clever and elegant solution❤
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🙏
At 6:30 that Perfect Square theorm of (an)^2=a^2-2ab+b can explain piece by piece how it fits bcuz it Don’t seem correct!
(a-b)^2 = (a-b)*(a-b) = a^2 - ab - ab + b^2 = a^2 - 2ab + b^2
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The side length of the yellow square is (2.sqrt(23))/sqrt(2) = sqrt(46).We note c the side length of the green square.
Now we use an orthonormal center A and first axis (AF). We have C(0; c), D(c + sqrt(46)); sqrt(46)), VectorCD(c + sqrt(46); sqrt(46) - c)
So: CD^2 = c^2 + 2.c.sqrt(46) + 46 + c^2 -2.c.sqrt(46) + 46 = 2.(c^2) + 92. As CD^2 = 10^2 = 100, we have that 2.(c^2) + 92 = 100
That gives that c^2 = 4 and c= 2. Finally AB = c.sqrt(2) = 2.sqrt(2). (Was easy)
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Bom dia Mestre
Acertei graças as vossas aulas
Obrigado Mestre
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You are very welcome!😀
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Triangle DCE is perpendicular. e²=d²+c² Diagonals of the square are equal
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side of green square : x
side of yellow square : 2√23/√2=√46
(x+√46)²+(√46-x)²=10² x²+46+x²+46=100
2x²-8=0 x²-4=0 (x+2)(x-2)=0 x>0 , x=2 AB²=2²+2²=8 AB=2√2
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AB=CE
Pitágoras a CDE
100=(2√(23))^2+(CE)^2
100=92+(CE)^2
100-92=(CE)^2
(CE)^2=8
CE=√8
CE=2√2
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L=√46..(L+l)^2+(L-l)^2=100..92+2l^2=100..l=2..AB=2√2
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AB=√[10²-(2√23)²] =2√2.
Gracias y saludos.
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I saw the construction of adding CE ( = AB) and then calculating CE directly using right △CED. I was amazed to spot the simple solution since I'm usually so slow and feeble-minded with math.
As for Mr. Premath's solution, I would not have been smart enough to extend CB like that.
Thanks ❤️🙏
Have a happy, safe, and prosperous New Year!
AB=2√2
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Chia sẻ hay rất mong cùng🤝🔔
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Essa foi fácil, professor !
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il valore della diagonale del quadrato verde è: √8 = 2√2
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Сторона жёлтого квадрата √(4*23/2)=√2*23=√46. Верхний отрезок даёт прямоугольный треугольник, основанием длиннее на сторону зелёного квадрата, а высотой на неё же короче. Обозначим за х, тогда (√46+х)²+(√46-х)²=10²=100. Двухчлен сокращается остаётся только два раза по квадрату числа и переменной, т. е. 2(46+х²)=100⇔46+х²=50⇔х²=4. Отсюда сторона квадрата 2, а, следовательно, диагональ 2√2.
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Let the side length of green square ACBE be x. As AB is a diagonal of ACBE, then AB = √2x.
As the diagonal of the yellow square is 2√23, then the side length of the yellow square is 2√23/√2 = √46.
Extend CB to T on FD. As CT is parallel to AF, then it is perpendicular to FD.
Triangle ∆CTD:
CT² + TD² = DC²
(x+√46)² + (√46-x)² = 10²
x² + 2√46x + 46 + 46 - 2√46x + x² = 100
2x² = 100 - 92 = 8
x² = 8/2 = 4
x = √4 = 2
AB = √2x
[ AB = 2√2 ]
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Sir, as green box and yellow box are square, angle CED will be 90 degree and by applying Pythagoras theorem on triangle ACD , we can calculate CE.
CE and AB are equal
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🙏🎉😎😘👍🔥
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Shorter solution
1. Connect 'C' & 'E' so we know that the
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STEP-BY-STEP RESOLUTION PROPOSAL :
01) Finding Yellow Square Side :
02) Diagonal of any Square is equal to : D = Side * sqrt(2)
03) 2 * sqrt(23) = Yellow Square Side (YSS) * sqrt(2) ; YSS = (2 * sqrt(23)) / sqrt(2) ; YSS = (2 * sqrt(2) * sqrt(23)) / (sqrt(2) * sqrt (2)) ; YSS = (2 * sqrt(2) * sqrt(23)) / 2 ; YSS = (sqrt(2) * sqrt(23)) ;
YSS = sqrt(46) ; YSS ~ 6,8 lin un
04) Finding Green Square Side (X) :
05) (sqrt(46) - X)^2 + (sqrt(46) + X)^2 = 100
06) Solutions : X = -2 or X = 2
07) AB = X * sqrt(2)
08) AB = 2 * sqrt(2) lin un
Therefore,
ANSWER : Beyond any reasonable Doubt AB is equal to [2sqrt(2)] Linear Units. AB ~ 2,83 Linear Units.
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Label the side of the yellow and green square as A and a.
Extend CB intersecting DF at point B’.
We have CB’= A+a and DB’= A-a
--> sq (A+a) +sq (A-a)=sq CD
-> 2 (sqA+sqa) =100
--> sqA+sqa=50
Because A=sqrt46
-> sqa=50-46=4
-> a =2
-> AB= 2sqrt2😅😅😅
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Let's find the length:
.
..
...
....
.....
Let y be the side length of the yellow square and let g be the side length of the green square. Now we add point G on DF such that CDG is a right triangle. Then we apply the Pythagorean theorem step by step to the right triangles DEF, CDG and ABE:
DF² + EF² = DE²
y² + y² = (2√23)²
2y² = 92
CG² + DG² = CD²
(BC + BG)² + (DF − FG)² = CD²
(g + y)² + (y − g)² = CD²
g² + 2gy + y² + y² − 2yg + g² = CD²
2g² + 2y² = CD²
⇒ 2g² = CD² − 2y² = 10² − 92 = 100 − 92 = 8
AB² = AE² + BE² = g² + g² = 2g² = 8
⇒ AB = √8 = 2√2
Best regards from Germany
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Sqrt (8)
Took 5 minutes
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Solution:
The two boxes are square, so we'll calculate the yellow square side
EF = 2√23/√2
EF = 2√46/2
EF = √46
Let's label the green square side as "a" and extending a line from "C" to "P", such that "P" is on the line "DF", we have:
CP = a + √46
DP = √46 - a
CD = 10
Applying the Pythagorean Theorem
(a + √46)² + (√46 - a) = 10²
a² + 2√46a + 46 + 46 - 2√46a + a² = 100
a² + 46 + 46 + a² = 100
2a² + 92 = 100
2a² = 8
a² = 4
a = 2
AB = a√2
AB = 2√2 Square Units ✅
AB ≈ 2,8284 Square Units ✅
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Im still here.
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2.8284
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@ 8:20 you use the word "thus" but don't say it. That's okay though because the anti-intellectual crowd isn't gonna read it anyway. 😊
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It's easier with the insight that angle CED =90 degrees.