the three white triangles are similar, their respective hypotenuse = ✓10, ✓40 (= 2✓10) and ✓40 + ✓10 (= 3✓10) So their linear ratio = 1 : 2 : 3 their area ratio = 1 : 4 : 9 If the long leg of the smallest white triangle is x The short leg of the smallest white triangle = ✓(10 - x^2) The long and short legs of the medium triangle = 2x and 2✓(10 - x^2) The long and short legs of the largest triangle = 3x and 3✓(10 - x^2) The side of the big square = ✓(10 - x^2) + 3x = 3✓(10 - x^2) + 2x x = 2✓(10 - x^2) x^2 = 40 - 4x^2 x = 2✓2 Side of the big square = 3✓(10 - x^2) + 2x = 3✓2 + 2✓8 = 7✓2 small white triangle area = (x✓(10 - x^2))/2 = 2 Green area = (7✓2)^2 - 10 - 40 - 2 - 4(2) - 9(2) = 98 - 50 - 2 - 8 - 18 = 48 - 28 = 20 cm^2
Lado cuadrado pequeño =c=√10 ---> Proyecciones ortogonales de los lados de la figura: AD=a+b+b+a+a ; AB=a+a+b+a+b=3a+2b ; BC=a+b+b+b=a+3b---> AB=BC---> 2a=b---> En el triángulo rectángulo de lados a/b/√10 = a/2a/√10: a²+(2a)²=(√10)²---> a²=2--->a=√2---> b=2√2 ---> Área sombreada =2*a²+2*ab+4*ab/2 =2*2+2*4+4*2 =4+8+8 =20 cm². Gracias y saludos.
At first, I get stuck after the calculation of the three triangles. Then after you said CD = BC , I got it. I knew how to solve. Sometimes I just need a little hint to be able to solve
If the top side of the 40 cm² square is extended to the left, it will intersect the corner point A of the blue square and divide the green area into 2 congruent right triangles, with sides √(10) and √(40), 10 cm² each, 20 cm² total. However, you must prove that point A is intersected in order for this solution to be valid, and you haven't done that.
@@jimlocke9320 Тебе какое доказательство нужно если квадраты дают смежные углы, которые в сумме равняются 90 градусов, значит треугольники подобные, а накрест лежащие. ещё и равные! Russia!
@@sergeyvinns931 When the top side of the 40 cm² square is extended to the left, we need to prove that it passes through point A. It could intersect AD between A and D or intersect AB between A and B, instead. The extension must pass through point A to divide the green area into 2 right triangles with sides of length √(10) and √(40). Also, once establishing that the extension goes through point A, we need to prove that the other side of the triangle with side √(40) has length √(10) and the other side of the triangle with side √(10) has length √(40).
@@jimlocke9320 Ничего она не может больше пересекать, так как все смежные углы в вписанных прямоугольных треугольниках в сумме равны 90 градусов, то есть, по периметру все треугольники подобны с разным коэффициентом подобия. самый маленький из подобных со сторонами \/2, 2\/2. \/10, второй 2\/2, 4\/2, 2\/10, третий, 5\/2/2, 5\/2. 5\/10/2, четвёртый, 3\/2, 5\/2, 3\/10. Это Пифагор, а не твоё хухры-мухры! Russia!
Side of square: S = √40cosα+(√40+√10)sinα S = (√40+√10)cosα+√10sinα Equalling: √40cosα+(√40+√10)sinα = (√40+√10)cosα+√10sinα √40cosα - (√40+√10)cosα =√10sinα-(√40+√10)sinα -√10cosα = √40sinα tanα = √10/√40 = 1/2 If tanα = 1/2 then top side of red square is collinear to top left vertex "A" of big square Therefore, green area is equal to area of two identical triangles, with base s₁=√10 and height s₂=√40 So, area of those two green triangles, is equal to area of half red square, or area of twice small square.
Let s be the side length of blue square ABCD and r and b the side lengths of the red and black squares. Red square: A = r² 40 = r² r = √40 = 2√10 Black square: A = b² 10 = b² b = √10 Let J and K be the intersection points between the black square and AB and BC respectively and M and N the intersection points between the red square and DA and CD respectively. Let ∠NMD = α, and as ∠MDN = 90°, let ∠DNM = β, where α and β are complementary angles that sum to 90°. As ∠MNK = 90° and CD is a straight line, ∠KNC = 180°-(90°+β) = 90°-β = α. As ∠NCK = 90°, ∠CKN = 90°-α = β. ∠NMD = ∠KNC = α and ∠DNM = ∠CKN = β, ∆MDN and ∆NCK are similar triangles. By similar steps, we can determine that ∆KBJ is also similar to ∆NCK and ∆MDN. Let KB = y and BJ = x. MD/NM = KB/JK MD/2√10 = y/√10 MD = 2y DN/NM = BJ/JK DN/2√10 = x/√10 DN = 2x NC/KN = KB/JK NC/3√10 = y/√10 NC = 3y CK/KN = BJ/JK CK/3√10 = x/√10 CK = 3x DC = CB = s DN + NC = CK + KB 2x + 3y = 3x + y x = 2y Triangle ∆KBJ: KB² + BJ² = JK² y² + x² = (√10)² y² + (2y)² = 10 y² + 4y² = 10 5y² = 10 y² = 10/5 = 2 y = √2 x = 2y = 2√2 s = 3x + y = 3(2√2) + √2 = 7√2 The green area is equal to the area of blue square ABCD minus the areas of the red and black squares and the three similar triangles. Green shaded region: A = s² - 40 - 10 - 2x2y/2 - 3x3y/2 - xy/2 A = (7√2)² - 50 - (xy/2)(4+9+1) A = 98 - 50 - (2√2(√2)/2)14 A = 48 - 2(14) = 48 - 28 = 20 cm³
The green area is 20 cm squared. I am wonfering at the 4:25 mark, is it possible for three triangles to be similar by HL??? That is definitely that first time that I have learned of that and I better use that for practice. And maybe there should playlists of problems that involve HL similarity and/or AA similarity. Just saying.
Let's find the area: . .. ... .... ..... First of all we calculate the side lengths s(red) and s(black) of the red and black square, respectively: s(red) = √A(red square) = √(40cm²) = (2√10)cm s(black) = √A(black square) = √(10cm²) = (√10)cm Now let's label E, F, G and H the points where the squares intersect AB, BC, CD and DA, respectively. From a look at the interior angles in the right triangles BEF, CFG and DGH we obtain: ∠EBF = 90° ∠BEF = α ∠BFE = 90°−α ∠FCG = 90° ∠CFG = α ∠CGF = 90°−α ∠GDH = 90° ∠DGH = α ∠DHG = 90°−α Therefore all these triangles are similar and we can conclude: BF:CG:DH = BE:CF:DG = EF:FG:GH = s(black):s(black)+s(red):s(red) = √10:(√10+2√10):2√10 = 1:3:2 s(blue) = BC = CD BF + CF = CG + DG BF + 3*BE = 3*BF + 2*BE ⇒ BE = 2*BF Now we apply the Pythagorean theorem to the right triangle BEF: EF² = BE² + BF² (√10)²cm² = (2*BF)² + BF² 10cm² = 4*BF² + BF² 10cm² = 5*BF² 2cm² = BF² ⇒ BF = (√2)cm ⇒ BE = 2*BF = (2√2)cm Now we are able to calculate the area of the green region:A(blue square) = s²(blue) = BC² = (BF + 3*BE)² = (√2 + 3*2√2)²cm² = (7√2)²cm² = 98cm² A(triangle BEF) = (1/2)*BE*BF = (1/2)*(2√2)*(√2)cm² = 2cm² A(triangle CFG) = (1/2)*CF*CG = (1/2)*(3*BE)*(3*BF) = 9*(1/2)*BE*BF = 9*2cm² = 18cm² A(triangle DGH) = (1/2)*DG*DH = (1/2)*(2*BE)*(2*BF) = 4*(1/2)*BE*BF = 4*2cm² = 8cm² A(triangles) = A(triangle BEF) + A(triangle CFG) + A(triangle DGH) = 2cm² + 18cm² + 8cm² = 28cm² A(green) = A(blue square) − A(red square) − A(black square) − A(triangles) = 98cm² − 40cm² − 10cm² − 28cm² = 20cm² Best regards from Germany
STEP-BY-STEP (LATE) RESOLUTION PROPOSAL : 01) My answer is that The Green Shaded Region equal 20 Square Centimeters. 02) Divide the Red Triangle in 4 equal Squares of 10 sq cm. 03) Call the Point between AD Point E (Corner of Red Square), and call the Point between AB, Point F (Corner of Black Square). 04) Draw a Parallel Line to AB (Horizontal Line) from E. Draw a Parallel Line to AD (Vertical Line) from F. The interception Point is Point G. And Point G belongs to Red Square. Point G is in the Middle of the Side of Red Square. 05) Draw a Square [AEGF] with Sides (S) ; S^2 = (sqrt(40))^2 + (sqrt(10))^2 ; S^2 = 40 + 10 ; S^2 = 50 ; S = sqrt(50) 06) Area of Square (A) = 50 07) Observe that the Area of that Square wich is not Green equal 30 sq cm 08) Green Shaded Region = 50 sq cm - 30 sq cm ; GSA = 20 sq cm Therefore, MY ANSWER : Green Shaded Region, as stated before, equal to 20 Square Centimeters.
Good lesson on proper grammar. The Blue, Red and Black boxes are the squares! Not the Blue, Red and Black box are the squares! Nothing wrong with how the problem was stated, just sumptin how my English teacher would've turned a math problem into a grammar question cuz she's not a mathematician. 🙂
You have increased my limits in maths😊
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Keep it up😀
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Muy buena presentación y descripción del ejercicio. Es importante que el dibujo de la figura este claro en cuanto a sus medidas.
Saludos.
the three white triangles are similar, their respective hypotenuse = ✓10, ✓40 (= 2✓10) and ✓40 + ✓10 (= 3✓10)
So their linear ratio = 1 : 2 : 3
their area ratio = 1 : 4 : 9
If the long leg of the smallest white triangle is x
The short leg of the smallest white triangle = ✓(10 - x^2)
The long and short legs of the medium triangle = 2x and 2✓(10 - x^2)
The long and short legs of the largest triangle = 3x and 3✓(10 - x^2)
The side of the big square = ✓(10 - x^2) + 3x = 3✓(10 - x^2) + 2x
x = 2✓(10 - x^2)
x^2 = 40 - 4x^2
x = 2✓2
Side of the big square = 3✓(10 - x^2) + 2x = 3✓2 + 2✓8 = 7✓2
small white triangle area = (x✓(10 - x^2))/2 = 2
Green area = (7✓2)^2 - 10 - 40 - 2 - 4(2) - 9(2) = 98 - 50 - 2 - 8 - 18 = 48 - 28 = 20 cm^2
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@@PreMath Thank YOU for the fun question 🙂
Wow
That’s very very nice
Thanks Sir
Thanks PreMath
I liked this method
Good luck with my respects
❤❤❤❤❤
So nice of you dear
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Nicely explained sir🙏
Glad to hear that!
Thanks for the feedback ❤️
شكرا لكم على المجهودات
√40=2√10 √10
2√10+√10=3√10
2y+3x=3y+x y=2x
x^2+y^2=(√10)^2 5x^2=10 x^2=2
ABCD=7x*7x=49x^2=98
x*y/2=2x^2/2=x^2=2
2x*2y/2=4x^2=8
3x*3y/2=9x^2=18
Green shaded region :
98 - (40 + 10 + 2 + 8 + 18) = 20(cm^2)
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A = 2.A₁ = ½.A₂
A = 2*10 = ½*40
A = 20cm² ( Solved √ )
Thank you! This one was a stretch for me
You're so welcome!
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Lado cuadrado pequeño =c=√10 ---> Proyecciones ortogonales de los lados de la figura: AD=a+b+b+a+a ; AB=a+a+b+a+b=3a+2b ; BC=a+b+b+b=a+3b---> AB=BC---> 2a=b---> En el triángulo rectángulo de lados a/b/√10 = a/2a/√10: a²+(2a)²=(√10)²---> a²=2--->a=√2---> b=2√2 ---> Área sombreada =2*a²+2*ab+4*ab/2 =2*2+2*4+4*2 =4+8+8 =20 cm².
Gracias y saludos.
At first, I get stuck after the calculation of the three triangles. Then after you said CD = BC , I got it. I knew how to solve.
Sometimes I just need a little hint to be able to solve
Green area is two equal triangle with base sqrt(10) and height sqrt(40). So sqrt(10) * sqrt(40) = 20
How
If the top side of the 40 cm² square is extended to the left, it will intersect the corner point A of the blue square and divide the green area into 2 congruent right triangles, with sides √(10) and √(40), 10 cm² each, 20 cm² total. However, you must prove that point A is intersected in order for this solution to be valid, and you haven't done that.
@@jimlocke9320 Тебе какое доказательство нужно если квадраты дают смежные углы, которые в сумме равняются 90 градусов, значит треугольники подобные, а накрест лежащие. ещё и равные! Russia!
@@sergeyvinns931 When the top side of the 40 cm² square is extended to the left, we need to prove that it passes through point A. It could intersect AD between A and D or intersect AB between A and B, instead. The extension must pass through point A to divide the green area into 2 right triangles with sides of length √(10) and √(40). Also, once establishing that the extension goes through point A, we need to prove that the other side of the triangle with side √(40) has length √(10) and the other side of the triangle with side √(10) has length √(40).
@@jimlocke9320 Ничего она не может больше пересекать, так как все смежные углы в вписанных прямоугольных треугольниках в сумме равны 90 градусов, то есть, по периметру все треугольники подобны с разным коэффициентом подобия. самый маленький из подобных со сторонами \/2, 2\/2. \/10, второй 2\/2, 4\/2, 2\/10, третий, 5\/2/2, 5\/2. 5\/10/2, четвёртый, 3\/2, 5\/2, 3\/10. Это Пифагор, а не твоё хухры-мухры! Russia!
Thank you!
The similarity part is tricky. After that, the problem became a piece of cake.
True!
Thanks for the feedback ❤️
S=20 cm²
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Thanks for sharing ❤️
A= 2 A₁ = 2 (½b.h) = b.h
A = √40.√10 = √400
A = 20cm² ( Solved √ )
A = A₂ / 2 = 40 / 2
A = 20cm² ( Solved √ )
A = 2 . A₁ = 2 . 10
A = 20cm² ( Solved √ )
Side of square:
S = √40cosα+(√40+√10)sinα
S = (√40+√10)cosα+√10sinα
Equalling:
√40cosα+(√40+√10)sinα = (√40+√10)cosα+√10sinα
√40cosα - (√40+√10)cosα =√10sinα-(√40+√10)sinα
-√10cosα = √40sinα
tanα = √10/√40 = 1/2
If tanα = 1/2 then top side of red square is collinear to top left vertex "A" of big square
Therefore, green area is equal to area of two identical triangles, with base s₁=√10 and height s₂=√40
So, area of those two green triangles, is equal to area of half red square, or area of twice small square.
Let s be the side length of blue square ABCD and r and b the side lengths of the red and black squares.
Red square:
A = r²
40 = r²
r = √40 = 2√10
Black square:
A = b²
10 = b²
b = √10
Let J and K be the intersection points between the black square and AB and BC respectively and M and N the intersection points between the red square and DA and CD respectively.
Let ∠NMD = α, and as ∠MDN = 90°, let ∠DNM = β, where α and β are complementary angles that sum to 90°. As ∠MNK = 90° and CD is a straight line, ∠KNC = 180°-(90°+β) = 90°-β = α. As ∠NCK = 90°, ∠CKN = 90°-α = β. ∠NMD = ∠KNC = α and ∠DNM = ∠CKN = β, ∆MDN and ∆NCK are similar triangles.
By similar steps, we can determine that ∆KBJ is also similar to ∆NCK and ∆MDN. Let KB = y and BJ = x.
MD/NM = KB/JK
MD/2√10 = y/√10
MD = 2y
DN/NM = BJ/JK
DN/2√10 = x/√10
DN = 2x
NC/KN = KB/JK
NC/3√10 = y/√10
NC = 3y
CK/KN = BJ/JK
CK/3√10 = x/√10
CK = 3x
DC = CB = s
DN + NC = CK + KB
2x + 3y = 3x + y
x = 2y
Triangle ∆KBJ:
KB² + BJ² = JK²
y² + x² = (√10)²
y² + (2y)² = 10
y² + 4y² = 10
5y² = 10
y² = 10/5 = 2
y = √2
x = 2y = 2√2
s = 3x + y = 3(2√2) + √2 = 7√2
The green area is equal to the area of blue square ABCD minus the areas of the red and black squares and the three similar triangles.
Green shaded region:
A = s² - 40 - 10 - 2x2y/2 - 3x3y/2 - xy/2
A = (7√2)² - 50 - (xy/2)(4+9+1)
A = 98 - 50 - (2√2(√2)/2)14
A = 48 - 2(14) = 48 - 28 = 20 cm³
The green area is 20 cm squared. I am wonfering at the 4:25 mark, is it possible for three triangles to be similar by HL??? That is definitely that first time that I have learned of that and I better use that for practice. And maybe there should playlists of problems that involve HL similarity and/or AA similarity. Just saying.
Excellent!
Thanks for the feedback ❤️
Let's find the area:
.
..
...
....
.....
First of all we calculate the side lengths s(red) and s(black) of the red and black square, respectively:
s(red) = √A(red square) = √(40cm²) = (2√10)cm
s(black) = √A(black square) = √(10cm²) = (√10)cm
Now let's label E, F, G and H the points where the squares intersect AB, BC, CD and DA, respectively. From a look at the interior angles in the right triangles BEF, CFG and DGH we obtain:
∠EBF = 90° ∠BEF = α ∠BFE = 90°−α
∠FCG = 90° ∠CFG = α ∠CGF = 90°−α
∠GDH = 90° ∠DGH = α ∠DHG = 90°−α
Therefore all these triangles are similar and we can conclude:
BF:CG:DH = BE:CF:DG = EF:FG:GH = s(black):s(black)+s(red):s(red) = √10:(√10+2√10):2√10 = 1:3:2
s(blue) = BC = CD
BF + CF = CG + DG
BF + 3*BE = 3*BF + 2*BE
⇒ BE = 2*BF
Now we apply the Pythagorean theorem to the right triangle BEF:
EF² = BE² + BF²
(√10)²cm² = (2*BF)² + BF²
10cm² = 4*BF² + BF²
10cm² = 5*BF²
2cm² = BF²
⇒ BF = (√2)cm ⇒ BE = 2*BF = (2√2)cm
Now we are able to calculate the area of the green region:A(blue square) = s²(blue) = BC² = (BF + 3*BE)² = (√2 + 3*2√2)²cm² = (7√2)²cm² = 98cm²
A(triangle BEF) = (1/2)*BE*BF = (1/2)*(2√2)*(√2)cm² = 2cm²
A(triangle CFG) = (1/2)*CF*CG = (1/2)*(3*BE)*(3*BF) = 9*(1/2)*BE*BF = 9*2cm² = 18cm²
A(triangle DGH) = (1/2)*DG*DH = (1/2)*(2*BE)*(2*BF) = 4*(1/2)*BE*BF = 4*2cm² = 8cm²
A(triangles) = A(triangle BEF) + A(triangle CFG) + A(triangle DGH) = 2cm² + 18cm² + 8cm² = 28cm²
A(green) = A(blue square) − A(red square) − A(black square) − A(triangles) = 98cm² − 40cm² − 10cm² − 28cm² = 20cm²
Best regards from Germany
STEP-BY-STEP (LATE) RESOLUTION PROPOSAL :
01) My answer is that The Green Shaded Region equal 20 Square Centimeters.
02) Divide the Red Triangle in 4 equal Squares of 10 sq cm.
03) Call the Point between AD Point E (Corner of Red Square), and call the Point between AB, Point F (Corner of Black Square).
04) Draw a Parallel Line to AB (Horizontal Line) from E. Draw a Parallel Line to AD (Vertical Line) from F. The interception Point is Point G. And Point G belongs to Red Square. Point G is in the Middle of the Side of Red Square.
05) Draw a Square [AEGF] with Sides (S) ; S^2 = (sqrt(40))^2 + (sqrt(10))^2 ; S^2 = 40 + 10 ; S^2 = 50 ; S = sqrt(50)
06) Area of Square (A) = 50
07) Observe that the Area of that Square wich is not Green equal 30 sq cm
08) Green Shaded Region = 50 sq cm - 30 sq cm ; GSA = 20 sq cm
Therefore,
MY ANSWER : Green Shaded Region, as stated before, equal to 20 Square Centimeters.
I did it the same way.
Good lesson on proper grammar. The Blue, Red and Black boxes are the squares! Not the Blue, Red and Black box are the squares! Nothing wrong with how the problem was stated, just sumptin how my English teacher would've turned a math problem into a grammar question cuz she's not a mathematician. 🙂
~1st comment🙋🏻♀️😌
Excellent!
Thanks ❤️
Good night sir
Same to you dear❤️
20
Excellent!
Thanks for sharing ❤️
=
My way of solution ▶
E ∈ [AB]
F ∈ [BC]
G ∈ [DC]
P ∈ [AD]
i) by considering the right triangle ΔEFB
∠BEF= α
∠EFB= β
∠FBE= 90°
[BE]= x
[FB]= y
[EF]= √10
ii) by considering the right triangle ΔFGC
∠CFG= α
∠FGC= β
∠GCF= 90°
[GC]= z
[CF]= a-y
[FG]= √10+√40
[FG]= 3√10
iii) by considering the right triangle ΔGPD
∠DGP= α
∠GPD= β
∠PDG= 90°
[PD]= t
[DG]= a-z
[GP]= √40
[GP]= 2√10
iv) we see that ΔEFB ~ ΔFGC
[FB]/[GC]= [BE]/[CF]= [EF]/[FG]
y/z = x/(a-y)= √10/3√10
y/z= x/(a-y)= 1/3
⇒
z= 3y
a-y= 3x
v) we see that ΔEGC ~ ΔGPD
[GC]/[PD]= [CF]/[DG]= [FG]/[GP]
[GC]=z
z= 3y
[CF]= a-y
a-y= 3x
⇒
3y/t = 3x/(a-z) = 3√10 / 2√10
⇒
a-z= 2x
t= 2y
vi) we see that ΔEFB ~ ΔGPD
[FB]/[PD]= [BE]/[DG]= [EF]/[GP]
y/t= x/a-z= √10 / 2√10
t= 2y
a-z= 2x
⇒
y/2y= x/2x= 1/2
1/2 = 1/2 = 1/2 ✅
⇒
The relations between the sides are correct !
vii) while it is a squaree
[CB] = [DC]
2x+3y= y+3x
2y= x
⇒
we can write the Pythagorean theorem
x²+y²= 10
⇒
(2y)²+y²= 10
5y²= 10
y²= 2
⇒
y= √2 cm
x= 2√2 cm
viii) Let's calculate the surfaces of each triangles
A(ΔEFB)= xy/2
A(ΔEFB)= 2√2*√2/2
A(ΔEFB)= 2 cm²
A(ΔFGC)= 3x*3y/2
A(ΔFGC)= 6√2*3√2/2
A(ΔFGC)= 18 cm²
A(ΔGPD)= 2x*2y/2
A(ΔFGC)= 4√2*2√2/2
A(ΔFGC)= 8 cm²
A(ΔEFB)+ A(ΔFGC) + A(ΔFGC)= 2+18+8
= 28 cm²
ix) find the side length a
a= y+3x
a= 2x+3y
⇒
a= y+3x
a= √2 +6√2
a= 7√2 cm
⇒
a²= 98 cm²
x) find the green area
Agreen= a² - A(ΔEFB)+ A(ΔFGC) + A(ΔFGC) - 40 - 10
Agreen= 98- 28 - 50
Agreen= 20 cm²
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Thanks for sharing ❤️
Area of the Qreen shaded region 2\/10*\/10=20. Для этого достаточно гкометрических постороений! Russia.
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