You can use the 3-4-5 pythagorean triplet to calculate the sides as a factor of X instead of the similar triangles method. So EF being the 3-side makes it 3/3X, that makes CF 4/3X and CE 5/3X. Likewise DE being the 5-side makes it 5/5X, then BD is 3/5X and BE is 4/5X.
If the triangles are in a 3-4-5 relationship then instead of assuming "x" is the side of the square, we make it equal to "5x". That is, DE=EF=FG=GD=5x. If we take triangle BDE we have that DE=5x; BE=4x and BD=3x. As BC=148 then EC=148-4x. Now, the angle at c is α so sinα=111/185=EF/EC=5x/(148-4x). We solve and we get x=12. Consequently, the side of the square is equal to 60 and its area is 3600.
Solution: ∆ ABC Area = ½ b . h 8,214 = ½ b . 111 b = 16,428/111 b = 148 Lets applying Pythagorean Theorem to calculate AC (148)² + 111² = AC² AC² = 21,904 + 12,321 AC² = 34,225 AC = √34,225 AC = 185 ∆ ABC is similar to ∆ DBE, and, like that, we have proportions 111/185 = DB/a DB = 111a/185 (÷37) DB = 3a/5 ∆ ABC is similar to ∆ ADG, and, like that, we have proportions a/(111 - 3a/5) = 148/185 185a = 148 (111 - 3a/5) 185a = 16,428 - 444a/5 (×5) 925a = 82,140 - 444a 1,369a = 82,140 a = 60 Yellow Square Area (YSA) = a² YSA = a² YSA = 60² YSA = 3,600 Square Units ✅
8214 x 2/111 = 148 √(111^2 + 148^2) = √(12321 + 21904) = √34225 = 185 The big triangle and the three white triangles are similar. If a is the hypotenuse length of the upper left triangle and b is the side of the square b/a = 148/185 (111 - a)/b = 111/185 111/b - 185/148 = 111/185 b = 111/(185/148 + 111/185) = 111(148)(185)/(185^2 + 111(148)) = 60 Square area = 3600
*Outra alternativa:* No ∆AGD: cos α = x/AD → AD = x/cos α No ∆DBE: sen α = DB/ x → DB = x sen α Daí, AD + DB = AB x/cos α + x sen α = 111 _x(1/cos α + sen α) = 111_ No ∆ ABC: sen α =111/185=3/5 e cos α=4/5 x(5/4 + 3/5 ) = 111 x(25 +12)/20 = 111 x = 111 × 20/37 = 3 × 20 → x = 60. Logo, a área é *3600 unidades quadradas.*
BC=2*8214/111=148---> 111=37*3 ; 148=37*4---> Si "a" es el lado del cuadrado: AB={5a/4}+(3a/5)=37a/20 ---> a=111*20/37=60---> Área del cuadrado amarillo =60²=3600. Gracias y saludos
I wouldn't wanna die at sea for impiety and so I stick with Pythagoras' logic when it comes too almost- isosceles triangles too avoid the wrath of Poseiden...unlike numerology where meaning in numbers like 111 333 666 999 is analogous too meaning of reading tea leaves clouds in my coffee guts of sacrificial animals or you know, any error in perception that random numbers have any meaning. In this problem that 111 stands out like a sore thumb and makes me spin counter clockwise craving angelic approval. 🙂 ...BTW, @ 8:33 determining x was completely awesome! 🙂
Triangle ∆ABC: A = bh/2 = BC(AB)/2 8214 = BC(111)/2 111BC = 8214(2) = 16428 BC = 16428/111 = 148 As AB = 111 = 37(3) and BC = 148 = 37(4), then ∆ABC is a 3-4-5 Pythagorean triple right triangle with a scale of 37:1 and CA = 37(5) = 185. As ∠DGA = ∠ABC = 90° and ∠A is common, then ∆DGA is similar to ∆ABC. As ∠CFE = ∠ABC = 90° and ∠C is common, then ∆CFE is similar to ∆ABC. Thus all three aforementioned triangles are similar. Let s be the side length of square DEFG. GA/DG = AB/BC GA/s = 111/148 = 3/4 GA = 3s/4 CF/FE = BC/AB CF/s = 148/111 = 4/3 CF = 4s/3 CA = CF + FG + GA 185 = 4s/3 + s + 3s/4 185 = (16+12+9)s/12 = 37s/12 s = 185(12/37) = 5(12) = 60 Yellow Square DEFG: [ A = s² = 60² = 3600 sq units ]
BE = (2.area of ABC)/AB = (2.8214)/111 = 148. Now AC^2 = AB^2 + BC^2 = 111^2 + 148^2 = 34225 = 185^2, so AC = 185. Now let's name t = angleBCA. We have sin(t) = AB/AC = 111/185 = 3/5, cos(t) = BC/AC = 148/185 = 4/5 and then tan(t) = 3/4. AngleDEB = t, so in triangle DBE we have DB/DE = sin(t) = 3/5, so DB = (3/5).DE = (3/5).c if we name c the side length of the square DEFG. AngleADG = t, so in triangle AGD we have DG/AD = cos(t) = 4/5, so AD = (5/4).DG = (5/4).c Now AB = 111 = DB + AD = (3/5).c + (5/4).c = (37/20).c. So we have c= 111/(37/20) = 60 and finallt the area of the square DEFG is 60^2 = 3600.
Sir my focus is the hypotenuse(AC) of triangle ABC! AC =AG + GF(x) +FC --(1)[ x be the side of square ] 2) In 🔺 ABC AB =111 units & area is 8214 sq units Hence BC = 8214/(2*111) = 148 units AB ঃ BC = 111ঃ 148=3 ঃ 4 (dividing by 37) According to Pythagorean Triplets AC = 37*5=185 units 3) Now 🔺 ADG & 🔺 EFC are similar to the big 🔺 ABC 4)So the ratio of their short leg to long leg will be 3ঃ 4 like the ratio of short leg and long leg of 🔺 ABC. 5) Hence in 🔺 ADG = AG ঃ DG =3ঃ 4 AG =3DG/4 = 3x /4(x be the side of the square) -- (2) 6 ) in 🔺 EFC in the same way EF ঃFC =3ঃ4 FC =4EF/3=4x/3--(3)(EF =x =side of the square) 7)from (1), (2).(3) we get 3x /4 + x +4x /3 =185 >37x /12 = 185 >x =60 Side of square =60 units Area of square = 60^2 =3600 sq units
Let's find the area: . .. ... .... ..... From the given area and the given side length of the right triangle ABC we can conclude: A(ABC) = (1/2)*AB*BC ⇒ BC = 2*A(ABC)/AB = 2*8214/111 = 148 Now let's assume that B is the center of the coordinate system and that BC is located on the x-axis. Then the line AC is represented by the following function: y = (yA − yC)*(x − xC)/(xA − xC) + yC = (111 − 0)*(x − 148)/(0 − 148) + 0 = −3*(x − 148)/4 The line AC is parallel to DE and perpendicular to EF. Therefore the lines DE and EF are represented by the following functions: DE: y = −3*(x − xE)/4 EF: y = 4*(x − xE)/3 AC and EF intersect at F. So we obtain: −3*(xF − 148)/4 = 4*(xF − xE)/3 −9*(xF − 148) = 16*(xF − xE) −9*xF + 1332 = 16*xF − 16*xE 16*xE + 1332 = 25*xF ⇒ xF = (16*xE + 1332)/25 ⇒ yF = −3*(xF − 148)/4 = −3*[(16*xE + 1332)/25 − 148]/4 = −3*(16*xE + 1332 − 3700)/100 = (−48*xE + 7104)/100 = (−12*xE + 1776)/25 (⇒ yF = 4*(xF − xE)/3 = 4*[(16*xE + 1332)/25 − xE]/3 = 4*(16*xE + 1332 − 25*xE)/75 = (−36*xE + 5328)/75 = (−12*xE + 1776)/25 ✓) The lengths of DE and EF can be calculated as follows: DE² = (xE − xD)² + (yE − yD)² = (xE − xD)² + [(−3/4)*(xE − xD)]² = (xE − xD)² + (9/16)*(xE − xD)² = (25/16)*(xE − xD)² ⇒ DE = (5/4)*(xE − xD) = (5/4)*(xE − 0) = (5/4)*xE EF² = (xF − xE)² + (yF − yE)² = (xF − xE)² + [(4/3)*(xF − xE)]² = (xF − xE)² + (16/9)*(xF − xE)² = (25/9)*(xF − xE)² ⇒ EF = (5/3)*(xF − xE) = (5/3)*[(16*xE + 1332)/25 − xE] = (5/3)*(16*xE + 1332 − 25*xE)/25 = (5/3)*(−9*xE + 1332)/25 = (−3*xE + 444)/5 Since DEFG is a square, we have DE=EF. So we can conclude: (5/4)*xE = (−3*xE + 444)/5 25*xE = 4*(−3*xE + 444) 25*xE = −12*xE + 1776 37*xE = 1776 ⇒ xE = 1776/37 = 48 ⇒ DE = (5/4)*xE = (5/4)*48 = 60 ∧ EF = (−3*xE + 444)/5 = (−3*48 + 444)/5 = (−144 + 444)/5 = 300/5 = 60 ✓ Now we are able to calculate the area of the yellow square: A(DEFG) = DE² = EF² = 60² = 3600 Best regards from Germany
❤❤❤
Many thanks, dear ❤️
You can use the 3-4-5 pythagorean triplet to calculate the sides as a factor of X instead of the similar triangles method. So EF being the 3-side makes it 3/3X, that makes CF 4/3X and CE 5/3X. Likewise DE being the 5-side makes it 5/5X, then BD is 3/5X and BE is 4/5X.
Yes it may be true
Until now this was the delight of my day.
If the triangles are in a 3-4-5 relationship then instead of assuming "x" is the side of the square, we make it equal to "5x". That is, DE=EF=FG=GD=5x. If we take triangle BDE we have that DE=5x; BE=4x and BD=3x. As BC=148 then EC=148-4x. Now, the angle at c is α so sinα=111/185=EF/EC=5x/(148-4x). We solve and we get x=12. Consequently, the side of the square is equal to 60 and its area is 3600.
Thank you! The use of proportions, knowing the larger triangle's side length. That was spectacular! That definitely goes into the toolbox!
Solution:
∆ ABC Area = ½ b . h
8,214 = ½ b . 111
b = 16,428/111
b = 148
Lets applying Pythagorean Theorem to calculate AC
(148)² + 111² = AC²
AC² = 21,904 + 12,321
AC² = 34,225
AC = √34,225
AC = 185
∆ ABC is similar to ∆ DBE, and, like that, we have proportions
111/185 = DB/a
DB = 111a/185 (÷37)
DB = 3a/5
∆ ABC is similar to ∆ ADG, and, like that, we have proportions
a/(111 - 3a/5) = 148/185
185a = 148 (111 - 3a/5)
185a = 16,428 - 444a/5 (×5)
925a = 82,140 - 444a
1,369a = 82,140
a = 60
Yellow Square Area (YSA) = a²
YSA = a²
YSA = 60²
YSA = 3,600 Square Units ✅
8214 x 2/111 = 148
√(111^2 + 148^2) = √(12321 + 21904) = √34225 = 185
The big triangle and the three white triangles are similar. If a is the hypotenuse length of the upper left triangle and b is the side of the square
b/a = 148/185
(111 - a)/b = 111/185
111/b - 185/148 = 111/185
b = 111/(185/148 + 111/185) = 111(148)(185)/(185^2 + 111(148)) = 60
Square area = 3600
Yellow Square Area equal 3.600 Square Units.
A = (0 ; 0)
B = (66,6 ; 88,8)
C = (185 ; 0)
D = (45 ; 60)
E = (105 ; 60)
F = (45 ; 0)
G = (105 ; 0)
105 - 45 = 60
A = 8214 cm² = ½.b.h
b = 2A/h = 2*8214/111 = 148 cm
tanα = 111/148 = 3/4
x/cosα + x.sinα = 111
x/(4/5) + x.(3/5) = 111
5/4 x + 3/5 x = 111
x = 111 / (5/4 + 3/5) = 60cm
x² = 3600 cm² ( Solved √ )
111*BC/2=8214 BC=148 37*3=111 37*4 =148 AC=37*5=185
111 : 148 : 185 = 3 : 4 : 5
DB=3x BE=4x DE=EF=FG=GD=5x AD=5x*5/4=25x/4
3x+25x/4=111 37x/4=111 37x=444 x=12 5x=60
Yellow Square area = 60*60 = 3600
*Outra alternativa:*
No ∆AGD:
cos α = x/AD → AD = x/cos α
No ∆DBE:
sen α = DB/ x → DB = x sen α
Daí,
AD + DB = AB
x/cos α + x sen α = 111
_x(1/cos α + sen α) = 111_
No ∆ ABC:
sen α =111/185=3/5 e cos α=4/5
x(5/4 + 3/5 ) = 111
x(25 +12)/20 = 111
x = 111 × 20/37 = 3 × 20 → x = 60.
Logo, a área é *3600 unidades quadradas.*
Small error x = 60, and area is 3600 sq unit
@easyart1882 verdade! Corrigido! Obrigado
55.5×55.5=3080.25 big squaare area
BC=2*8214/111=148---> 111=37*3 ; 148=37*4---> Si "a" es el lado del cuadrado: AB={5a/4}+(3a/5)=37a/20 ---> a=111*20/37=60---> Área del cuadrado amarillo =60²=3600.
Gracias y saludos
I wouldn't wanna die at sea for impiety and so I stick with Pythagoras' logic when it comes too almost- isosceles triangles too avoid the wrath of Poseiden...unlike numerology where meaning in numbers like 111 333 666 999 is analogous too meaning of reading tea leaves clouds in my coffee guts of sacrificial animals or you know, any error in perception that random numbers have any meaning. In this problem that 111 stands out like a sore thumb and makes me spin counter clockwise craving angelic approval. 🙂 ...BTW, @ 8:33 determining x was completely awesome! 🙂
Triangle ∆ABC:
A = bh/2 = BC(AB)/2
8214 = BC(111)/2
111BC = 8214(2) = 16428
BC = 16428/111 = 148
As AB = 111 = 37(3) and BC = 148 = 37(4), then ∆ABC is a 3-4-5 Pythagorean triple right triangle with a scale of 37:1 and CA = 37(5) = 185.
As ∠DGA = ∠ABC = 90° and ∠A is common, then ∆DGA is similar to ∆ABC. As ∠CFE = ∠ABC = 90° and ∠C is common, then ∆CFE is similar to ∆ABC. Thus all three aforementioned triangles are similar. Let s be the side length of square DEFG.
GA/DG = AB/BC
GA/s = 111/148 = 3/4
GA = 3s/4
CF/FE = BC/AB
CF/s = 148/111 = 4/3
CF = 4s/3
CA = CF + FG + GA
185 = 4s/3 + s + 3s/4
185 = (16+12+9)s/12 = 37s/12
s = 185(12/37) = 5(12) = 60
Yellow Square DEFG:
[ A = s² = 60² = 3600 sq units ]
I thank you for showing, thanks once again
4107 square unit ?
BE = (2.area of ABC)/AB = (2.8214)/111 = 148. Now AC^2 = AB^2 + BC^2 = 111^2 + 148^2 = 34225 = 185^2, so AC = 185. Now let's name t = angleBCA.
We have sin(t) = AB/AC = 111/185 = 3/5, cos(t) = BC/AC = 148/185 = 4/5 and then tan(t) = 3/4.
AngleDEB = t, so in triangle DBE we have DB/DE = sin(t) = 3/5, so DB = (3/5).DE = (3/5).c if we name c the side length of the square DEFG.
AngleADG = t, so in triangle AGD we have DG/AD = cos(t) = 4/5, so AD = (5/4).DG = (5/4).c
Now AB = 111 = DB + AD = (3/5).c + (5/4).c = (37/20).c. So we have c= 111/(37/20) = 60 and finallt the area of the square DEFG is 60^2 = 3600.
4108
3600
Sir my focus is the hypotenuse(AC) of triangle ABC!
AC =AG + GF(x) +FC --(1)[ x be the side of square ]
2) In 🔺 ABC
AB =111 units & area is 8214 sq units
Hence
BC = 8214/(2*111)
= 148 units
AB ঃ BC = 111ঃ 148=3 ঃ 4 (dividing by 37)
According to Pythagorean Triplets AC = 37*5=185 units
3)
Now 🔺 ADG & 🔺 EFC are
similar to the big 🔺 ABC
4)So the ratio of their short leg to long leg will be 3ঃ 4 like the ratio of short leg and long leg of 🔺 ABC.
5) Hence in 🔺 ADG
= AG ঃ DG =3ঃ 4
AG =3DG/4 = 3x /4(x be the side of the square) -- (2)
6 ) in 🔺 EFC in the same way EF ঃFC =3ঃ4
FC =4EF/3=4x/3--(3)(EF =x =side of the square)
7)from (1), (2).(3) we get
3x /4 + x +4x /3 =185
>37x /12 = 185
>x =60
Side of square =60 units
Area of square = 60^2
=3600 sq units
Let's find the area:
.
..
...
....
.....
From the given area and the given side length of the right triangle ABC we can conclude:
A(ABC) = (1/2)*AB*BC ⇒ BC = 2*A(ABC)/AB = 2*8214/111 = 148
Now let's assume that B is the center of the coordinate system and that BC is located on the x-axis. Then the line AC is represented by the following function:
y = (yA − yC)*(x − xC)/(xA − xC) + yC = (111 − 0)*(x − 148)/(0 − 148) + 0 = −3*(x − 148)/4
The line AC is parallel to DE and perpendicular to EF. Therefore the lines DE and EF are represented by the following functions:
DE: y = −3*(x − xE)/4
EF: y = 4*(x − xE)/3
AC and EF intersect at F. So we obtain:
−3*(xF − 148)/4 = 4*(xF − xE)/3
−9*(xF − 148) = 16*(xF − xE)
−9*xF + 1332 = 16*xF − 16*xE
16*xE + 1332 = 25*xF
⇒ xF = (16*xE + 1332)/25
⇒ yF = −3*(xF − 148)/4 = −3*[(16*xE + 1332)/25 − 148]/4 = −3*(16*xE + 1332 − 3700)/100 = (−48*xE + 7104)/100 = (−12*xE + 1776)/25
(⇒ yF = 4*(xF − xE)/3 = 4*[(16*xE + 1332)/25 − xE]/3 = 4*(16*xE + 1332 − 25*xE)/75 = (−36*xE + 5328)/75 = (−12*xE + 1776)/25 ✓)
The lengths of DE and EF can be calculated as follows:
DE²
= (xE − xD)² + (yE − yD)²
= (xE − xD)² + [(−3/4)*(xE − xD)]²
= (xE − xD)² + (9/16)*(xE − xD)²
= (25/16)*(xE − xD)²
⇒ DE = (5/4)*(xE − xD) = (5/4)*(xE − 0) = (5/4)*xE
EF²
= (xF − xE)² + (yF − yE)²
= (xF − xE)² + [(4/3)*(xF − xE)]²
= (xF − xE)² + (16/9)*(xF − xE)²
= (25/9)*(xF − xE)²
⇒ EF = (5/3)*(xF − xE) = (5/3)*[(16*xE + 1332)/25 − xE] = (5/3)*(16*xE + 1332 − 25*xE)/25 = (5/3)*(−9*xE + 1332)/25 = (−3*xE + 444)/5
Since DEFG is a square, we have DE=EF. So we can conclude:
(5/4)*xE = (−3*xE + 444)/5
25*xE = 4*(−3*xE + 444)
25*xE = −12*xE + 1776
37*xE = 1776
⇒ xE = 1776/37 = 48
⇒ DE = (5/4)*xE = (5/4)*48 = 60
∧ EF = (−3*xE + 444)/5 = (−3*48 + 444)/5 = (−144 + 444)/5 = 300/5 = 60 ✓
Now we are able to calculate the area of the yellow square:
A(DEFG) = DE² = EF² = 60² = 3600
Best regards from Germany