Great problem ! Another shortcut I would like to share is knowing the concept that says : " If a point inside a parallelogram is connected to its verticies creating 4 triangles , then the sum of the areas of any two opposite triangles is equal to half of that parallelogram " Applying it , we get that the area of the triangle ADC + area of the triangle FDE = half the are of the square ACEF After finding that x = 13 then applying the Pythagorean theorem, we get the area of the square is 845 , half of it is 422,5 The area of the triangle ADC = 169 , by symmetry. So the area of the triangle FDE = 422,5 - 169 = 253.5 Thanks for reading, have a nice time
1/ Note that the sums of the areas of the two opposite triangles in the square are equal (=half of the area of the square.) -> sqx= 169 sq AC= 5.(169) Area of the green triangle= 5/2. 169 -169 = 253.5 sq cm😊
Area of yellow triangle: A=½b.h= ½2x² = 169 cm² x = 13 cm Side of square: s²= b²+h² = (2x)²+x²= 5x² s = 29,0689 cm Height of yellow triangle: A = ½b.h = ½s.h h = 2A/s = 11,62755 cm Area of green triangle: A₁ = ½b.h₁= ½s.(s-h) A₁ = 253,5 cm² ( Solved √ )
My way of solution ▶ Let's calculate the area of the triangle: A(ΔABC)= 169 cm² 169= [AB]*[BC]/2 [AB]= 2x [BC]= x ⇒ 169= 2x²/2 169= x² x= √169 x= 13 cm [AC]²= [CD]²+[DA]² [AC]²= 4x²+x² [AC]= √5 x [AC]= 13√5 cm II) for the right triangle Δ(ACD): G ∈ [AC] [DG] ⊥ [AG] [DA]²= [AG]*[AC] [DA]=x [AG]= y [AC]= 13√5 ⇒ x²= y*13√5 the same equations for the [CD]: 4x²= (13√5-y)*13√5 if we divide these two equations into each other we get: 4x²= (13√5-y)*13√5 x²= y*13√5 ⇒ y= 13√5/5 cm According to the Pythagorean theorem x²= y²+h² 13²= (13√5/5)²+h² h²= 169 - 169/5 h= 26√5/5 ⇒ h₂= a-h h₂= 13√5 - 26√5/5 h₂= 39√5/5 A(ΔDEF)= a*h₂/2 A(ΔDEF)= 13√5*39√5/5/2 A(ΔDEF)= 2535/10 A(ΔDEF)= 253,5 cm²
Another very attractive problem thank you. I used a similar method except never calculated AC, just its square (845) which equates to the area of ACEF. I then concluded that the sum of areas of triangles ACD and DEF is half of the area of square ACEF (for any position of point D inside square ACEF). So the solution is (845/2)-169.
Area of right triangle=169cm^2 So 1/2(x)(2x)=169 So x=13 cm ACEF is a square AC=√13^2+26^2=13√5cm AC=CE=EF=FA=13√5 cm Connect G to H (G on DE and H on AC) In right triangle ACD 1/2(DH)(AC)=169 1/2(DH)(13√5)=169 So DH=26√5/5cm So DG=13√5-26√5/5=39√5/5cm Green shaded area=1/2(13√5)(39√5/5)=507/2 cm^2=253.5 cm^2.❤❤❤
We now area of yellow triangle, so we can find x and then sides. 2x•x/2=169 x²=169 x=13 13•2=26 So sides are 13 & 26, hypotenuse will be: √(13²+26²)=√845=13√5 Let's drop a height from opposite vertex of a rectangle to the hypotenuse and to the upper side of a square (that is also side of green triangle). Then we got segment made of two heights that is parallel and equal to the side of a square. Let height to the hypotenuse be h(1), and height of triangle h(2). We can find h(1) by formula: h(1)=13•26/13√5=26/√5=26√5/5 As the whole segment is equal to the side of square and is 13√5 (hypotenuse is also a side of square), h(2) will be: h(2)=13√5-26√5/5=(65√5-26√5)/5=39√5/5 Finally, we can find area of a triangle: S∆=13√5•(39√5/5)/2=507/2=253,5 Answer: 253,5 cm². Pin pls 🙏
Let's find the area: . .. ... .... ..... Since the triangle ABC is a right triangle, we can calculate the value of x as follows: A(ABC) = (1/2)*AB*BC 169cm² = (1/2)*(2*x)*x 169cm² = x² ⇒ x = √(169cm²) = 13cm Now we are able to calculate the side length s of the square ACEF by applying the Pythagorean theorem: s² = AC² = AB² + BC² = (2*x)² + x² = 4*x² + x² = 5*x² ⇒ s = √5*x = (13√5)cm A line through point D, which is parallel to AF and CE, may intersect AC at point G and EF at point H. Obviously we have DG+DH=GH=AF=s. The triangle ACD is obviously congruent to the triangle ABC and has therefore the same area. Since DG is the height of this triangle according to its base AC, we can conclude: A(ACD) = A(ABC) = (1/2)*AC*h(AC) = (1/2)*s*DG Now we are able to calculate the area of the green triangle: A(DEF) = (1/2)*EF*h(EF) = (1/2)*EF*DH = (1/2)*EF*(GH − DG) = (1/2)*s*(s − DG) = (1/2)*s² − (1/2)*s*DG = (1/2)*s² − A(ABC) = (1/2)*[(13√5)cm]² − 169cm² = (5/2)*(169cm)² − 169cm² = (3/2)*(169cm)² = 253.5cm² Best regards from Germany
An alternate solution to the last part is by "reverse" cross-multiplication. Since we know a+b= 169 (845/2)= 169+c+d 845= 338+2(c+d) 507= 2(c+d) 507/2= c+d 253.5 = c+d c+d= A∆(green) A∆(green)= 253.5 cm²
The first bit looks straightforward, but trickier thereafter. 2x^2 = 338 --> x^2 = 169 --> x = 13. 676 + 169 = 845, so AC = sqrt(845). ACEF area = 845. Make point G on the line AC, such that DGA is 90deg. DGA is similar to ABC. 13, 26, sqrt845) is similar to a, 2a, 13. Find DG. Call it x.: 13/(DG) = (sqrt(845))/26 338 = x*sqrt(845) 338/(sqrt(845)) = x (338/sqrt(845)) * (sqrt(845)/sqrt(845) (338*sqrt(845))/845 = x I prefer leaving decimal to the very end but... x = 11.62755348299891 (rounded). sqrt(845) - 11.62755348299891 = 17.44133022449836 sqrt(845) * 17.44133022449836 = 2* area of green triangle = 507 un^2. This was the exact answer displayed in the calculator - surprisingly, an integer. Green area: 507/2 = 253.5 un^2. Different method, but at least I got the right answer. If you think I made that difficult for myself, you should have seen my first attempt LOL.
Solving in terms of X, and using cartesian system [D(0,0)], I found Green Triangle is an isosceles triangle. EF=ED=Sqrt(5)•X FD=Sqrt(2)•X Height: Sqrt(4.5)•X Area=½Sqrt(2)•Sqrt(4.5)•X²
Yellow area = (1/2).x.(2.x) = x^2 = 169, so x = 13. Now AC^2 = x^2 + (2.x)^2 = 5.(x^2) = 5.169, so AC = 13.sqrt(5), it is the side length of the square ACEF. We us an orthonormal center A and first axis (AB). We have C(26; 13), VectorAC(26; 13), VectorCE(-13; 26), so E(13; 39), VectorEF(-26; -13), so F(-13; 26). Equation of (EF): (x +13).(1) - (y -26).(2) = 0 or x -2.y +65 = 0, Then distance from D to (EF) = abs(0 -2.13 + 65)/sqrt((1)^2 + (-2)^2) = 39/sqrt(5) The area of the green triangle is then: (1/2).EF.distance from D to (EF) = (1/2).(13.sqrt(5)).(39/sqrt(5)) = (13.39)/2 = 507/2.
Perfect. I love it challenging. My way : Area of triangle ADC=169 Area of triangle AFD=RAD.845×RAD.845×h/2 Area of triangle EDC= RAD.845×RAD.845-h/2 So , areas of ADC+EDC+DAF- 845= 235.5 THE AREA OF GREEN TRIANGLE
Let the side length of square ACEF be s. Triangle ∆ABC: [ABC] = bh/2 = AB(BC)/2 169 = 2x(x)/2 x² = 169 x = √169 = 13 AB² + BC² = CA² (2x)² + x² = s² s² = 4x² + x² = 5x² s = √(5x²) = √5x = 13√5 Draw MN, where M is the point on EF and N is the point on AC where MN is perpendicular to AC amd EF, parallel to FA and CE, and passes through D. As MN is perpendicular to AC amd EF, MN = CE = FA = s. Let MD be h, so DN = s-h = 13√5-h. As AC is the diagonal of rectangle ABCD, triangles ∆ABC and ∆CDA are congruent. Triangle ∆CDA: [CDA] = bh/2 = AC(MD)/2 169 = 13√5(13√5-h)/2 338 = 845 - 13√5h 13√5h = 845 - 338 = 507 h = 507/13√5 = 39/√5 Triangle ∆EFD: [EFD] = bh/2 = EF(MD)/2 [EFD] = 13√5(39/√5)/2 = 507/2 = 253.5 cm²
x(2x)/2 = 169 x = 13 diagonal of the square = side of the square = ✓(13^2 + 26^2) = 13✓(1 + 4) = 13✓5 the height of the left triangle = 13(13/13✓5) = 13/✓5 the height of the right triangle = 26(26/13✓5) = 52/✓5 green area = (13✓5)^2 - (13✓5)(13/✓5)/2 - (13✓5)(52/✓5)/2 - 169/2 = 5(13^2) - (13^2)/2 - 13(52)/2 - 169/2 = 5(13^2) - (13^2)/2 - 2(13^2) - (13^2)/2 = 2(13^2) = 338 cm^2
STEP-BY-STEP RESOLUTION PROPOSAL : 01) 169 * 2 = 2X * X ; 338 = 2X^2 ; X^2 = 338 / 2 ; X^2 = 169 ; X = sqrt(169) ; X = 13 02) AC^2 = X^2 + 4X^2 ; AC^2 = 5X^2 ; AC^2 = 5 * 169 ; AC^2 = 845 ; AC = sqrt(845) ; AC = 13sqrt(5) ; AC ~ 29,07 03) Big Square Side = 13sqrt(5) lin un 04) Big Square Area = 845 sq un 05) Green Shaded Region Area = (Big Square Area - Rectangle Area) / 2 06) GSRA = (845 - 338) / 2 07) GSRA = 507 / 2 sq un 08) GSRA ~ 253,5 sq un Therefore, OUR BEST ANSWER : The Green Shaded Region Area must be equal to 507/2 Square Units
Even though on Friday the 13th I walked under a ladder ...didn't knock on wood...broke a mirror, ...the curse wasn't enough too drain the energy of my passion and success from doing math. 🙂
Thanks Sir
That’s wonderful method of solution
Thanks for your efforts
Good luck
❤❤❤❤
So nice of you dear
You are very welcome!
Thanks for the feedback ❤️
Great problem !
Another shortcut I would like to share is knowing the concept that says : " If a point inside a parallelogram is connected to its verticies creating 4 triangles , then the sum of the areas of any two opposite triangles is equal to half of that parallelogram "
Applying it , we get that the area of the triangle ADC + area of the triangle FDE = half the are of the square ACEF
After finding that x = 13 then applying the Pythagorean theorem, we get the area of the square is 845 , half of it is 422,5
The area of the triangle ADC = 169 , by symmetry.
So the area of the triangle FDE = 422,5 - 169 = 253.5
Thanks for reading, have a nice time
Excellent!
Thanks for sharing ❤️
Thank you for sharing!
1/ Note that the sums of the areas of the two opposite triangles in the square are equal (=half of the area of the square.)
-> sqx= 169
sq AC= 5.(169)
Area of the green triangle= 5/2. 169 -169 = 253.5 sq cm😊
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Area of yellow triangle:
A=½b.h= ½2x² = 169 cm²
x = 13 cm
Side of square:
s²= b²+h² = (2x)²+x²= 5x²
s = 29,0689 cm
Height of yellow triangle:
A = ½b.h = ½s.h
h = 2A/s = 11,62755 cm
Area of green triangle:
A₁ = ½b.h₁= ½s.(s-h)
A₁ = 253,5 cm² ( Solved √ )
Excellent!
Thanks for sharing ❤️
My way of solution ▶
Let's calculate the area of the triangle:
A(ΔABC)= 169 cm²
169= [AB]*[BC]/2
[AB]= 2x
[BC]= x
⇒
169= 2x²/2
169= x²
x= √169
x= 13 cm
[AC]²= [CD]²+[DA]²
[AC]²= 4x²+x²
[AC]= √5 x
[AC]= 13√5 cm
II) for the right triangle Δ(ACD):
G ∈ [AC]
[DG] ⊥ [AG]
[DA]²= [AG]*[AC]
[DA]=x
[AG]= y
[AC]= 13√5
⇒
x²= y*13√5
the same equations for the [CD]:
4x²= (13√5-y)*13√5
if we divide these two equations into each other we get:
4x²= (13√5-y)*13√5
x²= y*13√5
⇒
y= 13√5/5 cm
According to the Pythagorean theorem
x²= y²+h²
13²= (13√5/5)²+h²
h²= 169 - 169/5
h= 26√5/5
⇒
h₂= a-h
h₂= 13√5 - 26√5/5
h₂= 39√5/5
A(ΔDEF)= a*h₂/2
A(ΔDEF)= 13√5*39√5/5/2
A(ΔDEF)= 2535/10
A(ΔDEF)= 253,5 cm²
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l=13√5...h=13√5-338/13√5=507/13√5=39/√5...Agreen=lh/2=13√5*39/√5/2=507/2
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Another very attractive problem thank you. I used a similar method except never calculated AC, just its square (845) which equates to the area of ACEF. I then concluded that the sum of areas of triangles ACD and DEF is half of the area of square ACEF (for any position of point D inside square ACEF). So the solution is (845/2)-169.
Very accurately my way of solving is similar to you.Area of opposite portions in a square are equal(845/2)🙏
Nice work!
You are very welcome!
Thanks for sharing ❤️
Thank you!
You are very welcome!
Thanks for the feedback ❤️
Area of right triangle=169cm^2
So 1/2(x)(2x)=169
So x=13 cm
ACEF is a square
AC=√13^2+26^2=13√5cm
AC=CE=EF=FA=13√5 cm
Connect G to H (G on DE and H on AC)
In right triangle ACD
1/2(DH)(AC)=169
1/2(DH)(13√5)=169
So DH=26√5/5cm
So DG=13√5-26√5/5=39√5/5cm
Green shaded area=1/2(13√5)(39√5/5)=507/2 cm^2=253.5 cm^2.❤❤❤
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Thanks for sharing ❤️
We now area of yellow triangle, so we can find x and then sides.
2x•x/2=169
x²=169
x=13
13•2=26
So sides are 13 & 26, hypotenuse will be:
√(13²+26²)=√845=13√5
Let's drop a height from opposite vertex of a rectangle to the hypotenuse and to the upper side of a square (that is also side of green triangle).
Then we got segment made of two heights that is parallel and equal to the side of a square.
Let height to the hypotenuse be h(1), and height of triangle h(2).
We can find h(1) by formula:
h(1)=13•26/13√5=26/√5=26√5/5
As the whole segment is equal to the side of square and is 13√5 (hypotenuse is also a side of square), h(2) will be:
h(2)=13√5-26√5/5=(65√5-26√5)/5=39√5/5
Finally, we can find area of a triangle:
S∆=13√5•(39√5/5)/2=507/2=253,5
Answer: 253,5 cm².
Pin pls 🙏
Pin pin 💪
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Thanks for sharing ❤️
Let's find the area:
.
..
...
....
.....
Since the triangle ABC is a right triangle, we can calculate the value of x as follows:
A(ABC) = (1/2)*AB*BC
169cm² = (1/2)*(2*x)*x
169cm² = x²
⇒ x = √(169cm²) = 13cm
Now we are able to calculate the side length s of the square ACEF by applying the Pythagorean theorem:
s² = AC² = AB² + BC² = (2*x)² + x² = 4*x² + x² = 5*x²
⇒ s = √5*x = (13√5)cm
A line through point D, which is parallel to AF and CE, may intersect AC at point G and EF at point H. Obviously we have DG+DH=GH=AF=s. The triangle ACD is obviously congruent to the triangle ABC and has therefore the same area. Since DG is the height of this triangle according to its base AC, we can conclude:
A(ACD) = A(ABC) = (1/2)*AC*h(AC) = (1/2)*s*DG
Now we are able to calculate the area of the green triangle:
A(DEF)
= (1/2)*EF*h(EF)
= (1/2)*EF*DH
= (1/2)*EF*(GH − DG)
= (1/2)*s*(s − DG)
= (1/2)*s² − (1/2)*s*DG
= (1/2)*s² − A(ABC)
= (1/2)*[(13√5)cm]² − 169cm²
= (5/2)*(169cm)² − 169cm²
= (3/2)*(169cm)²
= 253.5cm²
Best regards from Germany
Excellent!
Thanks for sharing ❤️
An alternate solution to the last part is by "reverse" cross-multiplication.
Since we know a+b= 169
(845/2)= 169+c+d
845= 338+2(c+d)
507= 2(c+d)
507/2= c+d
253.5 = c+d
c+d= A∆(green)
A∆(green)= 253.5 cm²
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АС = х√5. Height in ▲ADC = h = 2х/√5. Height in ▲FDE = H = х√5 - 2х/√5 = 3x/√5.
S(FDE)/S(ADC) = H/h = (3x/√5)/(2х/√5) = 3/2. S(FDE) = 169 × 1,5 = 253,5.
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The first bit looks straightforward, but trickier thereafter.
2x^2 = 338 --> x^2 = 169 --> x = 13.
676 + 169 = 845, so AC = sqrt(845).
ACEF area = 845.
Make point G on the line AC, such that DGA is 90deg.
DGA is similar to ABC.
13, 26, sqrt845) is similar to a, 2a, 13.
Find DG. Call it x.:
13/(DG) = (sqrt(845))/26
338 = x*sqrt(845)
338/(sqrt(845)) = x
(338/sqrt(845)) * (sqrt(845)/sqrt(845)
(338*sqrt(845))/845 = x
I prefer leaving decimal to the very end but...
x = 11.62755348299891 (rounded).
sqrt(845) - 11.62755348299891 = 17.44133022449836
sqrt(845) * 17.44133022449836 = 2* area of green triangle = 507 un^2. This was the exact answer displayed in the calculator - surprisingly, an integer.
Green area: 507/2 = 253.5 un^2.
Different method, but at least I got the right answer. If you think I made that difficult for myself, you should have seen my first attempt LOL.
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Thanks for sharing ❤️
2x*x/2=169 x^2=169 x=13
AC=√[13^2+26^2]=√845
ADC=169cm^2
square ACEF=√845*√845=845cm^2
area of Green shaded region :
845/2 - 169 = 422.5 - 169 = 253.5cm^2
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Solving in terms of X, and using cartesian system [D(0,0)], I found Green Triangle is an isosceles triangle.
EF=ED=Sqrt(5)•X
FD=Sqrt(2)•X
Height: Sqrt(4.5)•X
Area=½Sqrt(2)•Sqrt(4.5)•X²
Thanks for sharing ❤️
Aatec=845, Adec=338, Aadf=84,5, Aacd=169, Adec=845-338-169-84,5=253,5.
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Yellow area = (1/2).x.(2.x) = x^2 = 169, so x = 13. Now AC^2 = x^2 + (2.x)^2 = 5.(x^2) = 5.169, so AC = 13.sqrt(5), it is the side length of the square ACEF.
We us an orthonormal center A and first axis (AB). We have C(26; 13), VectorAC(26; 13), VectorCE(-13; 26), so E(13; 39), VectorEF(-26; -13), so F(-13; 26).
Equation of (EF): (x +13).(1) - (y -26).(2) = 0 or x -2.y +65 = 0, Then distance from D to (EF) = abs(0 -2.13 + 65)/sqrt((1)^2 + (-2)^2) = 39/sqrt(5)
The area of the green triangle is then: (1/2).EF.distance from D to (EF) = (1/2).(13.sqrt(5)).(39/sqrt(5)) = (13.39)/2 = 507/2.
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Perfect. I love it challenging.
My way :
Area of triangle ADC=169
Area of triangle AFD=RAD.845×RAD.845×h/2
Area of triangle EDC= RAD.845×RAD.845-h/2
So , areas of ADC+EDC+DAF- 845= 235.5 THE AREA OF GREEN TRIANGLE
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A(yellow) = 1/2 * 2x * x = x² = 169 cm²
A(square) = d² = 4x² + x² = 5x² = 5 * 169 cm² = 845 cm²
1/2 * A(square) = 422.5 cm² = 169 cm² + A(green)
A(green) = 422.5 cm² - 169 cm² = 253.5 cm²
The area is 507/2. That is definitely some clever geometry right there!!!
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Thanks for the feedback ❤️
Let the side length of square ACEF be s.
Triangle ∆ABC:
[ABC] = bh/2 = AB(BC)/2
169 = 2x(x)/2
x² = 169
x = √169 = 13
AB² + BC² = CA²
(2x)² + x² = s²
s² = 4x² + x² = 5x²
s = √(5x²) = √5x = 13√5
Draw MN, where M is the point on EF and N is the point on AC where MN is perpendicular to AC amd EF, parallel to FA and CE, and passes through D. As MN is perpendicular to AC amd EF, MN = CE = FA = s. Let MD be h, so DN = s-h = 13√5-h.
As AC is the diagonal of rectangle ABCD, triangles ∆ABC and ∆CDA are congruent.
Triangle ∆CDA:
[CDA] = bh/2 = AC(MD)/2
169 = 13√5(13√5-h)/2
338 = 845 - 13√5h
13√5h = 845 - 338 = 507
h = 507/13√5 = 39/√5
Triangle ∆EFD:
[EFD] = bh/2 = EF(MD)/2
[EFD] = 13√5(39/√5)/2 = 507/2 = 253.5 cm²
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x(2x)/2 = 169
x = 13
diagonal of the square = side of the square = ✓(13^2 + 26^2) = 13✓(1 + 4) = 13✓5
the height of the left triangle = 13(13/13✓5) = 13/✓5
the height of the right triangle = 26(26/13✓5) = 52/✓5
green area = (13✓5)^2 - (13✓5)(13/✓5)/2 - (13✓5)(52/✓5)/2 - 169/2 = 5(13^2) - (13^2)/2 - 13(52)/2 - 169/2 = 5(13^2) - (13^2)/2 - 2(13^2) - (13^2)/2 = 2(13^2) = 338 cm^2
253,5 cm^2
Thanks sir😊
STEP-BY-STEP RESOLUTION PROPOSAL :
01) 169 * 2 = 2X * X ; 338 = 2X^2 ; X^2 = 338 / 2 ; X^2 = 169 ; X = sqrt(169) ; X = 13
02) AC^2 = X^2 + 4X^2 ; AC^2 = 5X^2 ; AC^2 = 5 * 169 ; AC^2 = 845 ; AC = sqrt(845) ; AC = 13sqrt(5) ; AC ~ 29,07
03) Big Square Side = 13sqrt(5) lin un
04) Big Square Area = 845 sq un
05) Green Shaded Region Area = (Big Square Area - Rectangle Area) / 2
06) GSRA = (845 - 338) / 2
07) GSRA = 507 / 2 sq un
08) GSRA ~ 253,5 sq un
Therefore,
OUR BEST ANSWER :
The Green Shaded Region Area must be equal to 507/2 Square Units
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Much computation😢. x=13, AC^2=5×13^2, then the area is 1/2×5×13^2-169=3/2×169.😊
Thanks for the feedback ❤️
Even though on Friday the 13th I walked under a ladder ...didn't knock on wood...broke a mirror, ...the curse wasn't enough too drain the energy of my passion and success from doing math. 🙂
😅
😀😀
Thanks for sharing ❤️