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[@user-sk9oi9jl2g]

BF=a, BC=5√2. FC=√(a^2-50). DF=5√2-√(a^2-50). ΔEDF is isosceles. By the Pythagorean theorem a^2=2*(5√2-√( a^2-50))^2. We square both sides of the equation and solve the biquadrate equation. Choose a^2=400-200√3. S(ΔEBF)=0.5*(400-200√3)*√3/2=23.205. Approximate values are always difficult. It is better to round up at the end of the calculations.

[@user-sk9oi9jl2g]

@@rabotaakk-nw9nm СПАСИБО. ИСПРАВИЛА.

[@rabotaakk-nw9nm]

@@user-sk9oi9jl2g 👍🥰, умница!

[@brettgbarnes]

Draw a second diagonal line in the square Let a = side length of equilateral triangle Use the side proportionalities of 30-60-90 triangles (1:√3:2) and 45-45-90 triangles (1:1:√2) a(√3/2) + a/2 = 10 Area of equilateral triangle = a²√3/4

[@ptbx6986]

Neat!

[@timeonly1401]

That's how I did it! ❤ From your 1st equation, we solve for a: a(√3+1)/2 = 10 a = 20/(√3+1) Then we square to get a²: a² = [20/(√3+1)]² a² = 400/(4+2√3) And use the formula for area of equilateral triangle: A = (√3/4)a² A = (√3/4)[400/(4+2√3)] A = (100√3)/(4+2√3) A = (50√3)/(2+√3) A = (50√3)(2-√3) A = 100√3 - 150 A ≈ 23.2

[@gaylespencer6188]

Finding the length of the square's side is easy enough. Then you know that triangle EDF is an isosceles right triangle of 45-90-45, meaning that angle AEB is 75 degrees and angle ABE is 15 degrees. (5*2^.5) divided by cos15 = the hypotenuse which is also the length of each side of the equilateral triangle. Then use formula for area of an equilateral triangle to come up with 23.205...

[@RAG981]

We were taught, and I firmly believe, that the congruence you claim at 2:40 is RHS, for right angle, hypotenuse, side. Technically you do not have an angle other than the Right one to claim SAS, but once you know R, H and S the third sides must be the same also, defining congruence. Just saying.

[@ChuzzleFriends]

HL? Hypotenuse-Leg Theorem?

[@user-qd2bk5pc1n]

Yes, the initial congruence is by HL theorem, not SAS.

[@CloudBushyMath]

Superb!💡

[@alster724]

Another approach at the last part is using the standard formula for the area of the equilateral ∆ A= s²√3/4

[@phungpham1725]

1/ Let a be the side of the equilateral triangle. Draw the diagonal DB. Notice that DB is perpendicular bisector of EF and the triangle EDF is a right isosceles triangle. DB intersects EF at point H We have DB= DH+HB = a.sqrt3/2 + a/2 =10 -> a =20/(sqrt3+1) Area of the green triangle= sqa. (sqrt3)/4= 50sqrt3.(2-sqrt3) =23.20 sq units😅

[@redfinance3403]

yup, did it same way. it is simpler and doesn't require trigonometry!

[@hongningsuen1348]

It is difficult to prove directly DB is perpendicular bisector of EF. It is easy to prove triangle EDF is a right-angled isosceles triangle (ABE congruent to CBF hence AE = CE hence ED = FD). Diagonal DB bisects triangle EDF into 2 congruent triangles (RHS) hence EF is bisected by DB. DB = BH + HD(not HB)

[@phungpham1725]

@@hongningsuen1348 So sorry about my typo mistakes😂. 1/ The 2 right triangles ABE and BAF are congruent--> AE=CF-> DE=DF --> Triangle EDF is a right isosceles-EF//AC -> the diagonal BD is the perpendicular bisector of EF as well😅

[@hongningsuen1348]

@@phungpham1725 2nd thought. It is actually easy to prove directly diagonal BD is perpendicular bisector of triangle BEF. As triangles ABE and CBF are congruent, angles ABE and CBF = (90 - 60)/2 = 15. Perpendicular bisector of equilateral triangle BEF is also its angle bisector, bisecting 60 into 2 x 30. 15 + 30 = 45, the angle made by diagonal.

[@jamestalbott4499]

Thank you!

[@HuguesJacobs-qs7nr]

The is a simple and elegant way to proceed: Once you have noticed that BEF is symetric to AC, then you can mark O the center of EF. Let's consider EO = OF = x. By theorem, BO = x sqrt(3). By construction, EDF is rectangle isocele in O; DO = x. Therefore, AC = x(sqrt(3) +1) = 10 x = 5(sqrt(3) - 1) and area of EBF = x²sqrt(3) = 50(2sqrt(3)-3).

[@himo3485]

10/√2=5√2 EB=BF=FE=√2x ED=DF=x AE=FC=5√2-x (5√2-x)²+(5√2)²=(√2x)² 50-10√2x+x²+50=2x² x²+10√2x-100=0 (x+5√2)²-150=0 x+5√2=5√6 x=5√6-5√2=5√2(√3-1) Green Triangle area = √3/4*(√2x)² = √3x²/2 = √3*50*(4-2√3)/2 = 25√3*(4-2√3) = 100√3 - 150

[@AmirgabYT2185]

S=50(2√3-3)≈23,2

[@srirajan1933]

2 other methods to compute area after obtaining equilateral triangle side length = 7.32: 1. Heron’s Formula, since we know all 3 side lengths. 2. Drop perpendicular from corner B to bisect EF into 7.32/2 = 3.66. Use Pythagorean theorem to obtain height of triangle = 6.34. Base EF = 7.32, with height = 6.34. So, 1/2 * 6.34 * 7.32 = 23.2, as shown above using 1/2 * EF * BF * sin(60). Superb fun, Prof. thank you!!

[@wackojacko3962]

Anytime the world looks unfamiliar one can rely on the rigid constructed rules of math that gives potential too solving problems such as this. 🙂

[@giuseppemalaguti435]

Posto AE=a,risulta a^2+(5√2)^2=l^2..arccos(a/l)+60+arccos((5√2-a)/l)=180...dopo i calcoli risulta a=5√2(2-√3)...l^2=50+50(7-4√3)=400-200√3...Agreen=(1/2)l^2sin60=(200-100√3)√3/2=. =100√3-150=50(2√3-3)

[@allanflippin2453]

It can also be solved without resorting to looking up the cosine of 15 degrees. Looking at the angles, it's clear that EDF is an isosceles triangle. If EF is x, then ED = DF = x / sqrt(2). We can use Pythagorus to find AE^2 = x^2 - 50. Also AE = 5 * sqrt(2) - ED. Substituting for ED we find that AE = (10 - x) / sqrt(2). Plug the AE value into the pythagorean formula for ABE: (10-x)^2/2 + 50 = x^2. Rearranging, we get the quadratic equation x^2 + 20x - 200 = 0. Solving that: x = (-20 +/- sqrt(300))/2. To avoid a negative result, we have to choose "+" from the "+/-". Ultimately, x = 10 * (sqrt(3) - 1) or approximately 7.32. With side length x, the area of an equilateral triangle is x^2 * sqrt(3) / 4. Plugging in x, I get the triangle area is 50 * (2 * sqrt(3) - 3) which is about 23.2.

[@ChuzzleFriends]

This is pretty much similar to an older question on this channel, except we now know the diagonal length instead of the area. Nevertheless, it's still easy. d = s√2 10 = s√2 s = 10/(√2) = (10√2)/2 = 5√2 By definition of squares, ∠BAD, ∠BCD, & ∠D are right angles and AB = AC. So, because BE = EF & △BAE & △BCF are right triangles, △BAE ≅ △BCF by HL. So, AE = CF by CPCTC. Label AE = CF = x. Then DE = EF = 5√2 - x. Use the Pythagorean Theorem on △BAE & △EDF. a² + b² = c² (5√2)² + x² = c² 50 + x² = c² (5√2 - x)² + (5√2 - x)² = c² (50 - 10x√2 + x²) + (50 - 10x√2 + x²) = c² 100 - 20x√2 + 2x² = c² 2x² - 20x√2 + 100 = x² + 50 x² - 20x√2 + 50 = 0 Use the Quadratic Formula. a = 1, b = -20√2, c = 50 x = [-b ± √(b² - 4ac)]/2a {-(-20√2) ± √[(-20√2)² - (4 * 1 * 50)]}/(2 * 1) [20√2 ± √(800 - 200)]/2 (20√2 ± √600)/2 (20√2 ± 10√6)/2 x = (20√2 + 10√6)/2 or x = (20√2 - 10√6)/2 = 10√2 + 5√6 = 10√2 - 5√6 ≈ 26.39 ≈ 1.89 s ≈ 7.07 And x represents the length of a segment shorter than the side of square ABCD. So, x = 10√2 - 5√6. x² + 50 = c² (10√2 - 5√6)² + 50 = c² (200 - 200√3 + 150) + 50 = c² 200 - 200√3 + 200 = c² 400 - 200√3 = c² The hypotenuses are sides of △BEF, which is equilateral, so we can use a different area formula: A = (√3)/4 * c² = [√3 * (400 - 200√3)]/4 = (400√3 - 600)/4 = 100√3 - 150 So, the area of the green triangle is 100√3 - 150 square units, a. w. a. 50(2√3 - 3) square units (exact), or about 23.21 square units (approximation). The video doesn't show the exact answer, so it's right here.

[@almosawymehdi3416]

a better answer for the lenght of the green triangle is 10*sqrt(3) -10, then the area of the triangle is 100*sqrt(3) -150 square units. Have a good day.

[@marcgriselhubert3915]

*Side length of the square: 10/sqrt(2). *Side length of the equilateral triangle: c. *DF = c/sqrt(2) in right isosceles triangle EDF, then FC = DC - DF = (10/sqrt(2)) - (c/sqrt(2)) = (10 -c)/sqrt(2) *In triangle FCB: FC^2 + CB^2 = FB^2, so we have: ((10 -c)^2)/2 + 50 = c^2, or ((c^2)/2) -10.c +50 + 50 = c^2 or ((c^2)/2) + 10.c -100 = 0, or c^2 + 20.c -200 = 0 Deltaprime = 300, then c = -10 -10.sqrt(3), which is rejected as negative, or c = -10 +10.sqrt(3). *The area of the equilateral triangle is (sqrt(3)/4).(c^2) so it is ((sqrt(3)/4).100.(4 -2.sqrt(3)) = 25.sqrt(3).(4 -2sqrt(3)) = 100.sqrt(3) - 150. (which is about 23.2)

[@juanalfaro7522]

[BEF] - x^2 * sqrt (3)/4 where x is the side of the equilateral triangle. EF parallel to AC -> DEF=DFE = 45 -> BEA = 180-45-60=75 -> ABE=15 -> CBF = 90-15-60=15 -> CBF = 75. Now AC=10 -> AB= 5*sqrt (2) = BC=CD=AD. Now x = AB/cos (15) = 10 * [sqrt (3) -1] --> [BEF] = 10^2 * [sqrt (3) -1] ^2 * SQRT (3)/4 = 100* [4-2*sqrt (3)] * sqrt (3)/4 = [400*sqrt (3) - 600]/4 = 100*sqrt (3) - 150 = 173.2 - 150 = 23.2 sq. units

[@toninhorosa4849]

Great solution teacher. I solved a little different. AB=BC=CD=AD 😢= a AC = 10 BE=BF=EF = b ∆ABC applying Pythagoras: AC^2 = AB^2 + BC^2 10^2 = a^2 + a^2 2a^2 = 100 a^2 = 50 a = √(25*2) a = 5√2 I did ==>AE = CF = x ∆DEF applying Pythagoras EF^2 = DE^2 +DF^2 b^2 =(5√2-x)^2 + (5√2-x)^2 ∆ABE applying Pythagoras b^2 = x^2 + (5√2)^2 b^2 = x^2 + 50 Equating the two b^2 we get: x^2 + 50 = (5√2 - x)^2 + (5√2 - x)^2. x^2 + 50 = 50 - 10√2x + x^2 + 50 - 10√2x + x^2 x^2 - 20√2x + 50 = 0 x = (20√2 +-√(800- 200))/2 x = (20√2 +- 10√6)/2 x1 = 10√2 + 5√6 =26,38958 Rejected It is > than 5√2 = a x =10√2 - 5√6 = 1,8946869 Accepted✓ ∆ABE applying Pythagoras: b^2 = (5√2)^2 + (10√2-5√6)^2 b^2 = 50 + 350 - 200√3 b^2 = (400 - 200√3) Formula for the area of an equilateral triangle. A = b^2 * (√3)/4 A = (400 - 200√3)*(√3)/4 A = (400√3 - 600)/4 A = 100√3 - 150 ∆BEF area = 23,20508 unit^2 x

[@christianaxel9719]

From 5:02 : Recalling cos(15º)=(√6+√2)/4, then 1/cos(15º)=4/(√6+√2), rationalizing: 1/cos(15º)=4/(√6+√2)=4(√6-√2)/(6-2)=4(√6-√2)/4=√6-√2=√2(√3-1); then EB=5√2(√2)(√3-1)=10(√3-1). Finally Green Area=(√3/4)EB²=(√3/4)100(4-2√3)=25√3(4-2√3)=50√3(2-√3)=50(2√3-3)=23.2050807568...

[@ManojkantSamal]

*=read as square root AB=BC=CD=AD =AC/*2=10/*2=5.*2 The sides of the triangle intersects at the points P, Q, respectively to the sides AD, CD,of the square Now within triangle ABP & BQC AB=BC BP=CQ Angle BAP=Angle BCQ SO triangle ABP~BQC So, Angle ABP=angle CBQ=15(Angle PBQ=60,given ) Now, Angle APB=75degree In triangleAPB Sin75=AB/BP (*3+1)/2.*2= 5.*2/BP BP(*3+1)=5.*2.2.*2 =20 BP=20/(*3+1)=10(*3-1) Area of triangle =*3/4 ×{10(*3_1)}^2 (^=read as square ) =25×*3(*3_1)^2 =25.*3{4-2.*3) =100.*3-(50.*3.*3) =(100×1.732)-150 =173.2-150 =23.2(approximately )

[@nandisaand5287]

I got there without using trig, but a whole Lotta algebra: I call the bisected square side segments "X" and "5Sqrt(2)-X", and Triangle sides "S". This sets up 2 Pythagorean Theorem eqns: X²+[5•SQRT(2)]²=S² 2•(5•SQRT(2)-X)²=S² Set equal to each other, solve for X, X=1.89. Plug back into eqn 1: (1.89)²+(5•SQRT(2))²=S² S=7.31 Since triangle is isosceles: Area=1/2•S•(S/2)SQRT(3) =S²/4•SQRT(3) =(7.31)²/4•SQRT(3) =23.14

[@MrPaulc222]

DF is (10-h)*sqrt(2) EF is 20-2h Triangle area is h(10-h) M is the midpoint of EF EMB is a 30,60,90 triangle with sides (10-h), (10-h)*sqrt(3), and 20-2h Observe that h = (10-h)*sqrt(3) h^2 = 3(100 - 20h + h^2) 0 = 300 - 60h + 3h^2 - h^2 2h^2 - 60h + 300 = 0 h^2 - 30h + 150 = 0 (30+or-sqrt(900 - 4*1*150))/2 = h (30+or-sqrt(300))/2 = h (30+or-2*sqrt(75))/2 = h (15+or-sqrt(75))= h Strip out the positive result as it is >10 (the square's diagonal) h = 15 - sqrt(75) Area, as already noted, is h(10-h). so 10h - h^2 150 - 10*sqrt(75) - (15-sqrt(75))^2 ---> 150 - 10*sqrt(75) - (225 - 30*sqrt(75) + 75) ---> 150 - 10*sqrt(75) - 225 + 30*sqrt(75) - 75 ---> -150 + 20*sqrt(75) 20*sqrt(75) - 150 100*sqrt(3) - 150 173.21 - 150 Area = approx 23.21 un^2 I have now checked your video and see that you followed a different path from me. Although my noting that h = (10-h)*sqrt(3) led to some moderately fiddly calculations, I found that way a bit easier than yours - but I suppose we are all wired a bit differently. Thanks once again.

[@user-yx9kr8ur5q]

The square ABCD has a diagonal of 10 so each side is 10/sqrt(2). Let the triangle BEF have side lengths of "a" and let AE = x then ED = (10/sqrt(2) - x ) and also by symmetry around the diagonal BD we have DF = ED = (10/sqrt(2) - x ). Applying Pythagoras theorem to triangle ABE and then to triangle EDF we get: a^2 = x^2 + (10/sqrt(2))^2 and a^2 = (10/sqrt(2) -x)^2 +(10/sqrt(2) -x)^2. We can solve these equations to get x = 1.895 and a^2 = 53.59. We can find the area of the equilateral triangle BEF using the formula (sqrt(3)/4)*a^2 to be 23.20 square units.

[@devondevon4366]

23.21 let the side = a then a^2 + a^2 =100 2a^2=100 a^2 =50 a= sqrt 50 Since the green triangle is equilateral, then the triangle top is a 75, 15, 90 right triangle ( 90 -60=30/2 =15) To find the length of green, using the law of sine Let the length of the green = p, then p = sqrt 50 * sine 90 degrees/ sine 75 degrees p = sqrt 50*1.035276 p=7.32050807569 So the length of the equilateral is 7.32051 Hence, its area = 7.32051^2 * sqrt 3/4 = 23.21

[@JamesDavy2009]

The exact value of the area is (25/2)√3 sec²(15°)

[@lukeheatley4148]

Using triangles EDF and BCF, i worked out the side of the green triangle to be Sqrt[200 . (2 - Sqrt[3])] .... (or 7.32) height of equilateral triangle is easy using pythagorus on half a base and the side. area works out to be 50 * (2 * sqrt (3) - 3)

[@balalakshmiar9456]

we cant say ABE & CBF are congruent using SAS .

[@quigonkenny]

Let s be the side length of square ABCD and t be the side length of equilateral triangle ∆EBF. Triangle ∆ABC: AB² + BC² = CA² s² + s² = 10² = 100 s² = 100/2 = 50 s = √50 = 5√2 As EB = BF and AB = BC and ∠EAB = ∠BCF = 90°, then ∆EAB and ∆BCF are congruent right triangles. Let ∠ABE = ∠FBC = θ. As ∠EBF = 60° (being the internal angle of an equilateral triangle) and ∠ABC = 90° (being the internal angle of a square), then: θ + 60° + θ = 90° 2θ = 90° - 60° = 30° θ = 15° cos(15°) = cos(60°-45°) cos(15°) = cos(60°)cos(45°) + sin(60°)sin(45°) AB/BE = (1/2)(1/√2) + (√3/2)(1/√2) s/t = (√3+1)/2√2 t = 2√2s/(√3+1) t = 2√2s(√3-1)/(√3+1)(√3-1) t = 2s(√6-√2)/(3-1) t = 2s(√6-√2)/2 = s(√6-√2) t = 5√2(√6-√2) t = 5√2(√2(√3-1)) = 10(√3-1) Triangle ∆EBF: Aₜ = t²sin(60°)/2 Aₜ = (10(√3-1))²(√3/4) Aₜ = (100√3/4)(3-2√3+1) Aₜ = 25√3(4-2√3) = 25(4√3-6) Aₜ = 50(2√3-3) ≈ 23.205 units

[@nenetstree914]

50*(2*sqrt(3)-3)

[@kettlebellBob]

You've used SAS to prove triangles ABE & CBF as congruent. Then use congruent triangles to get angles ABE & CBF congruent. These are the included angles in SAS. You can't say the angles are congruent to prove the triangles congruent and then deduce the angles are congruent. Using hypotenuse-leg (HL) resolves this issue.

[@quigonkenny]

To be fair, hypotenuse-leg is basically SAS by a different name. The fact that ∆ABE and ∆CBF are both right triangles tells us that the cosine (and thus angle) of ∠ABE and ∠CBF are the same, as we know they're both equal to square side length (leg) over triangle side length (hypotenuse). It's why hypotenuse-leg works (and why it only works on right triangles).

[@PrithwirajSen-nj6qq]

Sir if we agree to ur we may also say it is a case of SSS/SAS congruency. As using Pythagorean theorem we may say AE = CF etc. But when we say of SAS congruency we must have to show the angle between the sides are equal. Here it was not done.

[@phungpham1725]

@@PrithwirajSen-nj6qq Honestly , I think it s a sss congruency. No need to calculate the angle. The 2 triangles are right angled ones and they have 2 long legs (= side of the square) and 2 hypothenuses (= side of the equilateral) so, the short legs must be equal. (Using Pythagorean theorem as proof)

[@prossvay8744]

ABCD is square Let side length=a So AB=BC=CD=DA AB^2+BC^2=AC^2 a^2+a^2=10^2 So 2a^2=100 So a=5√2 ∆ABE ~~∆BCF So AE=CF=x So DE=DF=5√2-x BR=BF=EF=b In ∆DEF DE^2+DF^2=EF^2 2(5√2-x)^2=b^2 100-20√2x+x^2=b^2 (1) In ABE x^2+(5√2)^2=b^2 x^2+50=b^2 (2) (1) and (2) 100-20√2x+x^2=x^2+50 So x=1.89 (2) (1.89)^2+50=b^2 So b=7.32 So Green triangle area=1/2(7.32)^2sin(60°)=23.2 square units.❤❤❤

[@devondevon4366]

Once you find out that the triangle on the top is 75, 90, 15, after subtracting 60 from 90, then dividing 30 by 2, then the most challenging part is over.

[@tellerhwang364]

1.△AEB:△CFB:△EDF:△BEF =1:1:2:(2sqrt3) 😊 2.ABCD=10·10/2=50 △AEB=50/(4+2sqrt3) =50-25sqrt3 3.△BEF=(50-25sqrt3)(2sqrt3) =100sqrt3-150😊

[@PrithwirajSen-nj6qq]

Sir u said triangles ABE and BCF are congruent over SAS theory. But their two sides are equal but the included angles are not proved equal. So SAS is not applicable. We may say these triangles are right angled triangle. Hence they will be congruent if the hypotenuse equal and any side is equal . Here it is seen that the hypotenuses are equal and one side of each them is equal. Hence the triangles are congruent over RHS congruent theory Then angle ABE = angle BCF may be said. Comment please

[@rabotaakk-nw9nm]

[ABCD]=AB²=½AC² => AB²=100/2=50 BE=AB/cos15°; BE²=AB²/cos²15°= =AB²/(1+cos30°)/2=50/(1+vʼ3/2)/2= =200/(2+vʼ3)=200(2-vʼ3) [BEF]=(vʼ3/4)BE²=(vʼ3/4)200(2-vʼ3)= =50(2vʼ3-3)≈23.205 😁

[@stvcia]

Exact result for cos 15° is (1 + sqrt(3))/(2 sqrt(2))

[@Mediterranean81]

Yes

[@jimlocke9320]

The 15°-75°-90° right triangle appears so frequently in geometry problems that it is worthwhile for all of us to know its properties. The ratio of sides (short):(long):(hypotenuse) is (1 - √3):(1 + √3):(2√2), from which cos(15°) can be readily derived to be (1 + √3)/(2√2), as stvcia points out.

[@Mediterranean81]

@@jimlocke9320 exactly

[@Lemda_gtr]

Why angles 15 deg r same?🧐

[@korogonianos]

Has any one Solved this without using trigonometry ?

[@misterenter-iz7rz]

Reluctant to do for trigonometric means cannot yield simple answeras usual, but I only can work in this way. First s^2=100/2=50, the answer is s^2-s^2 tan 15°-s^2(1-tan 15°)^2=50(1-tan 15-1-tan^2 15°+2tan 15°)=50(1+tan 15°+tan^2 15)😢😢😢😢😢😢😢😢.

[@JamesDavy2009]

I calculated the area as (25/2)√3 sec²(15°)

[@misterenter-iz7rz]

@JamesDavy2009 very nice 😁

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... From the given length of the diagonal we can directly calculate the side length s and the area A of the square: s = AC/√2 A = s² = AC²/2 = 10²/2 = 50 The triangles ABE and BCF are both right triangles. Additionally we have AB=BC=s and BE=BF=t (with t being the side length of the equilateral triangle). Therefore these two triangles are congruent and we can conclude: AE = CF AD − DE = CD − DF s − DE = s − DF ⇒ DE = DF Therefore the triangle DEF is an isosceles right triangle. By applying the Pythagorean theorem to the right triangles DEF and ABE we obtain: EF² = DE² + DF² t² = 2*DE² t²/2 = DE² ⇒ DE = t/√2 ⇒ AE = AD − DE = s − t/√2 BE² = AB² + AE² t² = s² + (s − t/√2)² t² = s² + s² − √2*s*t + t²/2 t²/2 + √2*s*t − 2*s² = 0 t² + (2√2)*s*t − 4*s² = 0 t = −√2*s ± √(2*s² + 4*s²) = −√2*s ± √(6*s²) = −√2*s ± √6*s Since t>0, the only useful solution is t=(√6−√2)*s. Now we are able to calculate the area of the green equilateral triangle: A(BEF) = (√3/4)*t² = (√3/4)*(√6 − √2)²*s² = (√3/4)*(6 − 2√12 + 2)*50 = (√3/4)*(8 − 4√3)*50 = √3*(2 − √3)*50 = 100√3 − 150 ≈ 23.21 Best regards from Germany

[@devondevon4366]

nice

[@unknownidentity2846]

@@devondevon4366 Thanks a lot for your kind feedback and best regards from Germany.

[@inyomansetiasa]

First, you can pin it?