Can you find area of the Blue Square? | (Fun Geometry Problem) |

[@parthtomar6987]

Very nice and easy solution sir🎉🎉🎉

[@PreMath]

Glad to hear that! Thanks for the feedback ❤️

[@devondevon4366]

1❤2​@@PreMath

[@devondevon4366]

1❤​@@PreMath

[@PreMath]

@@devondevon4366 Thanks dear 🌹

[@jimlocke9320]

At 2:40 extend CF up and drop a perpendicular to it from A. Label the intersection as point G. Note that AEFG is a rectangle, so FG = AE = 2 and AG = EF = 4. Construct AC, which is a diagonal of the square. Consider right triangle ΔAGC. AG = 4 and CG = CF + FG = 6 + 2 = 8. By the Pythagorean theorem, AC² = 4² + 8² = 16 + 64 = 80 and AC = √(80). Side length of square = diagonal divided by √2, so x = √(80)/√2 = √(40). Area of square = x² = (√(40))² = 40 cm², as PreMath also found. Alternatively, once you have the diagonal of the square, you can square it and divide by 2 to get the area, saving a step.

[@batavuskoga]

I also solved it this way.

[@sergioaiex3966]

Very good explanation !

[@JohnPeter-j1s]

Awesome

[@geoffreynorth9046]

Can you not get the length AC more easily just by sliding EF to the endpoint so you have a right triange with short length 4 and long length 8, giving the hypotenuse as 4 root 5...?

[@unknownidentity2846]

To say it with the famous words of Barack Obama: Yes we can.🙂 That was exactly the way I did it. Best regards from Germany

[@phungpham1725]

1/ Extend AE a segment EM such that EM= FC= 6 . We have EFCM a rectangle So sq AC= sq8+sq 4= 80-> AC= 4 sqrt5 The side of the square= a a .sqrt2 =AC= 4sqr5- > a= 4x (sqrt5/sqrt2) Area of the square= 16x(5/2)= 40 sq cm

[@quigonkenny]

Or just extend AE and CF to make a rectangle of width EF and height AE+CF, if "sliding" EF seems unorthodox. Given the inherent parallels and perpendiculars, either way you end up with a right triangle of dimensions 4×8 from which you can determine AC.

[@Waldlaeufer70]

@@unknownidentity2846 Me too.

[@Qwentar]

​@@someonespadre It doesn't have to be impressive. It needs to be easily understood.

[@robertlynch7520]

My dear professor PreMath … that was WAY too much work! The line EF merely needs another parallel line which is drawn from point A of length 4, then down to F. Those two then form a rectangle. It is becomes visually obvious that the overall diagonal length of the blue square is [1.1]  hypotenuse = √( (6 ⊕ 2)² ⊕ 4² ) [1.2]  hypotenuse = √( 64 + 16 ) [1.3]  hypotenuse = √( 80 ) Having that, and the generalized area-of-a-square-from-its-hypotenuse of [2.1]  area square = hypotenuse² ÷ 2 [2.2]  area square = √(80)² ÷ 2 [2.3]  area square = 80 ÷ 2 [2.4]  area square = 40 So… that's it! The area of a square, derived solely from its hypotenuse can be 'remembered' by the 1-1 square having √(2) as a hypotenuse. And a 1×1 square has area 1, and hypotenuse² of √(2)² is 2, which is 2× oversized. Thus 'divide by 2'. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

[@PreMath]

Great! You are awesome. Thanks for the feedback ❤️

[@santiagoarosam430]

EF=2*12/6=4; AE=2*4/4=2; Desplazamos paralelamente a sí mismo EF hasta que E coincida con A y se obtiene el triángulo rectángulo AFC cuya hipotenusa es la diagonal del cuadrado: 4^2 +(6+2)^2=d^2=80. Área del cuadrado =80/2=40 Gracias y saludos.

[@marcelowanderleycorreia8876]

I did this way as well.

[@jamestalbott4499]

Thank you!

[@chidinduuwaeziozi7360]

Great!!! Thank you! What tool do you use for annotations?

[@ChuzzleFriends]

Find the side length of square ABCD before finding the area. A = (bh)/2 12 = [6(EF)]/2 12 = 3(EF) EF = 4 4 = [4(AE)]/2 4 = 2(AE) AE = 2 Draw a point G outside square ABCD such that AEFG is a rectangle. So, AG = EF = 4 cm & FG = AE = 2 cm. So, CG = CF + FG = 6 + 2 = 8 cm. Draw diagonal AC. This is the hypotenuse of right △AGC (because AEFG is a rectangle, so ∠G is a right angle). Use the Pythagorean Theorem on △AGC. a² + b² = c² 4² + 8² = (AC)² 16 + 64 = (AC)² (AC)² = 80 AC = √80 = (√16)(√5) = 4√5 Diagonals of a square create two congruent isosceles right triangles. d = s√2 4√5 = s√2 s = (4√5)/(√2) = [(4√5)(√2)]/2 = (4√10)/2 = 2√10 Now find the area of square ABCD. A = s² = (2√10)² = 40 So, the area of the blue square is 40 square centimeters.

[@alster724]

After obtaining 4√5=x√2 I simply squared both sides... 16(5)= 2x² 80=2x² 40=x² And since I know that A= s² where s is a side of the square represented by x in this video, I stopped since I got the final answer ahead.

[@murdock5537]

This is awesome, many thanks, Sir! φ = 30°; ∎ABCD → AB = BC = CD = AD = a; AC = a√2; ∆ AEF → EFA = θ; AE = k; EF = n → kn/2 = 4 → n = 8/k; sin⁡(AEF) = sin⁡(3φ) = 1 ∆ CFE → ECF = δ; EF = n; CF = 6 → 3n = 12 → n = 4 = 8/k → k = 2 ∆ AEF → AF = 2√5; ∆ ECF → CE = 2√13 sin⁡(θ) = √5/5 → cos⁡(θ) = √(1 - sin^2(θ)) = 2√5/5; CFE = 3φ → sin⁡(3φ) = 1 → cos⁡(3φ) = 0 CFA = 3φ + θ → cos⁡(3φ + θ) = cos⁡(3φ)cos⁡(θ) - sin⁡(3φ)sin⁡(θ) = -sin⁡⁡(θ) = -√5/5 → ∆ ACF → (a√2)^2 = 20 + 36 - 2(2√5)6(-√5/5) = 80 → a^2 = 40

[@marcgriselhubert3915]

When EF = 4 and EA = 2 are (very easily) found, we use an orthonormal center E and first axis (EF). We have then A(0;2) and C(4; -6) and VectorAC(4;-8) Then AC^2 = 16 + 64 = 80, this is the square of the length of the diagonal of ABCD, so it is double of its area, so the area of ABCD is 80/2 = 40. Very quick!

[@Waldlaeufer70]

EF = 2 * 12 cm² / 6 cm = 4 cm AE = 2 * 4 cm² / EF = 8 cm² / 4 cm = 2 cm d²(square) = EF² + (AE + CF)² = 4² + (2 + 6)² = 16 + 64 = 80 cm² A(square) = 1/2 d² = 1/2 * 80 cm² = 40 cm²

[@spafon7799]

Easy way to solve: Extend segment AE down from E by 6 units. Call the end of this new segment G. Note that EFCG forms a rectangle that's 4x6. Now look at triangle AGC. It is a right triangle with the sides adacent to the right angle being AG=8 and GC=4. The third side AC of AGC happens to be the diagonal of the square, with length sqrt(2)*S. Applying pythagorean, we get 8^2+4^2=2*S^2. Thus 80=2*S^2 and 40=S^2. S^2 is the area of the square that was to be solved for, so the answer is 40 (cm^2).

[@MorgKev]

Create a rectangle with sides AE extended and CF extended. A&C are opposite vertices of the rectangle. Sides will be 8 and 4. The rectangle’s diagonal is the square’s diagonal. Result follows easily.

[@amazingjamaf]

What whiteboard app are you using to teach math. I want to also teach maths

[@AmirgabYT2185]

S=40 cm²

[@sergeyvinns931]

EF=4, AE=2. Drav the diagonal AC of a square, thrjugh point G on side EF. EG/GF=2/6. EG=1, GF=3. AG=\/5, GF=3\/5. АС=4\/5. Area of the Blue Square = AC^2/2= (4\/5)^2/2 = 40!

[@dugferd2266]

Instead of using Pythagorean formula the second time, because the triangles are similar wouldn't it be easy just to apply the proportion of the triangles (3x) to arrive at the 3√5 for the bottom triangle's portion of the diagonal of the square? 1,2, √5 to 3,6, 3√5

[@prossvay8744]

In ∆CEF 1/2(EF)(CF)=12 EF=24/CF=24/6=4cm In∆ AEF 1/2(AE)(EF)=4 AE=8/EF=8/4=2cm Connect.F to B Let BF=a and side of the square is x In ∆ ABE AB^2=AE^2+BE^2 BE=a+4 ; AB=x x^2=4+(a+4)^2 x^2=4+a^2+8a+16 x^2-a^2=8a+20 (1) in ∆ BCF BC^2=BF^3+CF^2 BF=a ; BC=x ; CF=6 x^2=a^2+6^2 x^2-a^2=36 (2) Compare (1) and (2) 8a+20=36 So a=2cm (2): x^2-2^2=36 So x^2=40cm^2 Blue square area=40cm^2.❤❤❤ Thanks sir. Best regards.

[@quigonkenny]

Triangle ∆CFE: A = bh/2 = EF(CF)/2 12 = EF(6)/2 = 3EF EF = 12/3 = 4 Triangle ∆AEF: A = bh/2 = EF(AE)/2 4 = 4AE/2 = 2AE AE = 4/2 = 2 Given that the goal is to find the area of blue square ABCD, either the side length or diagonal of the square needs ro be found. As AE and CF are both perpendicular to EF and thus parallel, if we extend AE and CF and draw perpendiculars from A (perpendicular to AE) and C (perpendicular ro CF) to meet those extensions at G and H respectively, we have a rectangle AGCH of width EF = AG = CH = 4 and height AE+EH = CF+FG = 8 that shares the same diagonal d with ABCD along AC. Triangle ∆AHC: HC² + AH² = AC² 4² + 8² = d² d² = 16 + 64 = 80 d = √80 = 4√5 Square ABCD: A = d²/2 = (4√5)²/2 = 80/2 = 40 cm²

[@cyrillelaurent8012]

Pourquoi vous n'utilisez pas la formule pour calculer la surface d'un losanges pour calculer la surface du carré : produit des diagonales divisé par 2. Soit AC²÷2

[@adogonasidecar1262]

How do we know that the blue square is a square? Not been demonstrated

[@StuartSimon]

It’s given in the question.

[@adogonasidecar1262]

@@StuartSimon and what says it actually exists? With such an overly meticulous presenter, it's a big miss. Now, it's not that difficult to demonstrate there is such a square, but it's needed

[@Waldlaeufer70]

AC has a certain length due to the measurements of the two yellow triangles. No matter how long AC is, its length can always be the diagonal of a square (whose area we are looking for).

[@adogonasidecar1262]

@@Waldlaeufer70 agreed but 1) needs to be stated and 2) the chart represents all points inscribed in the square; will that always happen? Does it matter (no, but needs to be discussed). It's sloppy not to examine that aspect. And then spend minutes dividing by 2...

[@nenetstree914]

40

[@RobertHorvat]

You are complicating alot. If you make rectangel from points AEFA', you get one line CA'=6+2=8 and second line A'A=4. Then 8*8+4*4=80. This is diagonal squared. So, x*x+x*x=80 is 2*x*x=80 is x*x=40. And this is your anwser

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... Since AEF and CEF are right triangles, we can easily calculate the areas: A(CEF) = (1/2)*CF*EF ⇒ EF = 2*A(CEF)/CF = 2*(12cm²)/(6cm) = 4cm A(AEF) = (1/2)*AE*EF ⇒ AE = 2*A(AEF)/EF = 2*(4cm²)/(4cm) = 2cm Now we reflect the triangle AEF along the side AF in order to get the new point G. Then we obtain the right triangle ACG, so we can apply the Pythagorean theorem: AC² = AG² + CG² = EF² + (FG + CF)² = EF² + (AE + CF)² = (4cm)² + (2cm + 6cm)² = (4cm)² + (8cm)² = 16cm² + 64cm² = 80cm² Since AC is the diagonal of the square, with s being the side length of the square we obtain: A(ABCD) = s² = (AC/√2)² = AC²/2 = 40cm² Best regards from Germany

[@LuisdeBritoCamacho]

Let's do it!! The Easiest part : 1) AE = 2 cm. 4 * AE = 8 ; AE = 8/4 ; AE = 2 cm 2) EF = 4 cm. 6 * EF = 24 ; EF = 24/6 ; EF = 4 cm The Hardest Part : 3) Rotate the Quadrilateral [AECF], with Side Lengths (AE + FC) = 8 cm and EF = 4 cm, clockwise until the Point E belongs to the Square Side AD. Extend Vertical Line AE passing through Point D, and obtain Point A'. Extend Vertical Line FC until it crosses Line AB and mark Point a'. Now we have a Rectangle [AA'CC']. 4) AE + FC = 2 + 6 = 8cm; and as stated before EF = 4 cm 5) AC = Diagonal of the Blue Square and the Diagonal of the Rectangle [AA'CC']. They share the same Length. 6) AC^2 = 4^2 + 8^2 ; AC^2 = 16 + 64 ; AC^2 = 80 ; AC = sqrt(80) cm ; AC = 4*sqrt(5) cm ; AC ~ 8,94 cm 7) As we know the Relationship between Side of Square and its Diagonal. 8) Diagonal = Side * sqrt(2) 9) 4*sqrt(5) = Side * sqrt(2) ; Side = 4*sqrt(5) / sqrt(2) ; Side = 4*sqrt(5)*sqrt(2) / 2 ; Side = 2*sqrt(10) cm 10) Area = Side^2 ; Area = [2*sqrt(10)]^2 ; Area = 4 * 10 ; Area = 40 square cm 11) My Best Answer : The Area of Blue Square is equal to 40 Square Centimeters. 12) THE END Note: I correct some mistakes but the Reasoning is the same.

[@Teamstudy4595]

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