Can you find area of the Yellow Square? | (inscribed in a Triangle) |

[@andreasproteus1465]

Easier if consider similar triangles ABC ~DEC, then the side of the square (a) can be found by: a/30 = (20-a)/20 => a =12

[@PreMath]

Thanks for the feedback ❤️

[@angeluomo]

Great and very simple method.

[@giuseppemalaguti435]

h=20...20:15=l:(15-l/2)...15l=20(15-l/2)=300-10l...25l=300..l=12..Ay=144

[@PreMath]

Excellent! Thanks for sharing ❤️ You are the best!

[@christianaxel9719]

Directly aply similarity over full triangles bases over heights on ABC and DEC : 30/20=l/(20-l) -> 3(30-l)=2l -> 60-3l=2l -> 60=5l -> 12=l -> Yellow square area = l²=144.

[@christianaxel9719]

Note: By Heron formula: if a triangle wih sides a,b,c has area A then similar triangle with sides ka, kb, kc has area k²A because s'=ks, s'-ka=k(s-a), s'-kb=k(s-b), s'-kc=k(s-c); S'²=(k²)²S² -> S'=k²S; then (kb)h'/2=k²bh/2 -> kh'=k²h -> h'=kh. Then if triangles are similar their heights are proporional too.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@pramodsingh7569]

Thanks

[@PreMath]

You are very welcome! Thanks ❤️

[@yalchingedikgedik8007]

Thanks Sir Very nice and useful Glades ❤❤❤

[@PreMath]

So nice of you dear🌹 Thanks for the feedback ❤️

[@TurquoizeGoldscraper]

Instead of cutting it in half, I combined ADF and BEG into a triangle with height = x and base = 30-x. Because of the angles, it would be similar to CDE with height = 20-x and base = x. (20 - x) / x = x / (30 - x) (20 - x) * (30 - x) = x * x 600 - 50x + x^2 = x^2 600 = 50x 12 = x Area = x^2 = 12^2 = 144.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@robertloveless4938]

Very logical. Makes the actual math (number crunching) quite simple.

[@HarHarMahadev-931]

Salute to premath 👑👑👑👑

[@PreMath]

Thanks dear ❤️

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... Since ABC is an isosceles triangle (AC=BC), we can calculate the height of the triangle ABC according to its base AB: A(ABC) = (1/2)*AB*h(AB) = (1/2)*AB*CM ⇒ CM = 2*A(ABC)/AB = 2*300/30 = 20 The triangles ADF and ACM are obviously similar. With s being the side length of the square we obtain: AF/DF = AM/CM (AM − FM)/DF = AM/CM (AB/2 − FM)/DF = (AB/2)/CM (AB/2 − s/2)/s = (AB/2)/CM (AB − s)/s = AB/CM (30 − s)/s = 30/20 = 3/2 30 − s = (3/2)*s 30 = (5/2)*s ⇒ s = 30*2/5 = 12 Finally we are able to calculate the area of the yellow square: A(DEGF) = s² = 12² = 144 Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing ❤️

[@LuisdeBritoCamacho]

🙂🙂🙂

[@PreMath]

Thanks ❤️😀

[@alster724]

After seeing the penultimate part of the video, I knew the area of the square right off the bat

[@PreMath]

Excellent! Thanks for sharing ❤️

[@AodhMacRaynall-dr1sf]

Thank you for these. I don't get to watch as many as I'd like, but I'm sure they're all good. I also want to commend you on for doing this. It's a much more useful occupation than many of your countrymen seek, who all seem to be want to be spam callers. Huzzah!!!

[@PreMath]

Glad you like them! Thanks for the feedback ❤️ Stay blessed!

[@sergeyvinns931]

CM=600/30=20. DF=DE=EG=FG=a. a/CM=GB/MB. GB=(30-a)/2. MB=15. a/20=(30-a)/15*2. 50a=600. a=12. a^2=144!.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@arthurschwieger82]

Very cool! I took a different approach and fortunately, got the same answer. ;-) I started the same and got a height of 20. I assigned X to each side of the yellow square. Then calculated the area of each of the white triangles using 1/2 * b * h. I had 1/2 * (15-X) * X as as there are two of these, multiplied by 2. the 1/2 and 2 cancel so I end up with 15X-X^2. Then add on the next white triangle: 1/2*X*(20-X) = 1/2*20X-X^2. The area of the yellow square is X^2. Adding all of these up and simplifying, I had 25X=300 or X=12.

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@mohabatkhanmalak1161]

I like solutions solved with the constant 'k'. Thanks for posting.🐞

[@himo3485]

30＊CM/2=300 CM=20 20 : 30 = 2 : 3 CP=2x DE=PM=3x 2x+3x=20 5x=20 x=4 3x=12 Yellow Square area : 12＊12=144

[@PreMath]

Excellent! Thanks for sharing ❤️

[@angeluomo]

The height of the triangle is 20. So if x is the length of each side of the square, then we can put together this equation: x^2 + (15-(x/2))*x + ((20-x)*x)/2 = 300, where (15-(x/2))*x = the total area of the two triangles ADF and BEG, and ((20-x)*x)/2 is the area of triangle DCE, and x^2 is the area of the square. This equation simplifies to 25x=300, so x=12 and the area = 144.

[@quigonkenny]

Given: AB = 30. ∠CAB = ∠ABC. Area of ∆ABC = 300. Observationally discernable: As ∠CAB = ∠ABC, ∆ABC is an isosceles triangle. As ∠CMB = 90° and ∆ABC is isosceles, CM bisects ∆ABC into two congruent right triangles ∆AMC and ∆CMB. As square DFGE is fully inscribed in ∆ABC and FG is collinear with AB, DFGE is symmetrical with ∆ABC about CM, and so CM also bisects DFGE into congruent rectangles DFMP and PMGE. As DE is parallel with AB and DC and CE are collinear with CA and BC respectively, ∠CDE and ∠DEC are congruent with ∠CAB and ∠ABC. Therefore ∆CDE is similar to ∆ABC and ∆DPC and ∆CPM are similar to ∆AMC and ∆CMB. As ∠AFD = 90° and ∠A is common, ∆AFD is similar to ∆AMC and all similar triangles. The same is true for ∆EGB by symmetry. Additional variables/assigned points: Let the side length of DFGE = 2x. A = bh/2 300 = 30CM/2 = 15CM CM = 300/15 = 20 As CM = 20 and MB = 30/2 = 15, ∆CMB (and by congruency, ∆AMC) is a 5:1 ratio 3-4-5 Pythagorean triple triangle and BC = (CA =) 5(5) = 25. In triangle ∆EGB, EG = 2x (being a side of the aquare), and as MG = 2x/2 = x and MB = 15, GB = 15-x. EG/GB = CM/MB 2x/(15-x) = 20/15 = 4/3 3(2x) = 4(15-x) 6x = 60 - 4x 10x = 60 x = 60/10 = 6 Square DFGE: Area = s² = (2x)² = (2(6))² = 12² = 144 sq units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@AmirgabYT2185]

S=144

[@PreMath]

Excellent! Thanks for sharing ❤️

[@santiagoarosam430]

Área FGED=2a*2a→ AF=(30/2)-a=15-a . MC=2*300/30=20→ Pendiente de AC: p=20/(30/2)=4/3→ p*AF=p(15-a)=2a→ (4/3)(15-a)=2a→ a=6→ Área FGED=4a²=4*36=144 ud². Gracias y un saludo cordial.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@Waldlaeufer70]

CM = 2 * 300 / 30 = 20 s = 2 (AB/2 x) = CM (1 - x) s = 2 (15 x) = 20 (1 - x) 30 x = 20 - 20 x 50 x = 20 x = 2/5 s = 2 * 15 * 2/5 = 4/5 * 15 = 12 units A(yellow) = s * s = 12 * 12 = 144 square units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marcgriselhubert3915]

Let's use an orthonormal, center M, first axis (AB). We have A(-15; 0) and C(0; 20) then VectorAC(15; 20) is colinear to VectorU(3; 4). The equation of AC) is then (x+15).(4) - (y).(3) = 0 or 4.x -3.y +60 = 0 or y = (4/3).x +20. Let 2.a be the side length of the square, then we have F(-a; 0) and D(-a; (-4/3).a +20) and FD = (-4/3).a + 20. Now, as FD is also the side length of the square, we have (-4/3).a +20 = 2.a or a = 6 and 2.a = 12 is the side length of the square. The area of the square is thes 12^2 = 144.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@michaelkouzmin281]

Just another solution: 1. MC =20; 2. Let x = DE; 3. PC = MC-x = 20-x; 4. Triangles ABC and DEC are similar due to AA term => DE/AB = PC/CM => x/30 = (20-x)/20; 5. 2x= 3(20-x); 5x=60; x=12 6. A(FDEG)=x^2 = 12^2 =144 sq.u.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@kennethstevenson976]

To easy, 20/15 = X/(15 - x/2) ; x = 12 ; x^2 = 144 square units. ( x = side of the square. )

[@georgebliss964]

Area of ABC = 300. 1/2 x 30 x CM = 300. CM = 300/15 = 20. Let side length of square = S. AM = 15. Therefore AF = 15 - 0.5S. Triangles AFD & AMC are similar. FD/AF = CM/AM. S/(15 - 0.5S) = 20/15. 20 (15 - 0.5S) = 15S. 300 -10S = 15S 300 = 25S. S = 12. Then area = 12 x 12 = 144.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@prossvay8744]

1/2(30)(h)=300 h=20 DE=DF=EQ=QF=2x ∆ABC~∆CDE DE/AB=CP/CM 2x/30=(20-2x)/20 So x=6 2x=2(6)=12 Area of the yellow square=12^2=144 square units.❤❤❤ Best regards.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@wackojacko3962]

Oh K !!!!! 🙂

[@PreMath]

Thanks ❤️

[@devondevon4366]

144 A different approach the height of the triangle = 600/30 = 20 Let the length of the square = x Since the square's length = the height of the two triangles (separated by the square), then there base = 3/4 x , since the both triangles are similar to the large triangle. Since there are two then the other is also 3/4 x Hence, the base of the large triangle = x + 3/4 x + 3/4 x Hence x + 3/4 x + 3/4x = 30 2.5 x = 30 x = 30/2.5 x = 12 Hence the area of the square = 12 * 12 = 144 Answer

[@PreMath]

Excellent! Thanks for sharing ❤️

[@alexniklas8777]

15/20=(15-а/2)/а; 25а=300; а=300/25=12; S=12^2=144

[@sorourhashemi3249]

Thanks easy. ∆CDP~∆CAB, CM =20, so 30/a= 20/20-a ===>a=12, area =144

[@allanflippin2453]

To me, adding "k" makes things confusing. I'll offer another approach, starting where we found out that CM = 15 and also that DF = 4/3 * AF. DF = S the square side. Therefore AF = 3/4S. FM = S/2. So 3S/4 + S/2 = 15. Simplifying, that becomes 5S/4 = 15 or S = 12 and area = 144.

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@MrPaulc222]

15h = 300, so h = 20 Square's sides are each 2x. 15/20 = (15-x)/2x 30x = 300-20x 50x = 300 x = 6 2x = 12 (2x)^2 = 144 Square's area is 144 un^2

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال AF=15-x/2 , DF=x tanBAC =20/15 =DF/AF ....... x=12 S=12^2

[@LuisdeBritoCamacho]

I am about to prove beyond any reasonable doubt that Yellow Square Area is Equal to 144 Square Units!! 1) (AB * CM) / 2 = 300 ; 30 * CM = 600 ; CM = 600 / 30 ; CM = 20 Linear Units. 2) h = 20 Linear Units 3) tan(alpha) = 20/15 ; tan(alpha) = 4/3 4) DE = DF = EG = FG = 2X Linear Units 5) DP = PE = FM = MG = X Linear Units 6) CM / MB = EG / GB = 2X / (15 - X) = 4/3 7) 6X = 60 - 4X ; 10X = 60 ; X = 60 / 10 ; X = 6 8) 2X = 12 9) (2X)^2 = 12^2 ; 4X^2 = 144 10) ANSWER : The Yellow Square Area is equal to 144 Square Units. QED.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@derphysikkanal]

144

[@PreMath]

Excellent! Thanks for sharing ❤️