Can you calculate area of the Green shaded Quadrilateral? | (Triangle) |

[@bigm383]

Merry Christmas, Professor. Thanks for a great year of mathematics!👍😀

[@PreMath]

Same to you my dear friend! Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@charlesmitchell5841]

Nice problem. Good geometry review. Thanks for the video lesson. 👍. Glad to be back doing pre-math!

[@PreMath]

Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@saumendrakumarnayak832]

Merry Christmas, team Premath.

[@PreMath]

Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@himo3485]

AD=DB=13 CD=√[13²-12²]=5 CB=5+13=18 12 : 18 = 2 : 3 ACB∞DEB DE=2x AE=EB=3x (2x)²+(3x)²=13² 13x²=169 x=√13 Greeen shaded area = 12*18*1/2 - 2√13*3√13*1/2 = 108 - 39 = 69

[@PreMath]

Excellent! Thanks for sharing ❤️ Merry Christmas! Wishing you a happy holiday season as well! 😊🙏

[@alster724]

Nice solution and Merry Christmas from The Philippines 🇵🇭 Premath Team!

[@PreMath]

Merry Christmas to you too! ❤️🙏 Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@sergioaiex3966]

Solution: Connecting A to D, we will form an isosceles triangle ABD and since AE is equal to EB, AD value is 13 Let's calculate CD by applying the Pythagorean Theorem AC² + CD² = AD² (12)² + CD² = (13)² 144 + CD² = 169 CD² = 25 CD = 5 BC = BD + CD BC = 13 + 5 BC = 18 Once again, let's calculate AB by applying the Pythagorean Theorem AC² + BC² = AB² (12)² + (18)² = AB² 144 + 324 = AB² AB² = 468 AB = 6√13 Since AB = 6√13, AE = 3√13 and BE = 3√13 Once again, let's calculate DE by applying the Pythagorean Theorem AE² + DE² = AD² (3√13)² + DE² = (13)² 117 + DE² = 169 DE² = 52 DE = 2√13 Green Shaded Area = ∆ ABC Area - ∆ BDE Area ... ¹ ∆ ABC Area = ½ 12 . 18 ∆ ABC Area = 108 ∆ BDE Area = ½ 3√13 . 2√13 ∆ BDE Area = ½ 3√13 . 2√13 ∆ BDE Area = 39 Replacing in ¹ Green Shaded Area = 108 - 39 Green Shaded Area = 69 Square Units ✅

[@PreMath]

Excellent! Thanks for sharing ❤️ Merry Christmas! Wishing you a happy holiday season as well! 😊🙏

[@veronicanicholas-robinson2330]

This way of repeating Pythagoras multiple times is a lot simpler imo

[@sergioaiex3966]

Yes, I agree, but obviously there are other ways to solve it.

[@marioalb9726]

Triangle ACD: Pytagorean triplet 5-12-13 --> b₁= 5cm Triangle ABC: b = b₁+b₂ = 5+13 = 18 cm A = ½bh = ½12*18 = 108cm² Pytagorean theorem: c²=12²+18² ---> c= 6√13cm ; c₁=½c Similarity of triangles: h₁/c₁ = 12/18 --> h₁= 2√13cm Triangle EDB: A₁= ½c₁h₁= ½*3√13*2√13 = 39 cm² Green shaded area: A₂= A- A₁= 108-39= 69 cm² ( Solved √ )

[@jamestalbott4499]

Thank you!

[@alexundre8745]

Bom dia Mestre Feliz Natal

[@PreMath]

Thanks dear❤️ Merry Christmas! Wishing you a happy holiday season as well! 😊🙏

[@DB-lg5sq]

Merci beaucoup pour votre effort

[@PreMath]

Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@Abdelfattah-hr8tt]

Merry Christmas 🎄🎁

[@PreMath]

Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@uwelinzbauer3973]

I calculated A=triangle areas AED+ACD, same result 69. Thanks and best regards!

[@PreMath]

Nice work! You are very welcome! Thanks for sharing ❤️ Merry Christmas! Wishing you a happy holiday season as well! 😊🙏

[@marioalb9726]

Triangle ACD: Pytagorean triplet 5-12-13 --> b₂= 5cm A₂= ½.b₂.h = ½*5*12=30 cm² Triangle ABC: b = b₁+b₂ = 13+5 = 18 cm A = ½bh = ½12*18 = 108cm² Triangle ADE: A₃ = ½(A-A₂) = ½(108-30)=39cm² Green shaded area: A= A₂+A₃= 30+39= 69 cm² ( Solved √ ) There's no need to calculate 'x' and 'h' as video does.

[@georgebliss964]

Area of triangle ABC = 108. Area of triangle ACD = 30. Therefore area of triangle ABD = 108 - 30 = 78. Then area of triangle EBD = 78 / 2 = 39. Thus green area = 108 - 39 = 69.

[@blogfilmes1134]

Gostei ! Muito bom !

[@PreMath]

Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... First of all we observe that the triangles ADE and BDE are congruent: AE = BE DE is common ∠AED = ∠BED = 90° Therefore we can conclude that AD=BD=13. Now we apply the Pythagorean theorem step by step to the right triangles ACD, ABC and BDE: AC² + CD² = AD² 12² + CD² = 13² 144 + CD² = 169 CD² = 25 ⇒ CD = √25 = 5 BC = BD + CD = 13 + 5 = 18 AB² = AC² + BC² = 12² + 18² = 6²*2² + 6²*3² = 6²*(2² + 3²) = 6²*(4 + 9) = 6²*13 ⇒ AB = √(6²*13) = 6√13 ⇒ BE = AB/2 = 3√13 BE² + DE² = BD² (3√13)² + DE² = 13² 9*13 + DE² = 13*13 DE² = 13*13 − 9*13 = (13 − 9)*13 = 4*13 = 2²*13 ⇒ DE = √(2²*13) = 2√13 Now we are able to calculate the area of the green quadrilateral: A(ACDE) = A(ABC) − A(BDE) = (1/2)*AC*BC − (1/2)*BE*DE = (1/2)*12*18 − (1/2)*(3√13)*(2√13) = 108 − 39 = 69 Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing ❤️ Merry Christmas! Wishing you a happy holiday season as well! 😊🙏

[@GetMatheFit]

Sehr schön gelöst. Alle unbekannten Längen berechnet. Vorbildlich 👌🔝👍 Schöne Feiertage Gerald

[@unknownidentity2846]

@@GetMatheFit Vielen Dank für die Blumen. Ich hoffe, dass du und deine Familie ein schönes Fest hatten und der Weihnachtsmann auch fließig war.

[@yalchingedikgedik8007]

That’s very good Very good Thanks Sir It is wonderful method Again thanks for your efforts ❤❤❤❤

[@PreMath]

Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@sorourhashemi3249]

Thanks easy

[@PreMath]

Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@marcgriselhubert3915]

*Let's name x = CD. Triangles BED and BCA are similar (same angles), so DE/12 = BE/(13 + x) = 13/(2.BE), so 2.(BE^2) = 13.(13 + x) and 4.(BE^2) = 338 + 26.x (eq 1) In triangle ABC: AB^2 = 4.(BE^2) = 12^2 + (13 + x)^2, so 4.(BE^2) = (x^2) + 26.x + 313 (eq 2) With (eq 1) and (eq 2) we get: (x^2) + 26.x +313 = 338 + 26.x, and that gives x^2 = 25 and x = 5. *Area of triangle ABC = (1/2).BC.CA = (1/2).(13 + 5).12 = 108 *We had 2.(BE^2) = 13.(13 + x), so 2.(BE^2) = 13.18 = 234, so BE^2 = 117 and BE = sqrt(117) = 3.sqrt(13) As DE/12 = (BE/(13 + x), we get DE = (12.(3.sqrt(13))(13 + 5) = 2.sqrt(13), and the area of triangle BED is (1/2).BE.DE = (1/2).(3.sqrt(13)).(2.sqrt(13)) = 39 Finally by difference the green shaded area is 108 - 39 = 69.

[@PreMath]

Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@zawatsky]

AD=DB, значит стороны ▲ACD 12 и 13, т. е. CD=5 (пифагорова тройка). Теперь мы знаем пропорции катетов - 18 к 12 или 3 к 2. Площадь большого 18*12/2=6*18. Катеты малого: 9x²+4x²=13x²=13², откуда х²=13. Его площадь 3*2х²/2=3х²=3*13. Считаем разницу. 6*18-3*13=3(2*18-13)=3(36-13)=3*23=69.

[@PreMath]

Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@reynaldowify]

Merry Xmas. Just draw a perpendicular from BC to point E, now you got two similar triangles. The righmost is ABC area divided by 4, and the remaining little triangle has base 6, and height 4 by similitude

[@PreMath]

Merry Christmas! That's a great way to think about it! 😊🎄 Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@phungpham1725]

1/ AD = BD= 13 --> CD= 5 ( the triangle ACD is a 5/12/14 triples) 2/ Area of the green area = Area of triangle ACD+ Area of triangle ADE Because triangles 8:04 ADE and BDE are congruent Focus on the triangle BDE. From E drop the height EH to the base BD, we have EH//= 1/2 AC= 6 So area of the green area = 1/2( 5x12 + 6x13) = 30+ 39= 69 sq units😅😅😅

[@PreMath]

Typo! 5-12-13 Thanks for sharing ❤️ Merry Christmas! Wishing you a happy holiday season as well! 😊🙏

[@phungpham1725]

Thank you for correcting my typo😊

[@phungpham1725]

@@PreMathMerry Christmas to you from Texas

[@MrPaulc222]

Similar triangles ABC and BED. Solution is ABD - BED. AB is 2a. Corresponding sides on the triangles are: b, a, 13 12, x+13, 2a I suspect the key is in seeing that ABD is isosceles. ACD is x, 12,13, so CD (or x) = 5 due to 5,12,13 being a Pythagorean triple. ABC: 12^2 + 18^2 = (AB)^2, so (AB)^2 is 468. AB = 2*sqrt(117), so a = sqrt(117). Update the sides of the similar triangles: b, sqrt(117), 13 12, 18, 2*sqrt(117) Forgot to calculate b: 169 - 117 = 52, so b = 2*sqrt(13). Now the areas: ABC: (18*12)/2 = 108 The 2's cancel, so sqrt(117)*sqrt(13) BED: sqrt(117)*2*(sqrt(13))/2 The 2's cancel, so sqrt(117)*sqrt(13) = sqrt(1521) which is a convenient 39. Green quadrilateral is 108-39 = 69 un^2

[@santiagoarosam430]

Si EA=EB---> DB=DA=13---> CD=√(13²-12²)=5---> DCB=5*12/2=30 ; ACB=12(5+13)/2=108---> DEA=(108-30)/2=39---> ACDE=30+39=69 u². Gracias y saludos

[@wasimahmad-t6c]

18×12=216-78=138 grin area

[@AmirgabYT2185]

S=69 square units

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@cyruschang1904]

Triangles DEB and ACB are similar If CD = a & AE = EB = b b/13 = (13 + a)/2b 2b^2 = 13(13 + a) 12^2 + (13 + a)^2 = 4b^2 = 26(13 + a) a^2 + 12^2 - 13^2 = 0 a^2 = 13^2 - 12^2 = 25 => a = 5 18/2b = b/13, b^2 = 9(13) => b = 3√13 Green area = (12)(18)/2)(1 - (13/6√13)^2) = 108(1 - 13/36) = 108(23/36) = 69

[@PreMath]

Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@cyruschang1904]

@@PreMath Thank you. May the New Year bring you and your love ones happiness and success.

[@sivaramanr6625]

உங்களது சேவை விலை மதிக்க மூடியாதது! ❤❤❤

[@PreMath]

மிக்க மகிழ்ச்சி அன்பே❤️🙏 உங்கள் தொடர்ந்த அன்புக்கும் ஆதரவுக்கும் நன்றி! ஆசீர்வதிக்கப்பட்டிருங்கள் 😀

[@michaelkouzmin281]

a bit more ridiculous solution with a respective bit of trig )))): 1. let a = angle ABC; x = BE = EA; h = DE; 2. cos(a) = BE/DB = x/13; sin(a) = AC/AB = 12/2x=6/x; 3. sin^2(a)+cos^2(a)=1 => (6/x)^2+(x/13)^2 = 1 2 positive roots ( there are 2 negative as well but let us drop them now) x1 = 2*sqrt(13) should be rejected as respective CB will be equal to 8 what is less than CD=13; x2 = 3*sqrt(13) - accepted; 3. h = sqrt(13^2-x^2)= sqrt(13^2-(3*sqrt(13))^2)= 2*sqrt(13) 4. A(DEC) = x*h/2 = 39 sq.units 5. AB = 2x= 6*sqrt(13) 6. CB = SQRT(AB^2-AC^2) = sqrt (( 6*sqrt(13))^2-12^2) = 18 7. A(ABC) = CA*CB/2 = 12*18/2 = 108 sq units; 8. A(ACDEgreen) = A(ABC)-A(DEC)= 108 - 39 = 69 sq units.

[@PreMath]

Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL : 01) Green Shaded Area = Triangle [ABC] Area - Triangle [BDE] Area 02) Length of Line AD = Length of Line BD = 13 03) Triangle [ABD] is an Isosceles Triamgle 04) 13^2 - 12^2 = CD^2 ; CD^2 = 169 - 144 ; CD^2 = 25 ; CD = 5 05) Triangle [ABC] Area = (12 * 18) / 2 = 216 / 2 = 108 sq un 06) AB^2 = 18^2 + 12^2 ; AB^2 = 324 + 144 ; AB^2 = 468 ; AB = sqrt(468) ; AB ~ 21,6 07) AE = BE ~ 10,82 lin un 08) DE^2 = 169 - (468 / 4) ; DE^2 = 169 - 117 ; DE^2 = 52 ; DE = sqrt(52) ; DE ~ 7,2 09) Triangle [BDE] Area = (7,2 * 10,8) / 2 ~ 38,88 10) Green Shaded Area (GSA) = 108 - 39 ; GSA ~ 69 sq un Therefore, OUR BEST ANSWER IS : Green Shaded area equal to 69 Square Units.

[@adept7474]

S(green) = S(ACE) + S(CDE) = [1/2 + (5/18)•(1/2)] • S(ABC) = (23/36)•108 = 69.

[@PreMath]

Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@alexniklas8777]

S(ACDE)=108-39=69 Thanks sir!😊

[@PreMath]

Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@wackojacko3962]

FAQ'S: What does SSS SAS ASA AAS MIS ISS IPP I mean? ...these are questions asked by three types of people in the world - those who can count and those who can't. I'm one of em! 😊

[@jimlocke9320]

SSS SAS ASA AAS ways of proving two triangles are congruent, referencing the corresponding theorems. S = side, A = angle, so SAS is an equal side, then angle, then side. These are common abbreviations used in high school geometry in the USA. I am not familiar with MIS ISS IPP. Maybe the same theorems in other languages?

[@Slimmo_09]

@@jimlocke9320 These are just other scientific abbreviations that we use: ISS = International Space Station; IPP = Inflatable Penile Prosthesis; MIS = Meteorological Impact Statement (think Victorian bushfires currently). The above are what I am familiar with, the "S" in MIS could well be software in some cases. Given how common abbreviations are now, there could well be engineering versions for these.

[@wackojacko3962]

Almost... MISS ISS IPP I is a river​. 3 types of people but only account for two... just a play on words! I tease a lot, I mean I do the math but then gotta have some fun is all . 😊

[@PreMath]

Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!

[@syedmdabid7191]

Hoc est 65 sq. Unis, spondeo Responsi. 😅

[@PreMath]

Thanks ❤️🙏 Have a happy, safe, and prosperous New Year!