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.
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😅😅😅
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.
You can make a point J lies on CB which is perpendicular with JE.JE =1/2AC(mid pt thm）.JE is height of Triangle DEB and DB is base.This will quick to finish the question
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
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
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.
*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.
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
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 ✅
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! 😊
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?
@@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.
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 . 😊
உங்களது சேவை விலை மதிக்க மூடியாதது! ❤❤❤
மிக்க மகிழ்ச்சி அன்பே❤️🙏
உங்கள் தொடர்ந்த அன்புக்கும் ஆதரவுக்கும் நன்றி!
ஆசீர்வதிக்கப்பட்டிருங்கள் 😀
Merry Christmas, team Premath.
That’s very good
Very good
Thanks Sir
It is wonderful method
Again thanks for your efforts
❤❤❤❤
Thank you!
Nice solution and Merry Christmas from The Philippines 🇵🇭 Premath Team!
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.
Gostei ! Muito bom !
Bom dia Mestre
Feliz Natal
Thanks dear❤️
Merry Christmas! Wishing you a happy holiday season as well! 😊🙏
Merci beaucoup pour votre effort
I calculated A=triangle areas AED+ACD, same result 69.
Thanks and best regards!
Nice work!
You are very welcome!
Thanks for sharing ❤️
Merry Christmas! Wishing you a happy holiday season as well! 😊🙏
Thanks easy
Merry Christmas 🎄🎁
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
Excellent!
Thanks for sharing ❤️
Merry Christmas! Wishing you a happy holiday season as well! 😊🙏
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
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😅😅😅
Typo! 5-12-13
Thanks for sharing ❤️
Merry Christmas! Wishing you a happy holiday season as well! 😊🙏
Thank you for correcting my typo😊
@@PreMathMerry Christmas to you from Texas
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.
S=69 square units
Excellent!
Thanks for sharing ❤️🙏
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 √ )
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.
Merry Christmas, Professor. Thanks for a great year of mathematics!👍😀
You can make a point J lies on CB which is perpendicular with JE.JE =1/2AC(mid pt thm）.JE is height of Triangle DEB and DB is base.This will quick to finish the question
S(green) = S(ACE) + S(CDE) = [1/2 + (5/18)•(1/2)] • S(ABC) = (23/36)•108 = 69.
12×15.754÷2=92.524full area-41.7=52.82green area
S(ACDE)=108-39=69
Thanks sir!😊
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
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
Hoc est 65 sq. Unis, spondeo Responsi. 😅
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.
*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.
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
Excellent!
Thanks for sharing ❤️
Merry Christmas! Wishing you a happy holiday season as well! 😊🙏
Sehr schön gelöst.
Alle unbekannten Längen berechnet.
Vorbildlich 👌🔝👍
Schöne Feiertage
Gerald
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 ✅
Excellent!
Thanks for sharing ❤️
Merry Christmas! Wishing you a happy holiday season as well! 😊🙏
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! 😊
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?
@@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.
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 . 😊