Thank u for doing hard work for us

@PreMath

28 күн бұрын

You are very welcome! Thanks dear❤️

There is no need to think outside the box. Drop a perpendicular from C to DE and label the intersection as point G, constructing right ΔDGC.

@tontonbeber4555

28 күн бұрын

Yep, did the same ...

@PreMath

28 күн бұрын

Excellent! Thanks for sharing ❤️

How about this way? [1.1]  3 / √(𝒉² - 3²) = 𝒔 ÷ 7 … cross multiply [1.2]  21 = 𝒔√(𝒉² - 9) … square both sides [1.3]  441 = 𝒔²(𝒉² - 9) Then note [2.1]  7² = 𝒉² - 𝒔² … rearrange [2.2]  𝒔² = 𝒉² - 49 Combine [1.3] and [2.1] [3.1]  441 = (𝒉² - 49)(𝒉² - 9) … substitute 𝒖 for 𝒉² [3.2]  441 = (𝒖 - 49)(𝒖 - 9) … then do the algebra [3.3]  441 = 𝒖² - 58𝒖 + 441 … ah, cancelling the 441 [3.4]  0 = 𝒖² - 58𝒖 … rearranging again [3.5]  𝒖² = 58𝒖 … dividing out 𝒖 [3.6]  𝒖 = 58 … and returning 𝒉² [3.7]  𝒉² = 58 Then remembering that the yellow △ is ½ of 𝒉² … [4.1]  Area yellow △ = ½ 58 [4.2]  Area yellow △ = 29 And that would be the solution, just as the other methods. It did NOT require either a calculator, nor drawing helper-triangles along the video's method. Just the proportionality of congruent sides in the given diagram. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

@PreMath

28 күн бұрын

Great! GoatGuy! Thanks for sharing ❤️

I solved it the same way as you, professor. This is a sign that my knowledge of mathematics is improving. Thanks to you.

Let's use an orthonormal, center E and first axis (EB). We have E(0; 0) A(-3; 0) B(7; 0) C(7; h) and D(0, l) where h and l are unknown positive reals. VectorCD(-7; l-h) and VectorAD(3; l). We have CD^2 = AD^2, so 49 + l^2 -2.l.h +h^2 = 9 + l^2 , or h^2 -2.l.h + 40 = 0, or 2.l.h = h^2 + 40 (equation 1) We have (CD) and (AD) perpendicular, so -21 +l^2 - l.h = 0, or l.h = l^2 - 21, or 2.l.h = 2.(l^2) -42. (equation 2) From these two equations we obtain that h^2 + 40 = 2.(l^2) - 42 , or h^2 = 2.l^2 -82, or h = sqrt(2.(l^2) -82), h beeing positive. We replace h by this value in equation 2, we get: 2.l.sqrt(2.(l^2) -82) = 2.(l^2) - 42, or l.sqrt(2.(l^2) -82) = l^2 -21. We square: (l^2).(2.(l^2) -82) = (l^4) -42.(l^2) +441, or 2.(l^4) -82.(l^2) = (l^4) -42.(l^2) + 441, or (l^4) -40.(l^2) -441 = 0, or (L^2) -40.L -441 = 0, with L = l^2 Deltaprime = 20^2 + 441 = 841 = 29^9, so L = 20 -29 = -9 which is rejected as negative, or L = 20 + 29 = 49, and then l = 7. We finish. We have then VectorAD(3, 7) and so AD^2 = 9 +49 = 58. So the area of the yellow triangle is the area of a semi square whose side length is AD, it is (AD^2)/2 , so it is 58/2 = 29.

@PreMath

28 күн бұрын

Excellent! Thanks for sharing ❤️

1) Triangle [ADE] = Triangle [DCC']. As they share the very same Diagonal. The Geometrical Figure [BC'DE] is a (7 * 7) Square. 2) C' is the Point of intersection of Vertical Line BC and the Horizontal Line passing through Point D. 3) As : DC' = EB = 7, one must conclude that : 4) DE = 7 5) As : DE = 7 6) AD^2 = AE^2 + DE^2 ; AD^2 = 3^2 + 7^2 ; AD^2 = 9 + 49 ; AD^2 = 58 7) So, Yellow Triangle Area = 58/2 ; YTA = 29 8) ANSWER : The Yellow Triangle Area is equal to 29 Square Units.

@PreMath

28 күн бұрын

Excellent! Thanks for sharing ❤️

Posto a=AD risulta per la legge del coseno a^2-9+49=a^2+(2a^2-100)-2a√(2a^2-100)cos(45+arcsin(10/√2a))...a^2=58...Ayellow=29

@PreMath

28 күн бұрын

Excellent! Thanks for sharing ❤️

with assume AD=a and CB=x using Pythagorean theorem=> AC^2=2*a^2=x^2+100 (1) in triangle ADE cos(A)=AE/AD= (3/a) using cos theorem in triangle ADB and triangle DCB DB^2=a^2+100-2*10*cos(A)=a^2+x^2- 2* a*x *cos(C) regarding that A+C=180 degree cos(C)=cos(180-A)=-cos(A) and cos(A)=(3/a) a^2+100-20*a*(3/a)=a^2+x^2+2*a*x*(3/a) => a^2+100-60= a^2+x^2+6*x =>x^2+6*x-40 =0 => x=4 or x=-10 only x=4 is allowable using equation (1) => 2*a^2=100+16=116 => a^2= 58 => Area of ADC= (a^2)/2 = 58/2=29

We can easily prove that ABCD is cyclic quadrilateral then angle DCA=angle DBA=45°. Therefore, Triangle DEB is isosceles right triangle with base DB. As a result ED=EB=7.In triangle DAE we have AD^2=AE^2+DE^2=9+49=58. Finally, the area of triangle ADC equal AD^2/2=29

@PreMath

28 күн бұрын

Thanks for the feedback ❤️

Goal is to determine the area of the yellow triangle ∆CDA. CD = DA and ∠CDA = 90°, so ∆CDA is an isosceles right triangle. Let CD = DA = s. If ∠DAE = α and ∠EDA = β, where α and β are complementary angles which sum to 90°, then as ∠CDA = 90°, then ∠CDE = α as well. Extend BC up to P and draw DP so that DP is perpendicular to both DE and BP. This will create the rectangle DEBP. As DEBP is a rectangle, DP = EB = 7. As ∠CDE = α, then as ∠PDE = 90°, ∠PDC = β. As CD = DA, by SAS, ∆AED and ∆CPD are congruent. Therefore CP = AE = 3, ED = BP = PD = 7, and DEBP is a square with side length 7. Triangle ∆AED: AE² + ED² = DA² 3² + 7² = s² s² = 9 + 49 = 58 s = √58 Yellow triangle ∆CDA: Area = s²/2 = 58/2 = 29 sq units.

@PreMath

28 күн бұрын

Excellent! Thanks for sharing ❤️

There are hidden two congruent right-angled triangle 3×7, so A^2=CD^2=3^2+7^2=9+49=5 0:43 8, therefore the answer is 58/2=29.😊

@PreMath

28 күн бұрын

Excellent! Thanks for sharing ❤️

Sir Premath, can you please find another problem but not Geometry? Thank you.

Let AD= DC = a, then AC = √2 a Let angle CAB = t. Angle DAE = 45° + t AD cos (angle DAE) = AE => a cos(45° + t) = 3 => a cos t - a sin t = 3√2 ...(1) Also, AC cos (angle CAB) = AB => √2 a cos t = 7+3 = 10 => a cos t = 5 √2 ...(2) Substituting the value of a cos t from (2) into (1) gives: a sin t = 2√2 ...(3) Squaring and adding (2) and (3), we get: a^2 = 4*2 + 25*2 = 58 Area of triangle ABC = 1/2 a^2 = 58/2 = 29 Substituting

@PreMath

28 күн бұрын

Thanks for sharing ❤️

connect D to F and C to F CE right DE ∆ADE~~∆CDF CF = AE=3 DF=BE=7 DC^2=CF^2+DF^2=3^2+7^2 DC=√9+49=√58 AD=DC (ADC issoles triangle So yellow triangle area=1/2(√58)^2=29 square units.❤❤❤

@PreMath

28 күн бұрын

Excellent! Thanks for sharing ❤️

Without a caiculator and orally, area yellow triangle = 29! AD^2/2, AD^2=7^2+3^2= 49+9=58. A=58/2=29!

@PreMath

28 күн бұрын

Excellent! Thanks for the feedback ❤️

Answer = 29 let DC= n, then DA=n The area of the yellow triangle = n^2/2 Draw a perpendicular line to the red vertical line point C. This line length = BE =7 Since BE= DE=7 and line DA= 3, then n = sqrt ( 7^2 + 3^2) Pythagorean = sqrt (49 + 9) = sqrt (58) Area = n^2/2 = sqrt 58 * sqrt 58 * 1/2 = 58 * 1/2 = 29 Answer

Con centro en D, giramos 90º, en sentido antihorario, el triángulo AED y obtenemos el cuadrado EBE´D, de lado =EB=7=E´D=ED→ AE²+ED²=AD²→ 3²+7²=58=AD²→ ACD es la mitad de un cuadrado de lado AD→ Área ACD=58/2=29 ud². Gracias y un saludo cordial.

@PreMath

28 күн бұрын

Excellent! Thanks for sharing ❤️

I was lazy. I drew a 7 x 10 rectangle, calculated the area as 70 units. Then I subtracted the area of the three triangles, ABC, DFC, and ADG. (F and G are two auxiliary points on the rectangle.) The combined areas is 41 square units. 70 units - 41 units = 29 square units.

@PreMath

28 күн бұрын

Thanks for the feedback ❤️

Good morning sir

4:15 [AED]=[CFD] !!! => [ACD]=[ABCD]-[ABC]= =[AED]+[BCDE]-[ABC]= =[CFD]+[BCDE]-[ABC]= =[BFDE]-[ABC]= =7²-½(7+3)(7-3)= =49-20=29 sq.un. 😁

@PreMath

28 күн бұрын

Thanks for sharing ❤️

ΠΕΡΑΣΤΙΚΑ ΑΑΑΑ

Second way, area ABCD-areaABC, CB=4, area ABCD=7*7=49? areaABC=4*10/2=20, 49-20=29!

@rabotaakk-nw9nm

28 күн бұрын

Тёзка, а как ты получил [ABCD]=7•7=49 ?! 🤔

@sergeyvinns931

28 күн бұрын

@@rabotaakk-nw9nm Если ты не заметил, то треугольник, образованный выстой, основанием 3 и катетом жёлтого прямоугольного равностороннего треугольника, можно перенести вверх катетом к катету, то получим квадрат, со сторонами 7, площадь которого равна 49, а площадь четырёхугольника ABCD, точно такая же, если ты не заметил, отсюда и СВ=4, но об этом, ты не спросил, значит знаешь, что красненькая высота, равна 7!

@rabotaakk-nw9nm

28 күн бұрын

​@@sergeyvinns931Я-то заметил, и всё тщательно указал в своём комменте - и время, когда @PreMath доказал, что ΔAED=ΔCFD а DE=7, CF=3, и то, что AB=10, BC=4. А твой "second way", к сожалению, выглядит не обоснованным. 😥

@PreMath

28 күн бұрын

Thanks for the feedback ❤️

29

Let's find the area: . .. ... .... ..... The triangles ACD and ABC are both right triangles. So according to Thales theorem A, C and D are located on the same circle and A, B and C are located on the same circle. With AC being the diameter of these two circles they become identical. In this case A, B, C and D are located on the same circle. Now let's assume that AB is parallel to the x-axis and that BC and DE are parallel to the y-axis. With O being the midpoint of AC and therefore also the center of the circle, we can assume the following coordinates: A: ( −5 ; yA ) B: ( +5 ; yA ) C: ( +5 ; yC ) D: ( −2 ; yD ) The triangle ACD is not only a right triangle, it is also an isosceles triangle (AD=CD). Therefore OD is the height of this triangle and the triangles OAD and OCD are congruent right isosceles triangles (OA=OC=OD=R is the radius of the circle). Since OD is perpendicular to OA, the product of their slopes is −1. With all these information we obtain: xA² + yA² = R² xD² + yD² = R² [(yO − yD)/(xO − xD)]*[(yO − yA)/(xO − xA)] = −1 (−5)² + yA² = R² (−2)² + yD² = R² [(0 − yD)/(0 + 2)]*[(0 − yA)/(0 + 5)] = −1 yA² = R² − (−5)² yD² = R² − (−2)² (−yD)/2*(−yA)/5 = −1 A² = R² − 25 yD² = R² − 4 yA*yD = −10 yD² − yA² = 21 yA*yD = −10 yD² − (−10/yD)² = 21 yD² − 100/yD² = 21 yD⁴ − 100 = 21*yD² yD⁴ − 21*yD² − 100 = 0 yD² = 21/2 ± √[(21/2)² + 100] yD² = 21/2 ± √(441/4 + 100) yD² = 21/2 ± √(441/4 + 400/4) yD² = 21/2 ± √(841/4) yD² = 21/2 ± 29/2 Since yD²>0 and yD>0, there is only one useful solution: yD² = 21/2 + 29/2 = 50/2 = 25 ⇒ yD = 5 ⇒ yA = −2 Now we are able to calculate the area of the yellow triangle: A(ACD) = (1/2)*AC*h(AC) = (1/2)*AC*OD = (1/2)*(2*R)*R = R² = xA² + yA² = (−5)² + (−2)² = 25 + 4 = 29 Best regards from Germany

@unknownidentity2846

28 күн бұрын

The method shown in the video is much smarter. 👍 But at the end I am also right and that counts.

@PreMath

28 күн бұрын

Excellent! Thanks for sharing ❤️

3nd

@PreMath

28 күн бұрын

Thanks ❤️

Seems to be not sufficient information to make a solution. 😂

@PreMath

28 күн бұрын

Thanks for the feedback ❤️