Can you find the area of the Yellow Triangle? | (Nice Geometry problem) |
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- เผยแพร่เมื่อ 21 พ.ค. 2024
- Learn how to find the area of the Yellow Triangle. Important Geometry and Algebra skills are also explained: Congruent Triangles; area of a triangle formula; isosceles triangles; Pythagorean Theorem. Step-by-step tutorial by PreMath.com
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Thank u for doing hard work for us
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.
Yep, did the same ...
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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 ✓ ≡=-⋅
Great! GoatGuy!
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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.
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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.
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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
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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
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.
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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.❤❤❤
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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
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Sir Premath, can you please find another problem but not Geometry? Thank you.
Good morning sir
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.😊
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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!
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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
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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.
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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.
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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. 😁
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29
Let's find the area:
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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
The method shown in the video is much smarter. 👍 But at the end I am also right and that counts.
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Second way, area ABCD-areaABC, CB=4, area ABCD=7*7=49? areaABC=4*10/2=20, 49-20=29!
Тёзка, а как ты получил [ABCD]=7•7=49 ?! 🤔
@@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!
@@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. 😥
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3nd
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Seems to be not sufficient information to make a solution. 😂
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