It's an awesome solution from Another awesome professor.. ..... ❤ ❤ 😊

@PreMath

2 ай бұрын

Glad to hear that! Thanks dear❤

ΔABE and ΔCDE are similar by angle-angle (

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

Let AC=a (x-2)^2+(x+3)^2=a^2 (2x-3)^2+(x-5)^2=a^2 x^2-4x+4+x^2+6x+9=4x^2-12x+9+x^2-10x+25 2x^2+2x+13=5x^2-22x+34 5x^2-22x+34-2x^2-2x-13=0 3x^2-24x+21=0 x^2-8x+7=0 (x-1)(x-7)=0 So x=7 area of the yellow triangle=1/2(11)(2)-1/2(2)(2.67)=8.33 square units.❤❤❤ Thanks.

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

1. For single unknown x, only 1 equation is needed for its solution. The equation is provided by the common hypotenuse of the 2 right-angled triangles using Pythagoras theorem. x is found to be 7. Hence AB = 5, CD = 2, AD = 11 and BC = 10. 2. To find area of yellow triangle, area-base side ratio for equal height triangles is the obvious method when the areas of the 2 right-angled triangles can be easily calculated. Hence knowing AE, ED, BE, EC is crucial. 4 equations are needed to solve these 4 unknowns. With the derived lengths of AD = 11 and BC = 10, and side ratio of similarity triangles AEB and CED (similarity confirmed by AAA) provided by AB/CD = 5/2. Four equations can be set up: (1) AE + ED = 11 (2) BE + EC = 10 (3) AE/CE = 5/2 (4) BE/DE = 5/2 For clarity, let AE = a, BE = b, CE = c, DE = d. The 4 equations are: (1) a + d = 11 (2) b + c = 10 (3) a/c = 5/2 (4) b/d = 5/2 Add (1) & (2) a + c + b + d = 21 Substitue a for c and b for d using (3) c = (2/5) a and (4) d = (2/5) b a + (2/5)a + b + (2/5)b = 21 a + b = 15 Substitute d for b using (4) b = (5/2) d a + (5/2)d = 15 d = (2/5)(15 - a) From (1) d = 11 - a Hence (11 - a) = (2/5)(15 - a) a = 25/3 3.For equal height triangles ACE and ACD, their base side ratio AE:AD = a: 11 = (25/3):11 = their area ratio As area of ACD = (2 x 11)/2 = 11 Hence area of ACE = [(25/3)/11] x 11 = 25/3.

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

(x-2)² + (x+3)² = (2x-3)² + (x-5)² (x²-4x+4) + (x²+6x+9) = (4x²-12x+9)+(x²-10x+25) 2x² + 2x + 13 = 5x² - 22x + 34 3x² - 24x + 21 = 0 x² - 8x + 7 = 0 (x-7)(x-1) = 0 x = 7 | x = 1 ❌ AB AB = (7) - 2 = 5 BC = (7) + 3 = 10 AD = 2(7) - 3 = 11 DC = (7) - 5 = 2 CA =√(5²+10²) = √125 = 5√5 Draw EF, where F is a point on CA where EF is perpendicular to CA. By SAS, ∆EFA is similar to ∆ADC and ∆CFE is similar to ∆ABC. Let CF = y. EF/FA = DC/AD EF/(5√5-y) = 2/11 EF = (2/11)(5√5-y) ----[1] FE/CF = AB/BC FE/y = 5/10 = 1/2 FE = y/2 ----[2] (2/11)(5√5-y) = y/2 10√5/11 - 2y/11 = y/2 10√5/11 = y/2 + 2y/11 = (11y+4y)/22 15y/22 = 10√5/11 y = (10√5/11)(22/15) = 4√5/3 FE = (4√5/3)/2 = 2√5/3 A = bh = 5√5(2√5/3)/2 = 25/3

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

No need to find AC or h. A,B,C,D are points of a circle. So use the chord theorem to find BE (or DE) and then subtract the areas of two right triangles.

@ybodoN

2 ай бұрын

It works! Simply set the following system of equations: {ab = cd, a + b = 11, c + d = 10, 5² + c² = a² or 2² + b² = d²}. Solutions: a = 125/7, b = −48/7, c = 120/7, d = −50/7 (rejected) or a = 28/3, b = 8/3, c = 20/3, d = 10/3 (accepted).

@PreMath

2 ай бұрын

Thanks for sharing ❤️

Truely exciting mixture of algebra and geometry concepts. ❤

Since we are dealing with two right triangles whose hypotenuse is in common, we can write the system of equation (x − 2)² + (x + 3)² = (2x − 3)² + (x − 5)² leading to 3x² − 24x + 21 = 0 and the solutions x = 1 or x = 7. Since 2x − 3 or x − 5 would be negative with x = 1, we must discard it. So we are left with x = 7 which gives AB = 3, BC = 10, AD = 11 and CD = 2. Then AC = 5√5, sin (CAD) = 2/5√5 and sin (ACB) = 5/5√5 = 1/√5. Now we are dealing with an ASA triangle. So the sinus of the third angle (AEC) is 3/5 and AE = 25/3 and CE = 10/3. Therefore the area of the yellow triangle is ½ (25/3) (10/3) (3/5) = 750/90 = 25/3 square units. Thank you PreMath! 🙏

@PreMath

2 ай бұрын

Excellent! You are very welcome! Thanks for sharing ❤️

(x-2)²+(x+3)²=(x-5)²+(2x-3)² 2x²+2x+13=5x²-22x+34 　　3x²-24x+21=0 x²-8x+7=0　　　　　(x-1)(x-7)=0 x=1 is rejected , x=7 AB=5 CD=2 BE=10 AD=11 ⊿ABE∞⊿CDE ⊿ABE=25s ⊿CDE=4s △AEC=x ⊿ABC=5*10/2=25 ⊿CDA=2*11/2=11 25s+x=25 4s+x=11 21s=14 s=2/3 x=25/3 x=△AEC=Yellow triangle area : 25/3

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

I began just as you did, I found x = 7 and AC = 5.sqrt(5) Then I noticed that triangles EDC and EBA are similar (proportion 5/2). I notes a = ED, then EC = sqrt(a^2 +4) in triangle EDC 10 = BE + EC = (5/2).a + sqrt(a^2 +4) gives that a = 8/3. At that point I have every side length. AE = 25/3, EC = 10/2 and AC = 5.sqrt(5) and I finished with the Heron formula with p = (5/6).(7 +3.sqrt(5)) the half perimeter. I found the area of triangle AEC = 25/3. My solution is longer than yours.

@PreMath

2 ай бұрын

Thanks for sharing ❤️

Very nice method.

@PreMath

2 ай бұрын

Excellent! Glad to hear that! Thanks for sharing ❤️

One interesting question could be added: What is the length BD? Answer: We use the Ptolemy theorem as ABCD is in the circle of diameter[A,C]: We have AD.BC = AB.CD + AC.BD. So 11.10 = 5.2 + 5.sqrt(5).BD We have then that BD = 100/(5.sqrt(5) = 20/sqrt(5) = 4.sqrt(5).

I got the solution with similar triangles but only after a long while, the other method is quite ugly but once you’ve gotten the value of x and the perpendicular heights of the triangles to AC, you can use coordinate geometry and find the intersection of the lines, giving you the value for the height and then find the area using bh/2. Really good problem! 👍

@PreMath

2 ай бұрын

Excellent! Glad to hear that! Thanks for sharing ❤️

I did it slightly differently. I solved for x and AC the same way as in the video. To get the area of ACE I used the areas of the two right trianges ABC and ADC. The areas of the triangles follow these equations: ABC=ABE+AEC =25 (1) ADC=DEC+AEC=11 (2) We note that ABE and EDC are similar triangles. Furthermore, segment DC is the side of EDC that corresponds to segment AB of triangle ABE. We know that DC=2 and AB=5. Thus the area of triangle DEC will be the square of the ratio of these sides compared to the area of triangle ABE. Thus, in terms of areas, DEC=(4/25)*ABE. So we modify equation 2 to get ADC=(4/25)*ABE+AEC = 11 or ABE+(25/4)AEC=(25/4)*11=275/4 (2a) Now take equation (2a)-(1) we get (21/4)*AEC=275/4-25=275/4-100/4=175/4 thus 21*AEC=175 AEC=175/21= 25/3

I like your method! However, as I've done in the past, I like to try the “intersecting line functions” method. It really isn't much different in terms of work. Define a few short variables [1.1]  𝒉 … the height of the '2' sided △ [1.2]  𝒚 … the height of the '5' sided △ [1.3]  𝒒 … the bit in from the margin of the 𝒉 vertical [1.4]  𝒎 … the bit in from the margin of the 𝒚 vertical [1.5]  𝒃 … the base of the whole thing [1.6]  𝒙 … the point along 𝒃 where the two △ cross [1.7]  𝒛 … the height of that crossing Since it makes no difference whether one solves it 'right-side-up' or 'upside down', I chose to flip it over. In any case, [2.1]  𝒃 = √(125) [2.2]  𝒃 = 5√5 [2.3]  𝒃 = 11.18 Y'all might not remember it, but worth remembering are the 'height of a right triangle internally' formulæ. [3.1]  𝒎 = 5² / 𝒃 ... short-side * short-side divided by hypotenuse [3.2]  𝒎 = 5 × 5 ÷ 5√5 [3.3]  𝒎 = 5 / √5 [3.4]  𝒎 = 5 × √5 / √5² [3.5]  𝒎 = √5 [3.6]  𝒎 = 2.2361 [4.1]  𝒚 = 5 × 10 / 𝒃 [4.2]  𝒚 = 5 × 2 × 5 ÷ 5√5 [4.3]  𝒚 = 2√5 [4.4]  𝒚 = 4.4721 [5.1]  𝒒 = 2²/5√5 [5.2]  𝒒 = 4 ÷ 5√5 [5.3]  𝒒 = 0.35777 [6.1]  𝒉 = (2 × 11)/5√5 [6.2]  𝒉 = 1.9677 Then, from these, defining the linear functions of the hypotenuses [7.1]  g(𝒙) = 𝒚𝒙 / (𝒃 - 𝒎) ⊕ 0 [7.2]  f(𝒙) = -𝒉𝒙 / (𝒃 - 𝒒) + ( 𝒉𝒃 / (𝒃 - 𝒒) ) Set those 2 equal to each other and determine what 𝒙 is [8.1]  𝒙 = 2.9814 … and of course plugging into [7.1] [8.2]  g(𝒙) = 1.4907 Well, now the area is kind of obvious [9.1]  Area = ½ base height [9.2]  Area = ½ 𝒃 g(𝒙) [9.3]  Area = 8.333333 ( which is 25⁄3 ) YAY? Completely different route, but produces satisfactory results directly. Very little geometry required. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

@PreMath

2 ай бұрын

Excellent! Glad to hear that! Thanks for sharing ❤️

Thank you!

@PreMath

2 ай бұрын

You are very welcome! Thanks ❤️

شكرا لكم على المجهودات Pythagore 2 fois ...x=7 EAB et ECD semblables S(EAB)=m^2S(ECD) m=CD/AB=2/5 S(ABE)+S(ACE)=25 S(CDE)+S(ACE)=11 S(ABE)=50/3 S(ACE)=25/3

@ 14:18 isolation of h could lead to depression and cognitive decline. ...not in this case though! 🙂

@PreMath

2 ай бұрын

😀 Thanks ❤️

Hi, thank you for this great question and answer of it. I solved it with similitary of ABE and CDE triangles after I found x by using pythagorean relation between ABC and ADC triangels. I thought it would be so easier with this method

Buenas tardes Sres. PreMath. Gracias por este interesante problema geométrico. Tengo una inquietud y necesito su valiosa opinión. Con las medidas obtenidas he construido la gráfica, pero noto que no se aprecian los ángulos rectos B y D = 90 grados. agradeciéndole la atención que se digne en prestarme. Éxitos.

Once x is calculated to be 7. Tan BCA = 5/10, BCA = 26.565 degrees. Tan DAC = 2/11, DAC = 10.305 degrees. AC^2 = 5^2 + 10^2 = 125. AC = 11.18. Angle AEC = 180 - 26.565 - 10.305 degrees. AEC = 143.13 degrees. Sine Law. 11.18/sin143.13 = EC/sin10.305. EC = 11.18 x sin10.305/sin143.13. EC = 3.33. Area yellow = 1/2 x 3.33 x 5. 8.33.

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

For finding height of triangle AEC, I would like to suggest a method to avoid using similar triangles which are prone to mistakes when constructing corresponding sides proportionality equations. Let angle CAD be A and angle BCA be B and height be h. From the found values of the sides, tanA = 2/11 and tanB = 5/10. AC = 5(sqrt 5) = h/tanA + h/tanB = 11h/2 + 10h/5 = 75h/10. Hence h = (2/3)(sqrt 5).

@PreMath

2 ай бұрын

Thanks for sharing ❤️

It is a bit interesting, but need plenty of calculations 😢, 5x^2-22x+34=2x^2+2x+13, 3x^2-24x+21=x^2-8x+7=0, (x-1)(x-7)=0, x=1, too small to reject, so x=7, so the diameter is sqrt(125)=5sqrt(5), and the radius is (5/2)sqrt(5), pairs of sides are 2, 11 and 10, 5, or 2, 2s, 11-2s. and 5, 5s, 10-5s, sqrt(4+4s^2)+5s=10, 4+4s^2=100-100s+25s^2, 21s^2-100s+96=(7s-24)(3s-4)=0, s=24/7, too large, rejected, so s= 4/3, then the triangles are 2×8/3 and 5×20/3, therefore the answer is (1/2×(2×11+5×10)-1/2×(2×8/3+5×20/3))/2=(36 -58/3)/2=50/6.😅😅😅

@PreMath

2 ай бұрын

Thanks for sharing ❤️

guess I'm going back to Pre PreMath

@PreMath

2 ай бұрын

No worries! 😀 Thanks ❤️

I divided this Problem in two Major Parts: A) Easy Part B) Difficult Part Let's go adventuring!! A) Easy Part 1) Quest for X. X must be bigger than 2; X > 2 2) AC^2 = (X - 2)^2 + (X + 3)^2 3) AC^2 = (2X - 3)^2 + (X - 5)^2 4) AC^2 = AC^2; so: 5) (X - 2)^2 + (X + 3)^2 = (2X - 3)^2 + (X - 5)^2 6) Two Solutions : X = 1 and X = 7 7) The only Possible Solution is X = 7 8) AC^2 = 25 + 100 = 125 9) AC^2 = 121 + 4 = 125 10) AC = sqrt(125) = 5*sqrt(5) B) Difficult Part 1) tan CAD = 2/11 2) tan ACB = 5/10 = 1/2 Using Analytic Geometry we have two Lines: Line AD and Line BC; and we want to find the Point of Interception by a System of Two Linear Equations with Two Unknowns. 3) Y = - 2X/11 4) Y = X/2 + sqrt(125)/2 5) As: Y = Y 6) - 2X/11 = X/2 + sqrt(125)/2 7) One Solution : X = - (11*sqrt(5)) / 3 9) Y = - 2X/11 10) Y = - 2 * [ - (11*sqrt(5)) / 3] / 11 ; Y = 22*sqrt(5)) / 33 ; Y = 2*sqrt(5)/3 Y ~ 1,4907 11) Base = 5*sqrt(5) 12) height = 2*sqrt(5)/3 13) Area = (Base * height) / 2 14) 2A = [5*sqrt(5)] * [2*sqrt(5)/3] ; 2A = (10*5) / 3 ; 2A = 50/3 ; A= 50/6 ; A = 25/3 ; A ~ 8,333(3) 15) So, my answer is that the Yellow Triangle Area is equal to 25/3 Square Units, or equal to approx. 8,333 Square Units.

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

Sir solve √(5-x)= 5-x^2 😊😊😊

@ybodoN

2 ай бұрын

5 − x = (5 − x²)² 5 − x = 25 − 10x² + x⁴ x⁴ − 10x² + x + 20 = 0 x = ½ (1 − √17), x = ½ (√21 − 1)

@rakeshkumarrkk5498

Ай бұрын

How u find the roots would u clarify

@ybodoN

Ай бұрын

​@@rakeshkumarrkk5498 Very simply: I used an equation solver... hence the mistakes in my quartic equation (now corrected) 😉 Manually, since we are dealing with a Double Multiplicity-2 (DM2) case, I would start by splitting it into (x² − x − 4) (x² + x − 5) = 0

Let's find the area: . .. ... .... ..... The triangles ABC and ACD are both right triangles, so we can apply the Pythagorean theorem for both of them: AC² = AB² + BC² ∧ AC² = AD² + CD² ⇒ AB² + BC² = AD² + CD² (x − 2)² + (x + 3)² = (2*x − 3)² + (x − 5)² x² − 4*x + 4 + x² + 6*x + 9 = 4x² − 12*x + 9 + x² − 10*x + 25 0 = 3x² − 24*x + 21 0 = x² − 8*x + 7 0 = (x − 7)*(x − 1) ⇒ x = 7 ∨ x = 1 For x=1 not all side lenghts are positive, so the only useful solution is: x = 7: AB = x − 2 = 7 − 2 = 5 AD = 2*x − 3 = 2*7 − 3 = 11 BC = x + 3 = 7 + 3 = 10 CD = x − 5 = 7 − 5 = 2 AC² = AB² + BC² = 5² + 10² = 25 + 100 = 125 ⇒ AC = √125 = 5√5 Now let's add the point F on AC such that AEF and CEF are both right triangles. In this case the triangles AEF and ACD are similar and the triangles CEF and ABC are similar as well. So we can conclude: EF/AF = CD/AD = 2/11 EF/CF = AB/BC = 5/10 = 1/2 (EF/AF)/(EF/CF) = (2/11)/(1/2) (EF/AF)*(CF/EF) = (2/11)*(2/1) CF/AF = 4/11 ⇒ CF = (4/11)*AF AC = AF + CF = AF + (4/11)*AF = (15/11)*AF ⇒ AF = (11/15)*AC ⇒ EF = (2/11)*AF = (2/11)*(11/15)*AC = (2/15)*AC Therefore the area of the yellow right triangle turns out to be: A(ACD) = (1/2)*AC*h(AC) = (1/2)*AC*EF = (1/2)*AC*(2/15)*AC = AC²/15 = 125/15 = 25/3 Best regards from Germany

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

My way of solution ▶ ΔABC is a right triangle with the hypotenuse AC ΔADC is a right triangle with the hypotenuse AC if we write the Pythagorean theorem for ΔABC, we get: AB= x-2 BC= x+3 AB²+BC²= CA² (x-2)²+(x+3)²= CA² if we write the Pythagorean theorem for ΔADC, we get: AD= 2x-3 DC= x-5 AD²+DC²= CA² (2x-3)²+(x-5)²= CA² while CA² is same in both eqautions: (x-2)²+(x+3)²= (2x-3)²+(x-5)² x²-4x+4+x²+6x+9= 4x²-12x+9+x²-10x+25 2x²+2x+13= 5x²-22x+34 3x²-24x+21=0 divided by 3 we get: x²-8x+7=0 Δ= 64-4*1+7 Δ= 64-28 Δ= 36 √Δ= 6 x₁= (8+6)/2 x₁= 7 x₂= (8-6)/2 x₂= 1 AB= x-2 if we put x= 1 AB= -1 ⇒ x= 7 AB= 5 BC= 10 AD= 11 DC= 2 ∠(BEA)= ∠(DEC) = α if α+ β= 90° ∠(EAB)= ∠ (ECD)= β so, ΔABE ~ ΔCDE both triangles are similar: BE/ED = EA/CE= AB/DC AD= 11 EA=y ED= 11-y BC= 10 BE= z CE= 10-z ⇒ z/11-y= 5/2= y/10-z 2y= 50-5z 2y+5z=50 2z= 55-5y 5y+2z= 55 10y+25z= 250 -10y-4z= -110 21z= 140 z= 140/21 z= 20/3 y= 25/3 Ayellow= A(ΔABC) - A(ΔABE) Ayellow= (AB*BC)/2 - (AB*BE)/2´ Ayellow= (5*10)/2 - (5*20/3)/2 Ayellow= 25- (50/3) Ayellow= 25/3 square units ✔

legal, sistema.