Can you find Perimeter and Area of the right triangle? | (Solve) |

[@The_Long_Walk_to_Freedom]

Very clear steps with explanations,thanks.

[@PreMath]

You are very welcome! Thanks for the feedback ❤️🙏

[@alster724]

Easy and solved it on my own. Only difference is that I substituted a²= x first before solving Given AC= 6a²+11 AB= 3a²-2 BC= 7a²-3 A=?, P=? a²= x (6x+11)²= (3x-2)²+(7x-3)² 22x² -186x-108= 0 11x²-93x-54= 0 11 6 1 -9 Cross multiplied then added the factors to obtain -93 (x-9)(11x+6) x= 9, -6/11 Since x > 0 in geometry x= 9 a²= 9 a= 3 AC (hypotenuse) = 65 AB (height)= 60 BC (base)= 25 Area= 750 units² Perimeter= 150 units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@unknownidentity2846]

Let's face this challenge: . .. ... .... ..... ABC is a right triangle, so we can apply the Pythagorean theorem: AB² + BC² = AC² (7a² − 3)² + (3a² − 2)² = (6a² + 11)² 49a⁴ − 42a² + 9 + 9a⁴ − 12a² + 4 = 36a⁴ + 132a² + 121 22a⁴ − 186a² − 108 = 0 11a⁴ − 93a² − 54 = 0 a² = {93 ± √[93² − 4*11*(−54)]}/(2*11) a² = [93 ± √(8649 + 2376)]/22 a² = (93 ± √11025)/22 a² = (93 ± 105)/22 Since a²>0, the only useful solution is: a² = (93 + 105)/22 = 198/22 = 9 ⇒ AB = 7a² − 3 = 7*9 − 3 = 60 ∧ BC = 3a² − 2 = 3*9 − 2 = 25 ∧ AC = 6a² + 11 = 6*9 + 11 = 65 Now we are able to calculate the area A and the perimeter P of the triangle: A(ABC) = (1/2)*AB*BC = (1/2)*60*25 = 750 P(ABC) = AB + BC + AC = 60 + 25 + 65 = 150 Best regards from Germany

[@PreMath]

Excellent job! Thanks for sharing ❤️

[@raghvendrasingh1289]

👍 After framing quadratic equation we will put a^2 = 9 t to reduce the coefficients 33 t^2 - 31 t - 2 = 0 t = 1 ( because it a^2 = 9 t) Now t = 1 gives a^2 = 9 sides are 25 , 60 , 65 area = 750 perimeter = 150

[@PreMath]

Excellent! Thanks for sharing ❤️

[@cyruschang1904]

Let x = a^2 (3x - 2)^2 + (7x - 3)^2 = (6x + 11)^2 22x^2 - 186x - 108 = 0 11x^2 - 93x - 54 = 0 (x - 9)(11x + 6) = 0 x = 9, a = 3 Perimeter = 16x + 6 = 150 Area = (3x - 2)(7x - 3)/2 = 750

[@michaelkouzmin281]

To my mind it is easier to START with substitution x=a^2 than to apply it in the middle ;))

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@quigonkenny]

Technically speaking, it's not even necessary to use substitution, as it's not really a you're solving for, but a². Whether a = 3 or a = -3 doesn't affect the side lengths of the triangle, as a never appears as a direct component of their lengths. It only appears as a².

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! Thanks for the feedback ❤️

[@Abdelfattah-hr8tt]

شكرا لك من المغرب

[@PreMath]

You are very welcome! Thanks for the feedback ❤️ Love and prayers from the USA! 😀

[@ekoi1995]

(7a^2-3)^2+(3a^2-2)^2=(6a^2+11)^2 a= 3 b=3(3)^2-2=25 h=7(3)^2-3=60 A= 1/2*25*60 = 750 c=6a^2+11=65 P = 25+60+65 = 150

[@Birol731]

My way of solution ▶ In this right triangle ΔABC, we have : [AB]= 7a²-3 [BC]= 3a² -2 [CA]= 6a²+11 By applying the Pythagorean theorem, we can write: [AB]² + [BC]² = [CA]² (7a²-3)² + (3a² -2)² = (6a²+11)² 49a⁴ - 42a² +9 + 9a⁴ - 12a² + 4 = 36a⁴ + 132a² + 121 58a⁴ - 54a²+ 13 = 36a⁴ + 132a² + 121 22a⁴ - 186a² - 108=0 Let's define a²= t ⇒ 22t² - 186t -108=0 both sides divided by 2, we get: 11t² - 93t - 54= 0 Δ= 93²-4*11*(-54) Δ= 11.025 √Δ= 105 t= (93+105)/2*11 t₁= 9 t= (93-105)/2*11 t₁= -6/11 ❌ t < 0 ⇒ t= 9 ⇒ a²= 9 a= 3 [AB]= 7a²-3 = 7*3²-3 = 60 [BC]= 3a² -2 = 3*3²-2 = 25 [CA]= 6a²+11 = 65 P(ΔABC)= 60 + 25 + 65 P(ΔABC)= 150 length units b) A(ΔABC)= [AB]*[BC]/2 A(ΔABC)= 60*25/2 A(ΔABC)= 750 square units

[@marcgriselhubert3915]

In triangle ABC: AC^2 = AB^2 + BC^2 or 36.(a^4) +132.(a^2) +121 = 9.(a^4) - 12.(a^2) +4 +49.(A^4) -42.(a^2) + 9, or 22.(a^4) -186.(a^2) -108 = 0, or 11.(a^4) -93.(a^2) - 54 = 0. Delta = 93^2 +4.11.54 = 11025 = 105^2, so a^2 = (93 +105)/22 = 9 (the other possibility is rejected as beeing negative) So a^2 = 9, and AC = 65, AB = 60, BC = 25. The perimeter of ABC is 65 + 60 + 25 = 150, and the area of ABC is (1/2).AB.BC = (1/2).60.25 = 750 (Easy.) (I am always surprised to see how you solve second degree equations!)

[@unknownidentity2846]

It is really impressive how Premath solves the quadratic equations every time. Of course, the abc-formula also works, but using this approach avoids handling more inconvenient numbers in between.

[@marcgriselhubert3915]

@@unknownidentity2846 See the present example: How is it possible to guess that it is a good idea to use the fact that -93 = -99 +6 whithout knowing in advance that (x - 9) will be a factor, or if you prefer that 9 will be a solution? (If 9 is an evident solution, no need to factorize, it is quicker to use the product of the solutions to find the other one.)

[@unknownidentity2846]

@@marcgriselhubert3915 You are absolutely right. I think that took a lot of practice. On this channel I found an interesting video called “How to Solve Quadratic Equations using Three Methods - When Leading Coefficient is Not One” published in 2018 that shows how this approach works. Best regards from Germany

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@sorourhashemi3249]

Thanks easy ❤

[@thewolfdoctor761]

I did it the same way except I substituted x for a^2 before using the Pythagorean theorem rather than after. Either way results in the same answer.

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@CherifaGhanouchi]

Thé périmètre of triangle ABC IS 7a^2-3+3a^2-2+6a^2+11 P=16a^2+6 Thé area of triangle ABC IS A=h×b/2 =(7a^2-3)(3a^2-2)/2 =7a^2-9a^2+6 A=-2a^2+6

[@AmirgabYT2185]

P=150 S=750

[@PreMath]

Excellent! Thanks for sharing ❤️

[@sergioaiex3966]

Solution: Perimeter = (7a² - 3) + (3a² - 2) + (6a² + 11) Perimeter = 16a² + 6 ... ¹ Area = ½ (3a² - 2) (7a² - 3) Area = ½ (21a⁴ - 9a² - 14a² + 6) Area = ½ (21a⁴ - 23a² + 6) ... ² Pythagorean Theorem [(3a² - 2)]² + [(7a² - 3)]² = [(6a² + 11)]² 9a⁴ - 12a² + 4 + 49a⁴ - 42a² + 9 = 36a⁴ + 132a² + 121 58a⁴ - 54a² + 13 = 36a⁴ + 132a² + 121 22a⁴ - 186a² - 108 = 0 (÷2) 11a⁴ - 93a² - 54 = 0 Let's consider a² = k 11k² - 93k - 54 = 0 ∆ = (-93)² - 4 . 11 . (-54) ∆ = 8.649 + 2.376 ∆ = 11.025 √(∆) = 105 k = (93 ± 105)/22 k' = -12/22 = - 6/11 Rejected k'' = 198/22 = 9 Accepted a² = k a² = 9 a = 3 ... ³ Replacing ³ in ¹ to find the Perimeter Perimeter = 16a² + 6 Perimeter = 16 (3)² + 6 Perimeter = 150 Units Replacing ³ in ² to find the Area Area = ½ [21. (3)⁴ - 23 . (3)² + 6)] Area = ½ (21. 81 - 23 . 9 + 6) Area = ½ (1.701 - 207 + 6) Area = ½ . 1.500 Area = 750 Square Units Thus: Perimeter = 150 Units ✅ Area = 750 Square Units ✅

[@PreMath]

Excellent! Thanks for sharing ❤️

[@quigonkenny]

Triangle ∆ABC: AB² + BC² = CA² (7a²-3)² + (3a²-2)² = (6a²+11)² (49a⁴-42a²+9) + (9a⁴-12a²+4) = (36a⁴+132a²+121) 58a⁴ - 54a² + 13 = 36a⁴ + 132a² + 121 22a⁴ - 186a² - 108 = 0 11a⁴ - 93a² - 54 = 0 11u² - 93u - 54 = 0 --- u = a² 11u² - 99u + 6u - 54 = 0 11u(u-9) + 6(u-9) = 0 (u-9)(11u+6) = 0 u = 9 | u = -6/11 ❌ u = a² ≥ 0 a² = 9 AB = 7(9) - 3 = 63 - 3 = 60 BC = 3(9) - 2 = 27 - 2 = 25 CA = 6(9) + 11 = 54 + 11 = 65 P = 60 + 25 + 65 [ P = 150 units ] A = bh/2 = BC(AB)/2 A = 25(60)/2 = 25(30) [ A = 750 sq units ] Note that if one wishes to know the value of a, then it can be either 3 or -3. The usual requirement for a positive value of a variable involved in the dimensions of a polygon is unnecessary, as a only ever appears squared in the values. This is why I did not bother to solve for a, but rather a².

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL : 01) Solving the Equation : 02) (7a^2 - 3)^2 + (3a^2 - 2)^2 = (6a^2 + 11)^2 03) Real Positive Solution : a = 3 04) Area = 60 * 25 / 2 ; A = 30 * 25 ; A = 3 * 250 ; A = 750 sq un 05) Perimeter = 60 + 25 + 65 ; P = 150 lin un Thus, OUR ANSWER : Area equal 750 Square Units and Perimeter equal 150 Linear Units.

[@PreMath]

Super work! Thanks for sharing ❤️

[@ilhammaharramov9927]

150

[@PreMath]

Thanks for sharing ❤️

[@EPaozi]

a^2 quel intérêt ?

[@misterenter-iz7rz]

Sorry, not my cup of tea.😢

[@wackojacko3962]

On the onset by inspection I automatically knew (a) around the perimeter was some form of the Number of the Beast, either 666 or 999. 😊

[@PreMath]

😀 That’s an interesting observation! 👍