Can you find Perimeter and Area of the right triangle? | (Solve) |
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- เผยแพร่เมื่อ 5 ก.พ. 2025
- Learn how to find the Perimeter and Area of the right triangle. Important Geometry and Algebra skills are also explained: Pythagorean Theorem; area of a triangle formula; perimeter. Step-by-step tutorial by PreMath.com
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Very clear steps with explanations,thanks.
You are very welcome! Thanks for the feedback ❤️🙏
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
Excellent!
Thanks for sharing ❤️
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
Excellent job!
Thanks for sharing ❤️
To my mind it is easier to START with substitution x=a^2 than to apply it in the middle ;))
Excellent!
Thanks for the feedback ❤️
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.
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
👍
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
Excellent!
Thanks for sharing ❤️
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
Thank you!
You are very welcome!
Thanks for the feedback ❤️
Thanks easy ❤
شكرا لك من المغرب
You are very welcome!
Thanks for the feedback ❤️
Love and prayers from the USA! 😀
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.
Excellent!
Thanks for the feedback ❤️
(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
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 ✅
Excellent!
Thanks for sharing ❤️
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.
Super work!
Thanks for sharing ❤️
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!)
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.
@@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.)
@@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
Excellent!
Thanks for the feedback ❤️
I know the Pythagorean formula is easier, But, can the problem be solved using Heron's formula to find the AREA of the triangle and find the value of (a^2) or (a) and use that to find the PERIMETER?
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
P=150
S=750
Excellent!
Thanks for sharing ❤️
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.
150
Thanks for sharing ❤️
a^2 quel intérêt ?
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. 😊
😀
That’s an interesting observation! 👍