Can you find Perimeter and Area of the triangle? | (Trigonometry) |
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- Опубликовано: 22 дек 2024
- Learn how to find the Perimeter and Area of the triangle. Important Geometry and Algebra skills are also explained: Law of Cosines; area of a triangle formula; perimeter. Step-by-step tutorial by PreMath.com
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💯👍🎄 Merry Christmas 🎁 Be happy and all the best to you, Professor!
Merry Christmas! Wishing you a happy holiday season as well! 😊🙏
Wow Sirr ❤❤ Very Interesting Video❤❤❤ Thanks for sharing ❤❤
Thank you!
Perimeter = 7.5 , area =1.623 Sq.units(15× root3÷16)
P=7,5
S=15√3/16≈1,624
*Let's name t = 3.x -4, The side lenthes of the triangle are now BA = t - 1 , BC = t and AC = t + 1.
The law of cosines in the triangle ABC gives: (t + 1)^2 = t^2 + (t - 1)^2 -2.t.(t - 1).cos(120°), or: t^2 +2.t + 1 = t^2 + t^2 -2.t + 1 +t^2 - t (as cos(120°) = -1/2).
That gives: 2.t^2 - 5.t = 0 and then t = 5/2 (t cannot be equal to 0 as the lengthes are positive). So t = 3.x -4 = 5/2 (and x = 13/6).
*The perimeter of the triangle is 3.t = 15/2.
*The area of the triangle (1/2).BA.BC.sin(120°) = (1/2).t.(t -1).(sqrt(3)/2) = (1/2).(5/2).(3/2).(sqrt(3)/2) = (15/16).sqrt(3).
Nice observation that simplifies the algebra!
Sir
1) we may write the side
3x -4 =a
Then 3x - 5=a -1
3x -3= a +1
2)Perimeter =3a
3)1/2*a*(a-1)sin120
4)
Cos 120 =[(a-1)^2+a^2 -(a+1)^2]/[2a(a-1)]
=(a^2-2a +1+a^2 -a^2-2a-1)/2a(a-1)
=(a^2-4a)/2a(a-1)
=(a -4)/(2a-2)
> - 1/2 =(a-4)/(2a -2)
> a=5/2 units
5)Perimeter =3a=3*5/2=15/2 units
6)Hence the legs of ang 120 degrees
are a=5/2 and a-1=5/2 -1=3/2
Area =1/2*5/2* 3/2*sin 120
=1/2*5/2*3/2*√3/2
=1*5*3*√3/2*2*2*2
=15√3/16 sq units
[ ***
please note that this solution does not have any - ve value of length and no question of rejecting the - -ve arose. ]
Yes, substituting a, (a-1) & (a+1) for the side lengths simplifies the Cosine Rule calculations and eliminates the need to solve for x.
Thanks chalenging. Draw a right line from A and mark it ad F. We have a triangle AFB in which B angle is 60 and A angle is 30. The side is FB is half of the cord AB. = 3x-5/2 focus on right triangle AFC and by phytagorus theorum AF^2+{( 3x-5/2/2)+(3x-4)}^2=(3x-3)
^2)===> X=2.16
I solved without trigonometry tracing the height AH and seeing that AHB is a right triangle of 30,60,90 degree whose hypotenuse is 3x-5. It leads to the same quadratic equation without the cosine law
I'm so glad to be one of your fallwers
Thanks dear ❤️🙏
You are the best! ❤️
Thanks for the feedback ❤️
Very nice, many thanks, Sir! Merry Christmas!
φ = 30° → cos(4φ) = -cos(6φ - 4φ) = -cos(2φ) = -sin(φ) = -1/2
sin(4φ) = sin(6φ - 4φ) = sin(2φ) = cos(φ) = √3/2
∆ ABC → ABC = 4φ; AB = 3x - 5; BC = 3x - 4; AC = 3x - 3; area & perimeter ∆ ABC = ?
3x - 3 ∶= a → 3x - 4 = a - 1 → 3x - 5 = a - 2 →
a^2 = (a - 1)^2 + (a - 2)^2 - 2(a - 1)(a - 2)cos(4φ) = (a - 1)^2 + (a - 2)^2 + (a - 1)(a - 2) →
a^2 - 9a + 7 = 0 → a1,a2 = (1/4)(9 ± 5) → a1 = 1 → a - 2 < 0 ≠ solution →
a2 = 7/2 = 3x - 3 → x = 13/6 → 3x - 3 = 7/2 → 3x - 4 = 5/2 → 3x - 5 = 3/2 →
perimeter ∆ ABC = 15/2 → area ∆ ABC = (1/2)sin(4φ)(a - 1)(a - 2) = 15√3/16
*Solução:*
Seja y = 3x - 5. Daí,
BC= y + 1 e AC = y + 2. Você pode usar a lei dos cossenos, porém , vamos construir uma perpendicular AD em relação ao lado BC. Assim,
O ângulo ABD = 60° e, usando a definição de seno e cosseno no triângulo retângulo ∆ABD, temos:
BD = y/2 e AD=(y√3)/2. Assim, DC = y/2 + y+1 = (3y+2)/2. Por Pitágoras no ∆ACD:
(y+2)² = [(y√3)/2]² + [(3y+2)/2]²
y²+4y+4 = 3y²/4+(9y²+12y+4)/4
y²+4y+4 = (12y²+12y+4)/4
y²+4y+4 = 3y²+3y+1
2y² - y -3 = 0, com y > 0. Resolvendo essas equação do 2° grau, obtemos y = 3/2, logo:
AB=3/2, BC=5/2, AC=7/2 e AD=3√3/4. Temos:
*_Perímetro=_* 3/2 + 5/2 + 7/2 = *15/2 U*
*_Área=_* AD×BC/2 =
= 5/2 × 3√3/8 = *15√3/16 U.Q*
I solved the problem using your method:
x= 13/6; Р=7,5; S= a×b×sin(60°)/2=
=(3/2×5/2×√3/2)/2=15√3/16
Thanks sir!❤
Bom dia Mestre
Irei usar a Lei dos cossenos e produtos notáveis p resolver essa questão
Grato
1/ Label AB= (3x-5)= a
-> AC= (3x-5) +2=(a+2) and BC=(a+1)
2/ Drop the height AH to BC-> AH = a sqrt3/2 and BH= a/2 ( the triangle AHB is a 30/90/60 one)
--> HC = a/2 + a+1= (3a+2)/2
By using the Pythagorean theorem
sq (asqt3/2) + sq((3a+2)/2))= sq( a+2)
--> 2sqa-a-3 = 0
-> a= 3/2 ( negative result rejected)
--> 3x -5 = 3/2
x= 13/6
Perimeter= 7.5 units
Area= 15sqr3/ 16😅😅😅
Perimeter = (3x - 3) + (3x - 4) + (3x - 5) = 9x - 12
Area = (height)(base)/2 = (Sin60°)(3x - 5)(3x - 4)/2 = (√3)(3x - 5)(3x - 4)/4
[(Sin60°)(3x - 5)]^2 + [(Cos60°)(3x - 5)]^2 = (3x - 5)^2
To find x
(3x - 3)^2 = (3x - 4)^2 + (3x - 5)^2 - 2(3x - 4)(3x - 5)(Cos120°)
18x^2 - 63x + 52 = 0
x = [63 +/- √(63)(63) - 4(18)(52)] / 36 = [21 +/-√(21)(21) - 4(2)(52)] / 12 = 13/6 or 4/3 (the base 3x - 4 cannot be zero)
Perimeter = 9(13/6) - 12 = 39/2 - 12 = 7.5
Area = (√3)(13/2 - 5)(13/2 - 4)/4 = (√3)(3/2)(5/2)/4 = (15√3)/16
Dropping a perpendicular from point A to meet extended line CB at point D.
Let BD = length Y.
Then AB = 2Y, since angle DAB = 30 degrees.
BC = 2Y + 1, AC = 2Y +2, DC = 3Y + 1.
In triangle ABD, AD^2 = (2Y)^2 - Y^2.
In triangle ACD, AD^2 = (2Y + 2)^2 - (3Y + 1)^2.
Equating two values for AD^2.
4Y^2 - Y^2 = 4Y^2 + 4 Y + 4 - 9Y^2 - 6Y -1.
8Y^2 - 2Y - 3 = 0.
(4Y - 3) (2Y + 1).
Y = 3/4 = 0.75.
Thus AB = 2Y = 1.5.
BC = 2Y + 1 = 2.5.
AC = 2Y + 2 = 3.5.
So perimeter = 7.5.
AD = AB cos 30 = 1.299.
Area = 0.5 x 2.5 x 1.299 = 1.624.
This is fine except that you could have used cos (30) = (√3)/2 to get the exact solution with the radical. Alternatively, since you have the lengths of all 3 sides, you could have used Heron's formula.
Your method is great for viewers who don't know the law of cosines!
Col teorema del coseno risulta 18x^2-63x+52=0..x=39/18,x=4/3(no)..per cui i lati sono 7/2,3/2,5/2...A=(1/2)(3/2)(5/2)sin120=15√3/16..P=15/2
Here there is another sol
Drop a perpendicular AD on extended CB
In 🔺 ADB is a special 🔺 of 30-60-90
So if BD =1 , AB =2
Hence AB =3x -5 = 2
> x =7/3
BC =3x -4 =3
AC = 3x -3 =4
Hence the sides of triangle ABC are 2,3,4
Perimeter = 2+3+5=9 units
Area =1/2*2*3*sin120
=3sin 120
=3*√3/2=3√3/2sq units
This have so many solutions as we take different values of BD
Solution:
First, we've to calculate x, by using the Law of Cosines:
(3x - 3)² = (3x - 4)² + (3x - 5)² - 2 (3x - 4) (3x - 5) cos 120°
9x² - 18x + 9 = 9x² - 24x + 16 + 9x² - 30x + 25 + 9x² - 15x - 12x + 20
9x² - 18x + 9 = 27x² - 81x + 61
18x² - 63x + 52 = 0
x = (63 ± √225)/36
x = (63 ± 15)/36
x' = 78/36 = 39/18 = 13/6 (Accepted)
x" = 48/36 = 12/9 = 4/3 (Rejected)
Therefore x = 13/6
Now we have to calculate the sides length
3x - 5 = 3 (13/6) - 5 = 39/6 - 5 = (39 - 30)/6 = 9/6 = 3/2
3x - 4 = 3 (13/6) - 4 = 39/6 - 4 = (39 - 24)/6 = 15/6 = 5/2
3x - 3 = 3 (13/6) - 3 = 39/6 - 3 = (39 - 18)/6 = 21/6 = 7/2
Area = ½ × 3/2 × 5/2 × sin 120°
Area = ½ × 3/2 × 5/2 × √3/2
Area = (15√3)/16 Square Units ✅
Area ≈ 1.6237 Square Units ✅
Perimeter = 3/2 + 5/2 + 7/2
Perimeter = 15/2 Units ✅
Perimeter = 7.5 Units ✅
This has so many solutions. May see and comment on the two solutions offered by me
Thanks professor for sharing this interesting video!
Merry Christmas to all who celebrate it.
I hope to do not wrong to wish a happy winter solstice to everyone 😊
To my opinion this is not political, not dependent on what kind of belief one has, including atheists, even of no significance if earth is flat or a globe.
It can be an occasion to send best wishes to anyone, like peace, health and happiness. For those who live in the north its the expectation of light and warmth will come back.
So kind of you dear🙏❤️
Merry Christmas! Wishing you a happy holiday season as well! 😊🙏
Stay blessed 😀
7.5×2.95÷2=11.0625
First and foremost as a US citizen I follow the Constitution of the United States of America. If anyone is offended by that, I don't care! Be offended...🤣. @ 7:43 , never say never! Things get very strange very fast in this 4-dimensional space-time continuum.😊
😀
Thanks for sharing ❤️
Acertei !!!!!
Excellent!
Glad to hear that!
Thanks for the feedback ❤️
Yhank you. I thought that x had to be intyeger.
Let's face this challenge:
.
..
...
....
.....
We should be able to find the value of x by applying the law of cosines. With y=3x−5 we obtain:
AC² = AB² + BC² − 2*AB*BC*cos(∠ABC)
(3x − 3)² = (3x − 5)² + (3x − 4)² − 2*(3x − 5)*(3x − 4)*cos(120°)
(y + 2)² = y² + (y + 1)² − 2*y*(y + 1)*(−1/2)
(y + 2)² = y² + (y + 1)² + y*(y + 1)
y² + 4y + 4 = y² + y² + 2y + 1 + y² + y
0 = 2y² − y − 3
0 = 2y² + 2y − 3y − 3
0 = 2y(y + 1) − 3(y + 1)
0 = (2y − 3)(y + 1)
First solution:
y + 1 = 0
⇒ y = −1
⇒ x = (y + 5)/3 = (−1 + 5)/3 = 4/3
⇒ AC = 3x − 3 = 3*(4/3) − 3 = 4 − 3 = 1
⇒ AB = 3x − 5 = 3*(4/3) − 5 = 4 − 5 = −1
⇒ BC = 3x − 4 = 3*(4/3) − 4 = 4 − 4 = 0
This is not a valid solution.
Second solution:
2y − 3 = 0
⇒ y = 3/2
⇒ x = (y + 5)/3 = (3/2 + 5)/3 = (3/2 + 10/2)/3 = (13/2)/3 = 13/6
⇒ AC = 3x − 3 = 3*(13/6) − 3 = 13/2 − 3 = 13/2 − 6/2 = 7/2
⇒ AB = 3x − 5 = 3*(13/6) − 5 = 13/2 − 5 = 13/2 − 10/2 = 3/2
⇒ BC = 3x − 4 = 3*(13/6) − 4 = 13/2 − 4 = 13/2 − 8/2 = 5/2
Now we are able to calculate the area and the perimeter of the triangle:
A(ABC) = (1/2)*AB*BC*sin(∠ABC) = (1/2)*(3/2)*(5/2)*sin(120°) = (15/8)*(√2/2) = (15/16)√2
P(ABC) = AB + BC + AC = 3/2 + 5/2 + 7/2 = 15/2
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