Find the Area of a Green Triangle in an Unequally Trisected Right Triangle | Fast & Easy Explanation

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  • เผยแพร่เมื่อ 5 ก.พ. 2025

ความคิดเห็น • 45

  • @Osama-js2gw
    @Osama-js2gw 3 ปีที่แล้ว +3

    Wonderful as usual

    • @PreMath
      @PreMath  3 ปีที่แล้ว +2

      Thank you so much Hanafy for taking the time to leave this comment. You are awesome 👍
      Take care dear and stay blessed😃
      Love and prayers from the USA!
      Jazakallah😀

    • @Osama-js2gw
      @Osama-js2gw 3 ปีที่แล้ว +1

      @@PreMath I am a teacher of English in Saudi Arabia. I Like math very much. I always visit your channel. I am always impressed by your methods of solution.

    • @PreMath
      @PreMath  3 ปีที่แล้ว +2

      @@Osama-js2gw
      I'm sure you are also an awesome teacher. Thank you so much my friend for your continued love and support. Take care dear and stay blessed😃

  • @philipkudrna5643
    @philipkudrna5643 3 ปีที่แล้ว +1

    Nice to feature Al-Kashi‘s law of cosines and the sine-Formula for the area of a triangle! Thank you!

    • @PreMath
      @PreMath  3 ปีที่แล้ว +1

      You're very welcome Philip!
      Thanks dear for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃

  • @ironman1518.
    @ironman1518. 3 ปีที่แล้ว +1

    Been MANY years since I studied that but you helped me remember the sine and cosine formulas for non right triangles. Thank you!

    • @PreMath
      @PreMath  3 ปีที่แล้ว +1

      Thanks Fred for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard

  • @aashsyed1277
    @aashsyed1277 3 ปีที่แล้ว +2

    This is the coolest Channel I have seen

    • @PreMath
      @PreMath  3 ปีที่แล้ว +2

      Thanks Syed dear for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard

    • @aashsyed1277
      @aashsyed1277 3 ปีที่แล้ว

      @@PreMath I will stay blessed thanks 😊😊😌 I feel relaxed 😌😊😊

  • @jerrymouse3420
    @jerrymouse3420 3 ปีที่แล้ว +2

    I know the formulae for sine and cosine rules, but wanted to solve this one just using Eucledian geometry,and without any trigonometric formulae.Glad that I did it using eucledian geometry only.Wonderful pblm..thanks for sharing.

    • @PreMath
      @PreMath  3 ปีที่แล้ว +2

      Well done!
      Thanks Jerry for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃

  • @mccannms
    @mccannms 3 ปีที่แล้ว +2

    Wow - Instead of taking the sqrt of BC in the first calculation, then squaring again in the second calculation, you just left as sqrt of 49. I never would have thought of that! :)

    • @PreMath
      @PreMath  3 ปีที่แล้ว +1

      Thanks Mark for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard

  • @johnbrennan3372
    @johnbrennan3372 3 ปีที่แล้ว +1

    Very nice method. Another way of doing this is to work on triangle PBC. Using cosine rule gives ||CB|=7.Then using sine rule find measure of angle CBP That gives the remaining two angles in triangle APB. Using sine rule in triangle APB gives x.Then area of green triangle equals half of x by 3 by sine of 120 degrees.

    • @PreMath
      @PreMath  3 ปีที่แล้ว +1

      Great tip!
      Thanks John for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard

  • @AmirgabYT2185
    @AmirgabYT2185 4 วันที่ผ่านมา +1

    S=99√3/8≈21,435≈21,44 🔥

  • @Okkk517
    @Okkk517 2 ปีที่แล้ว

    An alternative method is to use analytical geometry: find the coordinates of Point D by the intersection of Circles x^2 + y^2 = 9 and x^2 + (y-7)^2 = 25. You obtain D(15*sqrt(3)/14, 33/14). Then the height of the triangle is 33/14 and the base can be found using tan(120-theta) where tan theta = Dx/Dy=(15*sqrt(3)/14)/(33/14). Now AB= Dx + tan(120-theta) * Dy which leads to AB=18.19 and therefore the required area A=1/2 * AB * Dy=21.43.

  • @MathZoneKH
    @MathZoneKH 3 ปีที่แล้ว +1

    Pretty good sir!

    • @PreMath
      @PreMath  3 ปีที่แล้ว +1

      Thanks for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard

    • @MathZoneKH
      @MathZoneKH 3 ปีที่แล้ว

      Stay safe sir! Thanks anyway sir

  • @Awdcguk
    @Awdcguk 3 ปีที่แล้ว

    8:50 can some one explain?
    I learned that A=a×b× sin x ,x

  • @aashsyed1277
    @aashsyed1277 3 ปีที่แล้ว +2

    You are so awesome 😎😎😎😎

    • @PreMath
      @PreMath  3 ปีที่แล้ว +1

      You are the best Syed!

    • @aashsyed1277
      @aashsyed1277 3 ปีที่แล้ว

      @@PreMath thanks

  • @זאבגלברד
    @זאבגלברד 3 ปีที่แล้ว +1

    When you have BC you can find angle PBC then sin of PBA then area using the formula of the triangle's area, with one side and 3 angles. (With a calculator it is even easier).

    • @PreMath
      @PreMath  3 ปีที่แล้ว +1

      Thanks dear for the feedback. You are awesome.
      Cheers

    • @aashsyed1277
      @aashsyed1277 3 ปีที่แล้ว

      @@PreMath you are more awesome 😎💯😎 enjoy your life

  • @2csmoke
    @2csmoke 2 ปีที่แล้ว

    Elegant solution

  • @waltjhall
    @waltjhall 2 ปีที่แล้ว

    There should have been a comment that this is a one dimensional figure. The drawing makes it look 3 dimensional! I would never have thought the third angle was 120 degrees.

  • @India-jq7pi
    @India-jq7pi 3 ปีที่แล้ว +1

    Thank you sir

    • @PreMath
      @PreMath  3 ปีที่แล้ว +2

      Thanks for the visit! You are awesome Gowri👍 Take care dear and stay blessed😃 Kind regard

    • @aashsyed1277
      @aashsyed1277 3 ปีที่แล้ว

      @@PreMath you are awesome so much and the coolest Channel I have seen

  • @shadrana1
    @shadrana1 3 ปีที่แล้ว

    Use the Cosine Rule for triangle BPC,
    CB^2=CP^2+PB^2-2*CP*BP cos120
    = 5^2+3^2-2*5*3 cos 120
    =25+9-30*(-1/2)
    =25+9+15=49
    CB= sqrt(49)=7.
    Use the Cosine Rule again on triangle BPC
    CP^2=PB^2+CB^2-2*PB*CB*cos PBC
    5^2 =3^2+7^2-2*3*7*cos PBC
    25 =9+49-42 cos PBC
    -33 = -42 cos PBC
    cos PBC = 11/14
    Height of Green Triangle APB =PB*cosPB3*(11/14)
    =3*(11/14)=33/14
    Now,consider triangle APB,
    Angle PAB=180-120-arcsine(11/14)=60-arcsine(11/14)
    =60-51.7867893=8.2132107
    We now use the Sine Rule,
    PB/sin(8.2132107)=AB/sin(120)
    AB=PB*sin(120)/sin(8.2132107)
    =3*sqrt(3)/(2*0.1428571428)=18.18640154
    Area of green triangle APB=(AB*Ht. green triangle)/2
    ={(18.18640154*(33/14)}/2
    =(18.18640154*(33/28)
    =21.43397324 square units. and that is your answer.
    Assisted by calculator.
    This is a good place to stop.

  • @sravanijasti111
    @sravanijasti111 3 ปีที่แล้ว +1

    Wow 😍

    • @PreMath
      @PreMath  3 ปีที่แล้ว +1

      Thanks Sravani for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard

  • @فراسمعابره-ج5خ
    @فراسمعابره-ج5خ 2 ปีที่แล้ว

    هذا الشكل الهندسي والمطلوب إيجادها مساحة المنطقه الخضراء وحسب اليابان التي أجرتها قبل عدة أيام فيجب علينا إيجاد قبل الزاويه bap والتي مفارقتها حسابيا ضلع القاعدة بالزاويه 120 =٣ سم وهو يساوي تماما بالزاويه 90=2.77سم اي انه نقص
    .223 ملم على طول 1.321 سم
    فإن الحسابات تكون كمل يلي:
    الجزءالاول.2×1.2÷2+الجزء الثانيوحساباته كأآتي1.321×(30÷2.23)_1.321×2.77÷2
    +2.67×1.2÷2=مساحة المنطقه الخضراءسم

  • @theophonchana5025
    @theophonchana5025 3 ปีที่แล้ว

    #Cosine #Trigonometry

  • @theophonchana5025
    @theophonchana5025 3 ปีที่แล้ว

    Cosine 120°

  • @theophonchana5025
    @theophonchana5025 3 ปีที่แล้ว

    #LawofCosines

  • @theophonchana5025
    @theophonchana5025 3 ปีที่แล้ว

    cos 120° = -0.5

  • @theophonchana5025
    @theophonchana5025 3 ปีที่แล้ว

    Trigonometry

  • @theophonchana5025
    @theophonchana5025 3 ปีที่แล้ว

    Cosine