Can you find the area of the Green shaded region? | (Triangle and squares) |

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  • Опубліковано 9 січ 2025

КОМЕНТАРІ • 57

  • @nunoalexandre6408
    @nunoalexandre6408 Рік тому +3

    Love it!!!!!!!!!!!

    • @PreMath
      @PreMath  Рік тому +1

      Thank you so much ❤️

  • @ybodoN
    @ybodoN Рік тому +4

    △ABC is not necessarily isosceles. However, since DEGF is a square, DE is parallel to AB. Therefore, by the intercept theorem, △ABC ~ △DEC.
    So, if h is the height of △ABC, we have (h − √36) / √36 = h / √676 ⇒ h = 39/5 and the area of the green region is ½ (26) 39/5 − 36 = 327/5 cm².

  • @engralsaffar
    @engralsaffar Рік тому +4

    The way I did it is:
    Side of the red square = sqrt(676)=26
    Side of the blue square = sqrt(36)=6
    Base of the green triangle = (26-6)/2=10
    If we combine the 2 green triangles, we get a big green triangle similar to the small one in the top
    The base of the big combined green triangle =20, the height is=6
    The base of the small green triangle =6,the height =x
    x/6 = 6/20
    x=36/20=9/5
    Area of the big combined green triangle =0.5*20*6=60
    Area of the small green triangle =0.5*6*9/5=27/5=5.4
    Total green area =60+5.4=65.4 square units

  • @marcgriselhubert3915
    @marcgriselhubert3915 Рік тому +4

    This problem is simpler than usual.

  • @yalchingedikgedik8007
    @yalchingedikgedik8007 Рік тому

    Thanks Sir
    That’s very useful method
    Glades

  • @gaylespencer6188
    @gaylespencer6188 Рік тому +1

    Found arctan of angle CBA, then used that to find height of triangle.

  • @quigonkenny
    @quigonkenny Рік тому

    By similarities pink quadrangle is a square, so AB is the same length as all the other sides.
    Area of pink square = sₚ²
    676 = sₚ²
    sₚ = √676 = √(4)(169) = 2(13) = 26
    Area of blue square = sᵤ²
    36 = sᵤ²
    sᵤ = √36 = 6
    Let D be a point on AB where CD is perpendicular to AB and bisects the blue square, E be the point where the blue square intersects CB, and F be the bottom right corner of the blue square. ∆CDB is similar to ∆EFB, so CD is similarly proportional to EF as DB is to FB.
    CD/EF = DB/FB
    CD/6 = 13/(13 - 3)
    CD = (13/10)/6 = 78/10 = 7.8
    Green area = Area of ∆ABC - 36
    A = bh/2 - 36
    A = 26(7.8)/2 - 36
    A = 13(7.8) - 36
    A = 101.4 - 36 = 65.4

  • @MrPaulc222
    @MrPaulc222 Рік тому +1

    Before video: pink square sides are 26 and blue square side are 6.
    The two largest green triangles are (13-3)*6, so 10 by 6. Don't bother halving the base because there are two of them, so two equal green triangles = area of 60.
    Half of the top triangle is similar to one of the larger green triangles.
    Therefore, 3/x = 10/6.
    Cross multiply for 10x = 18, so x = 1.8.
    1.8*3=5.4.
    60+5.4=65.4cm^2.
    Just watched the video. This one didn't stretch me. I did it broadly the same way, but calculated the smaller triangles separately.
    Thank you. You are an education. Your explanations contain far more clarity than I see on other maths channels.

    • @PreMath
      @PreMath  Рік тому +1

      Super!
      Thanks ❤️🌹

  • @williamwingo4740
    @williamwingo4740 Рік тому +1

    No peeking, but I did use a calculator for the square root of 676.
    The side of the pink square is 26 cm. Therefore each of the lateral green triangles is 10 cm wide and 6 cm tall, and the two together total of 60 square cm.
    The uppermost small green triangle is made of two triangles each similar to lateral green triangles. Each of these will be 3 cm wide and (6/10)(3) cm tall, and the two together will be (3)(18/10), or a total of 5.4 square cm. So the total green area is 60 + 5.4 = 65.4 square cm.
    Coraggio. 🤠

  • @bigm383
    @bigm383 Рік тому +2

    Thanks, Professor, very succinct!❤

    • @PreMath
      @PreMath  Рік тому +1

      Thank you so much ❤️

  • @prossvay8744
    @prossvay8744 Рік тому +1

    ABDE is a square
    AB,=BD=DE=EA=√676=26cm
    Lenge of the small square=√36=6cm
    Small triangle is CFG
    Angle CFG ~ CAB (AA)
    h1/h1+6=6/26
    26h1=6(h1+6)
    26h1=6h1+36
    26h1-6h1=36
    20h1=36
    h1=36/20=9/5cm
    h=h1+6=9/5+6=(9+30)/5
    h=39/5=7.8cmcm
    So: Area of the green shaded region=1/2(26)(7.8)-36=65.4cm^2.thanks ❤❤❤

  • @Mascig83
    @Mascig83 Рік тому +2

    Good morning from Brazil. I really like your channel, but I would like there to be more diversity in the topics covered, that is, more different subjects, such as equations, potentiation, etc.

    • @PreMath
      @PreMath  Рік тому +1

      Good suggestion!
      Thanks ❤️🇺🇸

  • @wackojacko3962
    @wackojacko3962 Рік тому +2

    Similar triangles help with estimating distance. Using thumb and switch closing eyes focused on distant object and multiplied by ten a known width like a car. Cyclops had a distinct disadvantage! 🙂

    • @PreMath
      @PreMath  Рік тому +1

      Thanks for sharing!😀

  • @AmirgabYT2185
    @AmirgabYT2185 6 місяців тому +1

    S=65,4 cm²

  • @Copernicusfreud
    @Copernicusfreud Рік тому +1

    Yay! I solved the problem.

  • @LuisdeBritoCamacho
    @LuisdeBritoCamacho Рік тому +1

    Answer: 65, 4 cm^2
    x/13 = 6/10
    x = 78/10
    x = 7,8 cm
    (7,8 * 26) / 2 = 202,8 / 2 = 101,4
    101,4 - 36 = 65,4

  • @samerabdullah2042
    @samerabdullah2042 Рік тому +2

    How did you know that AC=CB ?

  • @murdock5537
    @murdock5537 Рік тому +1

    Nice! tan⁡(δ) = 3/5 = k/3 → k = 9/5 → (1/5)(3(169) - 180) = 327/5

  • @Ibrahimfamilyvlog2097l
    @Ibrahimfamilyvlog2097l Рік тому +1

    very nice video❤❤❤

    • @PreMath
      @PreMath  Рік тому

      Thank you so much ❤️

  • @ОльгаСоломашенко-ь6ы

    tg(A)=0,6, CQ=13*tg(A)=13*0,6=7,8. S=26*7.8/2-36=65,4

    • @PreMath
      @PreMath  Рік тому

      Thank you so much ❤️

  • @giuseppemalaguti435
    @giuseppemalaguti435 Рік тому +1

    I due triangoli laterali hanno area (6*10/2)*2=60..il triangolo in alto ha A=6*1,8/2=5,4...totale 65,4

    • @PreMath
      @PreMath  Рік тому

      Super!
      Thank you so much ❤️

  • @johnbrennan3372
    @johnbrennan3372 Рік тому

    13/ |cq| = 10/6=5/3 so |cq|= 39/5 etc.

  • @gulshanjoshi7682
    @gulshanjoshi7682 Рік тому

    This one was so simple that you can do it in your head.
    Also...too many find the area or length of geometry problems. Can you try something else please !!

  • @disraelidemon
    @disraelidemon Рік тому

    How do we determine that the blue square is located symmetrically? I abandoned my attempt to solve this problem because I realised I couldn't demonstrate symmetry, and then the proof skipped this step...

    • @ybodoN
      @ybodoN Рік тому +1

      This problem can be solved without assuming that there is symmetry 😉

  • @laxmikantbondre338
    @laxmikantbondre338 Рік тому

    Is it given that the blue square is in the exact center of top side of pink square?

    • @PreMath
      @PreMath  Рік тому

      Symmetry!
      Thanks ❤️

  • @kennethstevenson976
    @kennethstevenson976 Рік тому +1

    Got the answer on the first try.

    • @PreMath
      @PreMath  Рік тому

      Great!
      Thanks ❤️

  • @mibsaamahmed
    @mibsaamahmed Рік тому +1

    Coincidentally I just finished working on a previous problem from one of your videos lol

    • @PreMath
      @PreMath  Рік тому

      Great!
      Thanks ❤️

  • @bobbyheffley4955
    @bobbyheffley4955 Рік тому

    The side length of the pink square is 26 cm.

  • @alster724
    @alster724 Рік тому

    Easy

  • @JSSTyger
    @JSSTyger Рік тому +1

    65.4