Calculate the height h | Important Geometry and Algebra skills explained

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  • @bigm383
    @bigm383 Рік тому +7

    Thanks for another stepwise explanation!❤🥂

    • @PreMath
      @PreMath  Рік тому

      You are very welcome!
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  • @misterenter-iz7rz
    @misterenter-iz7rz Рік тому +19

    AD is 25, by pythagorean theorem, so DB is 25 in an isosceles triangle, then CB is 25+7=32, and thus AB is 40, by pythagorean theorem again, now EB is one half of AB, is 40/2=20, therefore h=15, also applying pythagorean theorem. 😃

    • @PreMath
      @PreMath  Рік тому +1

      Excellent!
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      You are awesome. Keep it up 👍
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    • @normanc918
      @normanc918 Рік тому

      I did the same.

    • @davidshen5916
      @davidshen5916 Рік тому

      BD*AC=CE*AB

  • @muttleycrew
    @muttleycrew Рік тому +4

    ABC is similar to DBE so once you know that ABC is in the proportions of a 3,4,5 triangle (it's 8x3, 8x4, 8x5) you know that DBE must also be in the proportions of a 3,4,5 triangle. You know that the hypotenuse is 5x5 = 25 so that leaves the other sides being 20 and 15. From similarity h must be the shortest side so that's h = 15.

    • @PreMath
      @PreMath  Рік тому

      Thanks for sharing! Cheers!
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  • @soli9mana-soli4953
    @soli9mana-soli4953 Рік тому +2

    Setting AE=EB=x and DB=y, I constructed two equations, the first by applying the Pythagorean theorem to the triangle ABC
    (24)^2+(7+y)^2=(2x)^2
    I found the second equation by applying the proportions to the two similar triangles ABC and EBD
    2x:y=(y+7):x hence 2x^2=y^2+7y
    whose solutions are x=20 and y=25
    Finally, to find h it is enough to apply the Pythagorean theorem again to the triangle EBD, the sides EB and DB being known

  • @BKNeifert
    @BKNeifert Рік тому +1

    This is how math should be taught. Geometry is a lot like poetry, when you're watching someone solve it. If I ever taught an English Class, I'd pull up one of these types of videos, and teach them to apply the same skills in this, to their poems. Nobody thinks like that, though, but that's also why we have a lot of people who can't think. They always want words to have ambiguity, but a good reader will follow the logic of the poem's words on the page, the same way they'd follow this guy's presentation. Like, poetry is exactly like this video. You're being walked step by step to the answer. And each verse breaks the logic down into simpler bits. Good poetry, anyway is like that.

  • @georgecaplin9075
    @georgecaplin9075 10 місяців тому

    I solved it, (probably the same way most others did), by connecting A and D, making that the hypotenuse of the triangle ACD, solving for the length AD, (which is also DB), which gives us BC, from which you can solve for AB, half of AB is BE, then you have all the values to “plug in to” the Pythagorean theorem.

  • @sail2byzantium
    @sail2byzantium Рік тому +3

    Excellent! Very memorable. I need to do a better job in remembering as you show here that one can actually MANIPULATE figures and shapes to solve them, rather than being just stuck with what is given to try to deductively figure it out. Thank you, PreMath!

    • @PreMath
      @PreMath  Рік тому

      Excellent!
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  • @AnonimityAssured
    @AnonimityAssured Рік тому +3

    An in-the-head quickie if one recognizes two familiar Pythagorean triples.
    Spoiler alert.
    An auxiliary line segment can be imagined linking points A and D.
    We can immediately recognize a 7-24-25 triangle in the top left, so AD = 25 units.
    Triangles ADE and BDE are congruent, so DB must also be 25 units.
    CB = 7 + 25 = 32 units.
    24 is three-quarters of 32, so ABC is a scaled 3-4-5 triangle.
    Therefore, AED is also a scaled 3-4-5 triangle, so:
    h = 25 (3 / 5) = 15 units.
    Edited to correct a small typo.

    • @PreMath
      @PreMath  Рік тому

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  • @latten531
    @latten531 Рік тому

    Thanks, it’s intersting. I used AREA to find out h value. Triangle ABD=300, AB=40, 40xhx1/2=300, 20h=300, h=15
    Just a comment from Tokyo, Japan.

  • @alster724
    @alster724 Рік тому +2

    Super easy. After seeing the Similarity Theorem, the problem became a piece of cake and solved it off the bat before I can finish the video.

    • @PreMath
      @PreMath  Рік тому

      Great job!
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  • @rudychan8792
    @rudychan8792 Рік тому +2

    More than one hour, I figured it out, h = 15! 😓
    I was connecting line CE = AE = BE = it's a Radius!* (Fail)
    Then, I connect AD, voila!
    Much-much Easier!
    It is Triple Phytagoras!!!
    Great Maths 🙂 👍

    • @PreMath
      @PreMath  Рік тому +1

      Great job!
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  • @spiderjump
    @spiderjump Рік тому +1

    Find AD by Pythagorean theorem , AD = 25.
    Triangles ADE and BDE are congruent. Hence DB =25.
    BC =32.
    AB can be found by Pythagorean theorem, AB= 40.
    EB=40/2=20
    DE can be found by Pythagorean theorem,DE= 15

    • @PreMath
      @PreMath  Рік тому

      Excellent!
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  • @samriddhadutta2104
    @samriddhadutta2104 Рік тому +1

    Yes!! I have solved this problem mentally.

    • @PreMath
      @PreMath  Рік тому

      Excellent!
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  • @juanpabloalvarado2721
    @juanpabloalvarado2721 Рік тому

    Thanks for the explanation, needs for me to see 100 times until I try by myself but I going to get it, Thanks again

  • @liliyakaloyanova377
    @liliyakaloyanova377 Рік тому +1

    AD=DB, because triangles ADE and EBD are equal. AD^2=24^2+7^2=576+49=625 AD=DB=25
    so CB=32. In triangle ABC AB^2= 24^2+32^2= 576+1024=1600 AB=40 , so DB=40/2=20 . In triangle DBE h^2= 25^2- 20^2= 625 - 400 = 225 h=V225=15

    • @PreMath
      @PreMath  Рік тому

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  • @calspace
    @calspace Рік тому +1

    I am terrible in geometry at knowing where to draw additional line. I used pure algebra.
    Let us call the length of DB = x and the length of EB = y (so AB = 2y).
    We have two right triangles, ABC and DBE. Because B is present in both of them, they are similar by AAA. The lengths of the sides of these two triangles (SL;LL;H) are 24;7+x;2y and h;y;x. Therefore, comparing the ratio of the long legs and hypotenuses, we get y:7+x=x:2y. When we multiply this, we get 2y^2 = (7+x)*x = x^2 + 7x.
    By the Pythagorean theorem, 24^2 + (7 + x)^2 = (2y)^2
    576 + 49 + 14x + x^2 = 4y^2
    If we multiply the first equation by 2 we get 4y^2 = 2x^2 + 14x.
    Therefore, 625 + 14x + x^2 = 2x^2 + 14x.
    Substracting x^2 + 14x from both sides, we get 625 = x^2, or x = 25.
    So now we know that the sides of ABC are 24; 7 + 25 = 32; 2y.
    Using the Pythagorean formula we get 576 + 1024 = 4y^2
    1600 = 4y^2
    400 = y^2
    20 = y
    Now we can either use the Pythagorean formula again to get x^2 - y^2 = x^2
    25^2 - 20^2 = h^2
    625 - 400 = h^2
    225 = h^2
    15 = h
    Or we can use the ratios of sides, either 24:h = 7+x:y
    24y = (7 + x)h
    24 * 20 = 32h
    480 = 32h
    480/32 = h
    15 = h
    or 24:h = 2y:x
    24x = 2y * h
    24 * 25 = 2 * 20 * h
    600 = 40h
    600/40 = h
    15 = h

    • @PreMath
      @PreMath  Рік тому

      No worries. We are all lifelong learners.
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  • @NoraBhsas
    @NoraBhsas Рік тому +1

    Me having an English degree and not having anything to do with Geometry actually listening and enjoying this explanation

  • @spywhale
    @spywhale Рік тому +1

    Thanx your first problem I could work out in my head all the practice is paying off. 🎂

    • @PreMath
      @PreMath  Рік тому

      Perfect!
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  • @DB-lg5sq
    @DB-lg5sq Рік тому +2

    شكرا
    يمكن استعمال cos c بصيغ مختلفة

    • @PreMath
      @PreMath  Рік тому

      You are very welcome!
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  • @rick57hart
    @rick57hart Рік тому +2

    I did the last step without pythagorean theorem:
    There are 2 ways to calculate the area of the triangle.
    (24/2*32) = (40/2*h) + (24/2*7)
    20*h = 12(32-7)
    20*h = 12*25
    20*h = 300
    h =15

    • @muttleycrew
      @muttleycrew Рік тому +1

      Yeah, nice.
      I did it another way, by knowing that if two sides of ABD are 24 and 32 that's 8x3 = 24, 8x4 = 32. It has to be in proportion to a 3,4,5 triangle since it is a right-angled triangle and the two shortest sides are in a 3:4 ratio. That means the hypotenuse of ABC *_must_* be 8x5 = 40.
      ABC is also similar to DBE so once you know that ABC is in the proportions of a 3,4,5 triangle you know that DBE must also be in the proportions of a 3,4,5 triangle. You know that the hypotenuse is 5x5 = 25 so that leaves the other sides being 4x5 = 20 and 3x5 = 15. From similarity of ABC with DBE, the length h must be the shortest side so that's h = 15.
      It's using 3,4,5 triangles and similarity arguments to get h=15. You only really need the Pythagorean theorem once, unless you happen to know that a 7,24,25 triangle is also a right-angled triangle in which case you would never need the Pythagorean theorem at all, at least not explicitly.

  • @davidbrown8763
    @davidbrown8763 Рік тому

    Thanks for a great challenge which gave me much pleasure to solve.

  • @vara1499
    @vara1499 Рік тому

    Great solution!.

  • @murdock5537
    @murdock5537 Рік тому

    Nice and awesome, many thanks, Sir!
    sin⁡(φ) = 3/5 = h/25 → h = DE = 15

  • @williamwingo4740
    @williamwingo4740 Рік тому +1

    A slightly different approach: integer Pythagorean triangles abound in this one.
    We could calculate that AD = 25 by adding 7^2 + 24^2 = 625 = 25^2; or we could just remember that 7-24-25 is an integer Pythagorean triangle. This gives us 25 for DB and 25 + 7 = 32 for CB.
    Hey, maybe we're onto something here....
    Similarly, the big triangle ABC is another integer Pythagorean, 24-32-40, so the base AB is 40 and half the base AE (or EB) is 40/2 = 20.
    And finally, triangle BDE (or ADE) is a 3-4-5 triangle multiplied by 5, so it's 15-20-25; and h = 15.
    No peeking, no calculators; just my table of integer-sided right triangles.
    Coraggio. 🤠

    • @PreMath
      @PreMath  Рік тому

      Thanks for sharing! Cheers!
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    • @williamwingo4740
      @williamwingo4740 Рік тому

      @@PreMath Thanks. I'm in the USA too, and we need all the love and prayers we can get.

  • @KAvi_YA666
    @KAvi_YA666 Рік тому +1

    Thanks for video.Good luck sir!!!!!!!!!

    • @PreMath
      @PreMath  Рік тому +1

      Thank you too
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  • @santiagoarosam430
    @santiagoarosam430 Рік тому +1

    AD=DB 》DB^2=24^2+7^2 》DB=25 》CB=25+7=32 》AB^2=24^2+32^2 》AB=40 》h^2=25^2-(40/2)^2 》h=15
    Gracias y un saludo.

    • @PreMath
      @PreMath  Рік тому

      Excellent!
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  • @eminwadiogbu1566
    @eminwadiogbu1566 Рік тому

    I just love this question. Keep up the good work sir.👍🏾

  • @HappyFamilyOnline
    @HappyFamilyOnline Рік тому +3

    Another great video👍
    Thanks for sharing😊😊

    • @PreMath
      @PreMath  Рік тому

      Thank you too
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      Love and prayers from the USA! 😀

  • @arvindhshennaiah5659
    @arvindhshennaiah5659 Рік тому

    Very nice explanation.

  • @kevi152
    @kevi152 Рік тому

    Tan BDE = h/20 =24/32 . So h =3/4 x 20 = 15 . Since triangle ABC is similar to triangle BDE !

  • @leslievincent20
    @leslievincent20 Рік тому

    Excellent I got this one.

  • @nandisaand5287
    @nandisaand5287 Рік тому +1

    I went trig route. Probably made it harder for myself:
    CosB=(7+Y)/2X=X/Y
    2X^2=Y(7+Y)
    Mr Pythagoras enters the chat:
    (7+Y)^2+(24)^2=(2X)^2
    =4X^2
    =2(2X^2)
    [Now substitute:]
    ...=2Y(7+Y)
    Drop the Algebra hammer:
    Y^2+14Y+49+576=2Y^2+14Y
    Combine:
    Y^2-625=0
    Y^2=625
    Y=25
    Solve for X:
    2X^2=7(25)+(25)^2
    2X^2=625+175=800
    X^2=400
    X=20
    SinB=24/2X=h/Y
    h=24Y/2X=24(25)/2(20)
    h=15

    • @PreMath
      @PreMath  Рік тому

      Thanks for sharing! Cheers!
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  • @mauriciosahady
    @mauriciosahady Рік тому

    AEDC é um quadrilátero inscritível, correto?

  • @Copernicusfreud
    @Copernicusfreud Рік тому +1

    Yay! I solved it.

    • @PreMath
      @PreMath  Рік тому

      Excellent!
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      You are awesome, Mark. Keep it up 👍
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  • @manojitmaity7893
    @manojitmaity7893 Рік тому +1

    Sir I request you to upload some videos on arithmetic.

    • @PreMath
      @PreMath  Рік тому

      Sure!
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  • @alexundre8745
    @alexundre8745 Рік тому +1

    BOM DIA PROFESSOR
    Agradeço ao SR por ter aprendido geometria
    DEUS lhe Abenço
    Grato

    • @PreMath
      @PreMath  Рік тому

      So nice of you, Alex
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      Stay blessed 😀

  • @WaiWai-qv4wv
    @WaiWai-qv4wv Рік тому +2

    Ok
    Very thanks

    • @PreMath
      @PreMath  Рік тому +1

      You are very welcome!
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      Love and prayers from the USA! 😀

  • @harrymatabal8448
    @harrymatabal8448 6 місяців тому

    Excellent

  • @micke_mango
    @micke_mango Рік тому

    If I only had recognized the first pythagorean triangle, I would have solved it in 20 seconds. Now it probably took two minutes for my slow brain to calculate the first hypotenuse in my head.
    Still a bit proud I didn't need pen and paper...

  • @RahmanPatwaryArifur
    @RahmanPatwaryArifur Рік тому

    how to do in case ED not in the middle of AB?

  • @davoodzamani4254
    @davoodzamani4254 Рік тому +1

    Nice job 🥂

    • @PreMath
      @PreMath  Рік тому

      Glad you think so!
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  • @gamingwithgamer3751
    @gamingwithgamer3751 Рік тому +1

    You make my calculations fast

    • @PreMath
      @PreMath  Рік тому

      Thanks for your feedback! Cheers!
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  • @giuseppemalaguti435
    @giuseppemalaguti435 Рік тому +1

    H=15,basta utilizzare la similitudine dei 2 triangoli rettangoli

    • @PreMath
      @PreMath  Рік тому

      Excellent!
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      Love and prayers from the USA! 😀

  • @gideonlapidus8996
    @gideonlapidus8996 Рік тому

    I love this one just think out the box

  • @alexniklas8777
    @alexniklas8777 Рік тому +1

    Suppose the triangle is Egyptian, then:
    AB= (24/3)×5= 40;
    BC= (24/3)×4= 32.
    Triangle ∆BDE~∆ABC:
    BE= AB/2= 40/2= 20;
    BD=BC-CD=32-7=25;
    DE=h=√(BD^2-BE^2)=
    =√(625-400)=√225=15.
    Examination.
    Let's draw the middle line ∆ABC, the line EF.
    BF=32/2=16;
    EF=√(BE^2-BF^2)=
    =√(20^2-16^2)= 12, i.e.
    EF=AC/2=24/2=12.

    • @PreMath
      @PreMath  Рік тому +1

      Excellent!
      Thanks for sharing! Cheers!
      You are awesome, Alex. Keep rocking 👍
      Love and prayers from the USA! 😀

    • @alexniklas8777
      @alexniklas8777 Рік тому

      @@PreMath Thanks sir 😊

  • @adgf1x
    @adgf1x Рік тому +2

    h/20=24/32=3/4=>h=20×3/4=15unit.

    • @PreMath
      @PreMath  Рік тому

      Excellent!
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      Stay blessed 😀

    • @adgf1x
      @adgf1x Рік тому

      @@PreMath thanks.

  • @victorfildshtein
    @victorfildshtein Рік тому

    let's assume that AE=BE=x
    △DEB and △ACB triangles are similar
    We make proportions:
    EB/BC = CE/AB ⇒ x/BC = (BC - 7)/2x ⇒
    2x² = BC² - 7BC (1)
    According to the Pythagorean theorem
    BC² = AB² - AC² = 4x² - 24² (2)
    Substitute (2) into (1)
    2x² = 4x² - 24² -7√(4x² - 24²)
    7√(4x² - 24²) = 2x² - 24²
    We get the biquadratic equation
    4x⁴ - 2500x² + 360000 = 0
    x₁² = 400; x₂² = 225
    x₁ = 20; x₂ = 15
    1)Consider the first option: x = 20
    AC = 40; BC = √(40² - 24²) = 32; EC = 25
    h = √(25² - 20²) = 15
    2) Consider the second option: x = 15
    AC = 30; BC = √(30² - 24²) = 18; EC = 11
    EC < EB the hypotenuse is smaller than the leg. This is impossible.
    Answer: h = 15

  • @jmlfa
    @jmlfa Рік тому +1

    Easy! (Again)

    • @PreMath
      @PreMath  Рік тому

      Thanks for your feedback! Cheers!
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  • @Reddogovereasy
    @Reddogovereasy Рік тому

    Why is the product of a square root always positive or negative? I've heard you say that hundreds of times but I cannot fathom a reason for that being so.

    • @nandisaand5287
      @nandisaand5287 Рік тому +2

      From problem above:
      Sqrt[625]=+/-25
      (+25)^2=(-25)^2=625
      Typically its easier to just think in terms of absolute value, thus positive, but technically it IS [Plus or Minus]
      Hope that helps.

  • @luisimtz
    @luisimtz 7 місяців тому

    I solved this problem with just mental calculations

  • @TDSONLINEMATHS
    @TDSONLINEMATHS Рік тому

    Great

  • @adgf1x
    @adgf1x 11 днів тому

    h^2=25^2-20^2=15^2=>h=15 unit.ans

  • @theoyanto
    @theoyanto Рік тому +1

    Have we seen this before?😉 Still great example though 👍🏻

    • @PreMath
      @PreMath  Рік тому +1

      Glad you think so!
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  • @adgf1x
    @adgf1x 5 місяців тому

    h=(25^2-20^^2)^1/2=15 unit.ans

  • @UvuvwevwevweOnyetenyevweUg-z7n

    So easy

  • @adgf1x
    @adgf1x 11 днів тому

    AB=40 and AE=BE=40/2=20

  • @jim2376
    @jim2376 Рік тому +1

    Fun problem. 25^2 - 20^2 = 225. √225 = 15.

    • @PreMath
      @PreMath  Рік тому

      Glad to hear that!
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      Stay blessed 😀

  • @seasea5938
    @seasea5938 Рік тому

    利用中垂線性質和畢氏組數。

  • @bigboicool101
    @bigboicool101 Рік тому

    👍👌

  • @esegueojogo
    @esegueojogo Рік тому

    The famous 3-4-5 triangle....TWICE

  • @simpleman283
    @simpleman283 Рік тому +1

    15 if u no ur 3,4,5 sets.
    edit:
    24/3 = 8
    8 x 5 = 40
    40/2 = 20
    20/4 = 5
    5 x 3 = 15

    • @PreMath
      @PreMath  Рік тому

      Thanks for sharing! Cheers!
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  • @RaniDevi-ur8gb
    @RaniDevi-ur8gb Рік тому

    ❤❤

  • @sreedharankp9878
    @sreedharankp9878 Рік тому

    👍

  • @rangaswamyks8287
    @rangaswamyks8287 Рік тому

    15 answer

  • @adgf1x
    @adgf1x 5 місяців тому

    h=15 unit.ans

  • @comdo777
    @comdo777 10 місяців тому +1

    asnwer=10cm isit

  • @Tom-os1qv
    @Tom-os1qv Рік тому

    Bruh, I solved it the exact way he did

  • @bobgoodman1451
    @bobgoodman1451 Рік тому +1

    Lots of 345 triangles is an easier solution, but not as robust

    • @PreMath
      @PreMath  Рік тому

      Thanks for your feedback! Cheers!
      You are awesome. Keep it up 👍
      Love and prayers from the USA! 😀

  • @Xyz-sd7ub
    @Xyz-sd7ub Рік тому +1

    h = 15

    • @PreMath
      @PreMath  Рік тому

      Excellent!
      Thanks for sharing! Cheers!
      You are awesome. Keep it up 👍
      Love and prayers from the USA! 😀

  • @samriddhadutta2104
    @samriddhadutta2104 Рік тому +1

    15

    • @PreMath
      @PreMath  Рік тому

      Thanks for sharing! Cheers!
      You are awesome. Keep it up 👍
      Love and prayers from the USA! 😀

  • @adilabusafa
    @adilabusafa Рік тому

    يتعذر الاشتراك ولا يوجد جرس-يتعذر التعليق-لا أعلم .

  • @holyshit922
    @holyshit922 Рік тому

    Only thing you have to know for this problem is Pythagorean Theorem

  • @-NguyenChiTai-A
    @-NguyenChiTai-A Рік тому

    This solution in VietNam is for grade 6 and you must be solved between 5 to 8 minutes 😅

  • @franciswalsh5469
    @franciswalsh5469 Рік тому

    h is 15

  • @n.662
    @n.662 Рік тому

    Only 1 line!)

  • @subhajitmukhopadhyay4780
    @subhajitmukhopadhyay4780 Рік тому

    If solved mathematically, it's ok; but when solved geometrically, it's completely wrong.

  • @antoniolorenzo7644
    @antoniolorenzo7644 Рік тому +1

    Intéressant
    Bon.fin

    • @PreMath
      @PreMath  Рік тому

      Excellent!
      Thanks for your feedback! Cheers!
      You are awesome. Keep it up 👍
      Love and prayers from the USA! 😀

  • @nusretgulmammadov5144
    @nusretgulmammadov5144 Рік тому +1

    h=15

    • @PreMath
      @PreMath  Рік тому

      Excellent!
      Thanks for sharing! Cheers!
      You are awesome, Nusret. Keep it up 👍
      Love and prayers from the USA! 😀
      Stay blessed 😀

  • @kostgenn1803
    @kostgenn1803 Рік тому

    15

  • @faritishmukhametov7063
    @faritishmukhametov7063 Рік тому +1

    h=15

    • @PreMath
      @PreMath  Рік тому

      Excellent!
      Thanks for sharing! Cheers!
      You are awesome. Keep it up 👍
      Love and prayers from the USA! 😀

  • @AmirgabYT2185
    @AmirgabYT2185 7 місяців тому +1

    h=15