How to find the length X | Circle inside a right angle triangle | Geometry Problem

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  • Опубликовано: 10 фев 2025
  • How to find the length X | Circle inside a right angle triangle | Geometry Problem
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Комментарии • 19

  • @dantallman5345
    @dantallman5345 Год назад +3

    Area of triangle, ht =15 and base =x, with an inscribed circle with radius 3.
    A=(1/2)rP. And r=3. A=(1/2)(15)(x)…..set A=A hence P=(5)x.
    P=15+12+x+(x-3)=24+(2)x…, hence (5)x=24+(2)x……x=8.

  • @marcelowanderleycorreia8876
    @marcelowanderleycorreia8876 Год назад

    Perfect Sir!!

  • @quigonkenny
    @quigonkenny 9 месяцев назад

    Let the points of tangency between the circle and triangle ∆ABC be as follows: M on AB, N on BC, T on CA. Let the center of the circle be O.
    Draw OM, ON, and OT. All 3 equal r, being radii of circle O, and As AB, BC, and CA are tangent to circle O at those three points, ∠OMB = ∠BNO = CTO = 90°. Therefore ∠NOM must equal 90° as well, so BNOM is a square with side length r = 3.
    Ad MB = 3 and AB = 15, AM = 15-3 = 12. As TA snd AM are tangents to circle O that intersect at A, TA = AM = 12. As BN = 3 and BC = x, NC = x-3. As NC and CT are tangents to circle O that intersect at C, CT = NC = x-3. CT = x-3 and TA = 12, so CA = x-3+12 = x+9.
    Triangle ∆ABC:
    AB² + BC³ = CA²
    15² + x² = (x+9)²
    225 + x² = x² + 18x + 81
    18x = 144
    x = 144/18 = 8

  • @muhammadayyaz3112
    @muhammadayyaz3112 Год назад

    if we find diagonal in rectangle
    and by adding it into radius and using pythagorean theorem
    we get x=6+3√2

  • @CanalMiTube
    @CanalMiTube Год назад

    Another more direct path would be to equalize areas. AB*BC/2 = AF*FC. 15x/2 = 12(x-3). x=8.
    Greetings.

  • @expertonnn
    @expertonnn Год назад +3

    X'2+15'2=(x+9)'2....x=8

    • @alexniklas8777
      @alexniklas8777 Год назад

      Аналогично решил

    • @WahranRai
      @WahranRai Год назад +2

      @@alexniklas8777 No, he used the variable c, lengthening the calculation !

  • @prbprb2
    @prbprb2 Год назад

    Call angle ACB = 2 phi. Then tan (2 phi) = 15/X , tan(phi) = 3/(X-3) . Eliminate phi: X =8

  • @PS-mh8ts
    @PS-mh8ts Год назад

    Or you can use the incircle radius formula like so:
    incircle radius r(of a right-triangle ABC with ∠B=90°)=(AB+BC-AC)/2 -- (i)
    AB=15, BC=x, and AC=√(15²+x²)=√(225+x²), and r=3. Using these in (i), we get:
    [15+x-√(225+x²)]/2=3
    i.e., 15+x-√(225+x²)=6
    => √(225+x²)=15+x-6=9+x
    Squaring both sides, 225+x²=81+18x+x²
    i.e., 144=18x or x=8

  • @nunoalexandre6408
    @nunoalexandre6408 Год назад

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

  • @channelsixtyeight068_
    @channelsixtyeight068_ Год назад

    By the time we got to C = 5, I forgot what the original problem was. 🤣 Love your videos.

  • @oxlq
    @oxlq Год назад

    Best solution; r=(a+b-c)/2
    Then we will get a equality and after solving this x=6

  • @wes9627
    @wes9627 Год назад

    x=15*tan[2*atan(1/4)]=8

  • @yakupbuyankara5903
    @yakupbuyankara5903 Год назад

    X=8

  • @skwest
    @skwest Год назад

    Lying in be, doin' it in my head... I'll go with:
    X = 8
    ?
    I'll check back later after some sleep.
    Cheers!

  • @adgf1x
    @adgf1x Год назад

    x=15^1/2