finding the radius of a circle inscribed in a right triangle

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  • Опубликовано: 24 дек 2024
  • In this video I show how to find the radius of a circle inscribed in a right triangle. This complex geometry problem involves ideas such as Pythagorean Theorem, similar triangles, circle geometry, radius of a circle, tangents drawn to a circle, tangents drawn to a common point.
    Write any topics you would like me to cover in the COMMENT section below. THANKS for watching!
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Комментарии • 33

  • @adlerbrietzke
    @adlerbrietzke Год назад +11

    Area of the circle inscribed in a right triangle:
    r = (a + b - c)/2
    Where a and b are base and height or height and base, and c is the hypotenuse
    Once the hypotenuse was not given, there was the additional effort to use the pythagorean theorem to solve for the hypotenuse

    • @jamesoluwole1475
      @jamesoluwole1475 6 месяцев назад

      Thank you for the formula, what if you're given the value of the base only, how will someone solve that kind of question?

    • @indianhorrorstories9146
      @indianhorrorstories9146 Месяц назад

      @@jamesoluwole1475 in that scenario you can not solve it by any method

  • @rayaanfuaad
    @rayaanfuaad 3 месяца назад +7

    Dude, I'm 15.. I solved it in about 5 second or so, I was pretty much sure I'm wrong.. but I clicked the video and skipped to the end and saw I got the same answer 😁

  • @sarmabudampati3131
    @sarmabudampati3131 Год назад +2

    If we join the vertices of the triangle with centre of the circle we get three small triangles with each side as base and radius as height.
    Now we may equate the area of original triangle=
    Sum of areas of three new triangles.
    1/2 ×27×36=1/2 × radius × (27+36+45)
    27×36 = r ×108
    Radius r = 9
    This is BSP sarma a retired Bank employee from Hyderabad,
    I hope you will appreciate the same.

    • @vinteachesmath
      @vinteachesmath  Год назад

      Thanks for the great comment! I appreciate a second approach to the question! I have to add more videos like this in the upcoming school year. This question is from an old Japanese Geometry math book. They have questions that are way more challenging!

    • @hongningsuen1348
      @hongningsuen1348 11 месяцев назад

      I've just watched a video on similar problem. The inscribed triangle can be any triangle, not just right-angled triangle. As lengths of 3 sides of the triangle are given, its area can be found by Heron's formula. Then using your method of dividing the triangle into 3 small triangles each with their height as the radius of the circle. Equating the sum of areas of the 3 triangles with that of the whole triangle can solve value of radius accordingly. Simple method is vital in examination.

  • @triole1
    @triole1 Год назад +2

    Hi, since the triangle legs are not the same length, are you sure that the line you draw on 1:03 is perpendicular to the hypothenuse/tangent? Imagine a square triangle whose legs would measure 20 and 65 (ex). The hypothenuse wouldn’t be a tangent to the circle in the same point. So it’s perpendicular line wouldn’t be the same.

    • @vinteachesmath
      @vinteachesmath  Год назад +2

      Yes, the concept of an inscribed circle is that it is tangent to the triangle at three locations. The radius drawn to those 3 locations forms a 90-degree angle

  • @rashidissa5887
    @rashidissa5887 11 месяцев назад +2

    I always argue that if a triangle is right angled,it should be specifically be mentioned. Presumption should be avoided even if it is drawn and looks like it

    • @vinteachesmath
      @vinteachesmath  11 месяцев назад

      Fair point!
      Would you say that it is enough to mention the triangle is a right triangle with legs measuring 36 and 27 units long? I believe that the information given in the original question establishes that we have a right triangle, with the right angle forming at the point where the legs meet.

  • @tushiglutbekh5453
    @tushiglutbekh5453 Год назад +3

    very helpfull video keep going !!!

    • @vinteachesmath
      @vinteachesmath  Год назад

      Thanks, will do! My next big video will be on triangle proofs.

  • @brynnfromscience
    @brynnfromscience 9 месяцев назад +1

    You’re doing good work man

    • @vinteachesmath
      @vinteachesmath  9 месяцев назад

      I appreciate that! I am trying to make as much quality content as possible. This question came from a book called Sacred Mathematics Japanese Temple Geometry. That book is loaded with excellent questions.

  • @pvpchampion9004
    @pvpchampion9004 Год назад +2

    Great video

  • @SiddhiSai-w3f
    @SiddhiSai-w3f Месяц назад +1

    Thanks for such a nice explanation 🙏👍its great👌

  • @applyinggaming9529
    @applyinggaming9529 2 месяца назад +1

    ohhh so this is how (a+b-c)/2 is derived. a- r + b - r = c. a+b-c=2r, (a+b-c)/2=r

    • @vinteachesmath
      @vinteachesmath  2 месяца назад

      It's cool to see where the formulas come from. Thanks for watching!

  • @adgf1x
    @adgf1x 25 дней назад

    27^2+36^=2025=45^2.hypotenuse=45.ar. of circle=27×18=486 sq.Radius b R.now 27+45+36=108.thus 54R=27×18=>R=27×18/54=18/2=9 unit.ans

  • @wasimahmad-t6c
    @wasimahmad-t6c 3 месяца назад

    41.21357 area cercel

  • @michaelempeigne3519
    @michaelempeigne3519 2 года назад +1

    54 - 45 = 9

  • @genesisanderson6052
    @genesisanderson6052 2 года назад +1

    Great video, all though it was a bit confusing

    • @vinteachesmath
      @vinteachesmath  2 года назад +1

      I am branching out and doing some challenge problems, this is definitely a stretch for a standard math class, but if you can do the challenge questions, the regular questions will seem easy!

  • @ShineMarn-u9n
    @ShineMarn-u9n 7 месяцев назад

    27-(45-36)/2=9

  • @adgf1x
    @adgf1x 6 месяцев назад

    hypotenuse=45unit.

  • @morbrakai8533
    @morbrakai8533 11 месяцев назад +1

    Bro... just divide 27 by 3

  • @adgf1x
    @adgf1x 6 месяцев назад

    😂radius(r)=9 unitans