Can you find the Radius of the circle? | (3 Methods) |

Поделиться
HTML-код
  • Опубликовано: 16 сен 2024
  • Learn how to find the Radius of the circle. Important Geometry and Algebra skills are also explained: Intersecting Chords theorem; perpendicular bisector theorem; Pythagorean theorem; Thales' Theorem; Euclid's Theorem. Step-by-step tutorial by PreMath.com.
    Today I will teach you tips and tricks to solve the given olympiad math question in a simple and easy way. Learn how to prepare for Math Olympiad fast!
    Step-by-step tutorial by PreMath.com
    • Can you find the Radiu...
    Need help with solving this Math Olympiad Question? You're in the right place!
    I have over 20 years of experience teaching Mathematics at American schools, colleges, and universities. Learn more about me at
    / premath
    Can you find the Radius of the circle? | (3 Methods) | #math #maths | #geometry
    Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
    #FindRadius #Circle #perpendicularBisectorTheorem #IntersectingChordsTheorem #Chords #GeometryMath #PythagoreanTheorem #Euclid'sTheorem #EuclideanTheorem
    #MathOlympiad #RightTriangle #RightTriangles #ThalesTheorem
    #PreMath #PreMath.com #MathOlympics #HowToThinkOutsideTheBox #ThinkOutsideTheBox #HowToThinkOutsideTheBox? #FillInTheBoxes #GeometryMath #Geometry #RightTriangles
    #OlympiadMathematicalQuestion #HowToSolveOlympiadQuestion #MathOlympiadQuestion #MathOlympiadQuestions #OlympiadQuestion #Olympiad #AlgebraReview #Algebra #Mathematics #Math #Maths #MathOlympiad #HarvardAdmissionQuestion
    #MathOlympiadPreparation #LearntipstosolveOlympiadMathQuestionfast #OlympiadMathematicsCompetition #MathOlympics #CollegeEntranceExam
    #blackpenredpen #MathOlympiadTraining #Olympiad Question #GeometrySkills #GeometryFormulas #Angles #Height #ComplementaryAngles
    #MathematicalOlympiad #OlympiadMathematics #CompetitiveExams #CompetitiveExam
    How to solve Olympiad Mathematical Question
    How to prepare for Math Olympiad
    How to Solve Olympiad Question
    How to Solve international math olympiad questions
    international math olympiad questions and solutions
    international math olympiad questions and answers
    olympiad mathematics competition
    blackpenredpen
    Andy Math
    math olympics
    olympiad exam
    olympiad exam sample papers
    math olympiad sample questions
    math olympiada
    British Math Olympiad
    olympics math
    olympics mathematics
    olympics math activities
    olympics math competition
    Math Olympiad Training
    How to win the International Math Olympiad | Po-Shen Loh and Lex Fridman
    Po-Shen Loh and Lex Fridman
    Number Theory
    There is a ridiculously easy way to solve this Olympiad qualifier problem
    This U.S. Olympiad Coach Has a Unique Approach to Math
    The Map of Mathematics
    mathcounts
    math at work
    Pre Math
    Olympiad Mathematics
    Two Methods to Solve System of Exponential of Equations
    Olympiad Question
    Find Area of the Shaded Triangle in a Rectangle
    Geometry
    Geometry math
    Geometry skills
    Right triangles
    imo
    Competitive Exams
    Competitive Exam
    Calculate the length AB
    Pythagorean Theorem
    Right triangles
    Intersecting Chords Theorem
    coolmath
    my maths
    mathpapa
    mymaths
    cymath
    sumdog
    multiplication
    ixl math
    deltamath
    reflex math
    math genie
    math way
    math for fun
    Subscribe Now as the ultimate shots of Math doses are on their way to fill your minds with the knowledge and wisdom once again.

Комментарии • 74

  • @Piyushkumar100-j3n
    @Piyushkumar100-j3n 3 месяца назад +5

    Nice explaination❤

    • @PreMath
      @PreMath  3 месяца назад +2

      Glad you liked it
      Thanks for the feedback ❤️

  • @unknownidentity2846
    @unknownidentity2846 3 месяца назад +2

    Let's find the radius:
    .
    ..
    ...
    ....
    .....
    First of all we calculate the length of AB:
    A(ABCD) = AB*AD
    ⇒ AB = A(ABCD)/AD = (540cm²)/(15cm) = 36cm
    Now let AE be the diameter of the circle with B located on AE. According to Thales theorem the triangle ACE is a right triangle, so we can apply the right triangle altitude theorem:
    BC² = AB*BE
    ⇒ BE = BC²/AB = AD²/AB = (15cm)²/(36cm) = (25/4)cm = 6.25cm
    Now we are able to calculate the radius R of the circle:
    R = AE/2 = (AB + BE)/2 = (36cm + 6.25cm)/2 = (42.25cm)/2 = 21.125cm
    Best regards from Germany

    • @PreMath
      @PreMath  3 месяца назад

      Excellent job!
      Thanks for sharing ❤️

  • @jimlocke9320
    @jimlocke9320 3 месяца назад +1

    For a fourth method, we note that ΔABC and ΔACE are similar by angle - angle (common angle

    • @PreMath
      @PreMath  3 месяца назад

      Excellent!
      Thanks for the feedback ❤️

  • @quigonkenny
    @quigonkenny 3 месяца назад +2

    Rectangle ABCD:
    A = hw
    540 = 15AB
    AB = 540/15 = 36
    First method:
    Triangle ∆OBC:
    OB² + BC² = CO²
    (36-r)² + 15² = r²
    1296 - 72r + r² + 225 = r²
    72r = 1521
    r = 1521/72 = 169/8 cm
    Second method:
    Extend AB to the circumference at E. Let BE = x. Extend CB to the circumference at F. As ∠OBC = 90°, BF = BC = 15, as if a radius of a circle intersects a chord perpendicularly, it bisects that chord.
    For any two intersecting chords of a circle, the products of their lengths on either side of the intersection point are the same. Thus:
    AB•BE = CB•BF
    36x = 15(15) = 225
    x = 225/36 = 25/4
    AE = AB + BE
    2r = 36 + 25/4 = 169/4
    r = (169/4)/2 = 169/8 cm
    Third method:
    Let G be the point where DC intersects with the circumference. Extend CB to F. As ∠OBC = 90°, BF = BC = 15, as if a radius of a circle intersects a chord perpendicularly, it bisects that chord.
    Draw radius OP so that OP intersects GC perpendicularly at H. By the same rule as above, GH = HC, as OP bisects the chord GC. As ∠OHC = ∠HCB = ∠CBO = 90°, ∠BOH must also equal 90°, and OHCB is a rectangle. Therefore GC = 2OB, as OB = HC = GH.
    Draw diameter GF. As G, F, and C, are point on the circumference and ∠GCF is a 90° angle connecting them all, G and F must be opposite ends of a diameter of the circle. Thus GF and O are collinear.
    Triangle ∆GCF:
    GC² + CF² = GF²
    (2(36-r))² + (2(15))² = (2r)²
    4(36-r)² + 4(15)² = 4(r²)
    (36-r)² + 15² = r²
    1296 - 72r + r² + 225 = r²
    72r = 1521
    r = 1521/72 = 169/8 cm

    • @PreMath
      @PreMath  3 месяца назад

      Excellent job!
      Thanks for sharing ❤️

  • @devondevon4366
    @devondevon4366 3 месяца назад +1

    21.125
    Line AB = 540/15 = 36
    Since AO = the radius= r
    then OB = 36- r
    Let's construct a triangle BC0
    BC=DA =15
    OC= r
    Hence, r^2 = (36-r)^2 + 15^2
    r^2 = 1,296 + r^2 -72r + 225
    r^2 = 1521 + r^2 - 72 r
    0 = 1521 - 72r
    72r = 1521
    r = 1521/72
    r = 21.125 Answer

    • @PreMath
      @PreMath  3 месяца назад

      Excellent!
      Thanks for sharing ❤️

  • @ChuzzleFriends
    @ChuzzleFriends 2 месяца назад

    A = lw
    540 = 15l
    l = 36
    So, the length of rectangle ABCD is 36 cm.
    Draw the radius of ⊙O containing vertex B. Label the point on the circle E. This forms a diameter AE of the circle.
    Extend side CB to another point F on the circle, such that the new segment, CF, is a chord of the circle.
    By the Perpendicular Chord Bisector Theorem, the diameter bisects the chord.
    Because BC = 15 (by the Parallelogram Opposite Sides Theorem), BF = 15 as well. Use the Intersecting Chords Theorem.
    36 * (2r - 36) = 15 * 15
    72r - 1296 = 225
    72r = 1521
    r = 1521/72
    = 169/8
    So, the radius of the circle is 169/8 centimeters (fraction), or 21.125 centimeters (decimal).

  • @MateusMuila
    @MateusMuila 3 месяца назад +1

    strategical solutions , Very amazing . Thanks Sir

    • @PreMath
      @PreMath  3 месяца назад

      Most welcome dear🌹
      Thanks for the feedback ❤️

  • @trumpetbob15
    @trumpetbob15 3 месяца назад +1

    It was interesting to see Euclid's Theorem right after your second method that basically shows a proof of it using the circle. I've never thought about it and why it works like that before.

    • @PreMath
      @PreMath  3 месяца назад

      Excellent!
      Thanks for the feedback ❤️

  • @MrPaulc222
    @MrPaulc222 3 месяца назад +1

    After looking at the diagram for a while and doing a few calculations in my head I came up with
    15*15 = 36(2r - 36) as intersecting chords.
    225 = 72r - 1296
    1521 = 72r
    They both divide by 3 so 24r = 507
    Again: 8r = 169
    So r= 169/8
    That is 21 and 1/8 so 21.125
    This shows me that even if something looks slightly complex, it is sometimes possible to work without a calculator

    • @PreMath
      @PreMath  3 месяца назад +1

      Excellent job!
      Thanks for the feedback ❤️

  • @phungpham1725
    @phungpham1725 3 месяца назад +1

    4th approach:
    1/ DC intersects the circle at point F.
    Let DF = x
    By using the tangent theorem:
    DF.DC= sq AD--> x= sq AD/DC= 225/36= 6.25
    If we drop the height FH to the diameter we have AH= x= 6.25
    Notice that the AFCE is a isosceles trapezoid so AH=BE= 6.25
    --> AE= 2r = 36+6.25
    r = 21.125 cm

    • @PreMath
      @PreMath  3 месяца назад

      Excellent!
      Thanks for sharing ❤️

    • @krishnaramachandran7722
      @krishnaramachandran7722 3 месяца назад

      When I looked at the rectangle and the tangent DA to the circle, I thought the first method would be the 4th method.

  • @hongningsuen1348
    @hongningsuen1348 3 месяца назад

    Point E is redunctant for solution as shown by the following method using trigonometric ratios:
    1. Right-angled triangle ABC has sides ratios of 5:12:13. (15 and 36 given, 39 found by Pythagoras theorem)
    For angle BAC, sin (BAC) = 5/13, cos(BAC) = 12/5
    2. Angle BOC = 2 x angle BAC (exterior angle of isosceles triangle AOC)
    Hence Sin (BOC) = 2 sin (BAC) cos (BAC) (double angle formula for sin function)
    = (2)(5/13)(12/13) = 120/169
    3. Radius = OC = BC/sin (BOC) = (15)/(120/169) = 169/8.

    • @PreMath
      @PreMath  3 месяца назад

      Excellent!
      Thanks for sharing ❤️

  • @sergeyvinns931
    @sergeyvinns931 3 месяца назад +1

    АВ=540:15=36; АО=ОС=R, OB=36-R, R^2=15^2+(36-R)^2, R^2=225+1296-72R+R^2, 72R=1521, R=21,125!

    • @PreMath
      @PreMath  3 месяца назад

      Excellent!
      Thanks for sharing ❤️

  • @marcgriselhubert3915
    @marcgriselhubert3915 3 месяца назад

    Let'a use an orthonormal, center P (the center of the rectangle), first axis parallel to (AB). O is the intersection of (AB) and of the mediatrice Delta of [A,C].
    VectorAC(36; 15) is orthogonal to Delta and Delta contains P, the equation of Delta is then 36.x +15.y = 0, or 12.x + 5.y = 0.
    The equation of (AB) is y = -15/2, then we have O(-15/2; 25/8). Then VectorOC(119/8; 15) and R^2 = OC^2 = 14161/64 + 225 = 28561/64, so R = 169/8.

  • @himo3485
    @himo3485 3 месяца назад +2

    Pythagorean Theorem :
    AB=540/15=36cm
    r²=(36-r)²+15² r²=1296-72r+r²+225 72r=1521 r=21.125cm
    Chords Theorem :
    AB=540/15=36cm
    15*15=36*(2r-36) 225=72r-1296 72r=1521 r=21.125cm

    • @PreMath
      @PreMath  3 месяца назад

      Excellent!
      Thanks for sharing ❤️

  • @santiagoarosam430
    @santiagoarosam430 3 месяца назад +1

    540/15=36=AB
    r^2-(36-r)^2=15^2 ; r=169/8=21,125.
    Gracias y un saludo cordial.

    • @PreMath
      @PreMath  3 месяца назад

      Excellent!
      Thanks for sharing ❤️

  • @hongningsuen1348
    @hongningsuen1348 3 месяца назад +1

    In case the problem is an MCQ in exam, a quick solution using calculator can be found in 30 seconds:
    1. angle BAC = arc tan (BC/AB) = arc tan (15/36) = 22.62
    2. angle BOC = 2 x angle BAC = 45.24 (exterior angle of triangle)
    3. radius = OC = BC/sin BOC = 15/0.71 = 21.125 = 169/8.

    • @hongningsuen1348
      @hongningsuen1348 3 месяца назад

      A more formal version of this method:
      1. Triangle ABC has sides ratios of 5:12:13.
      For angle BAC, sin (BAC) = 5/13, cos(BAC) = 12/5
      2. Angle BOC = 2 x angle BAC (exterior angle of isosceles triangle AOC)
      Hence Sin (BOC) = 2 sin (BAC) cos (BAC) (double angle formula for sin function)
      = (2)(5/13)(12/13) = 120/169
      3. Radius = OC = BC/sin (BOC) = (15)/(120/169) = 169/8.

    • @PreMath
      @PreMath  3 месяца назад

      Excellent!
      Thanks for sharing ❤️

  • @aljawad
    @aljawad 2 месяца назад

    I reached the same conclusion when solving the problem using the equation of the circle.

  • @spelunkerd
    @spelunkerd 3 месяца назад

    For some reason I always choose the longest path to the solution, or at least the only one needing a calculator. In this case I found angle BAC from arctan (15/36). From the inscribed angle theorem I knew angle EOC is double that angle, exactly the same as angle BOC. I calculated OC to be 15/(sin(angle BOC)).

  • @Waldlaeufer70
    @Waldlaeufer70 3 месяца назад

    4th method:
    Tangent-chord theorem:
    L = 540 cm² / 15 cm = 36 cm
    L * x = 15²
    x = 15² / 36 = 225 / 36 = 6.25 cm
    d = L + x = 36 cm + 6.25 cm = 42.25 cm
    r = d / 2 = 42.25 cm / 2 = 21.125 cm

    • @PreMath
      @PreMath  3 месяца назад

      Excellent!
      Thanks for sharing ❤️

  • @giuseppemalaguti435
    @giuseppemalaguti435 3 месяца назад +1

    AB=36...15^2+(36-r)^2=r^2...r=1521/72=169/8

    • @PreMath
      @PreMath  3 месяца назад

      Excellent!
      Thanks for sharing ❤️

  • @AbdulJaleel-rd5ul
    @AbdulJaleel-rd5ul 2 месяца назад

    For a triangle inscribed in a circle , intersecting chord theorem and Euclid's formula is same.

  • @CloudBushyMath
    @CloudBushyMath 3 месяца назад +2

    Cool👍

    • @PreMath
      @PreMath  3 месяца назад

      Glad to hear that!
      Thanks for the feedback ❤️

  • @LionelMaxwellCrKohli01237
    @LionelMaxwellCrKohli01237 3 месяца назад

    Angle at centre is twice than angle at circumference. Simple.

  • @misterenter-iz7rz
    @misterenter-iz7rz 3 месяца назад +1

    540÷15=36, (36-r)^2+15^2=r^2, 36^2-72r+15^2=0, r=(36^2+15^2)/72=(39×39)/72=(13×13)/8=169/8.😊

    • @PreMath
      @PreMath  3 месяца назад +1

      Excellent!
      Thanks for sharing ❤️

  • @alster724
    @alster724 3 месяца назад +1

    Straightforward problem

    • @PreMath
      @PreMath  3 месяца назад

      Excellent!
      Thanks for the feedback ❤️

  • @raya.pawley3563
    @raya.pawley3563 3 месяца назад +1

    Thank you

    • @PreMath
      @PreMath  3 месяца назад

      You are very welcome!
      Thanks Raya❤️

  • @LuisdeBritoCamacho
    @LuisdeBritoCamacho 3 месяца назад

    1) AB = 540 / 15 ; AB = 36 cm
    2) OA = R
    3) OB = (36 - R)
    4) OC = OA = R
    5) Pink Rectangle Area = 540 Sq cm
    6) Let the Point E be the a Vertical Line passing through O and intersecting Line CD.
    7) OC^2 = BC^2 + OB^2
    8) R^2 = 15^2 + (36 - R)^2
    9) R^2 = 225 + 1.296 - 72R + R^2
    10) 1.521 - 72R = 0
    11) 72R = 1.521
    12) R = 1.521 / 72 ; R = 507 / 24 ; R = 169 / 8 cm ; R = 21,125 cm
    13) ANSWER : The Radius of the Circle is equal to 169/8 Cm or equal to 21,125 Cm.

    • @PreMath
      @PreMath  3 месяца назад

      Excellent!
      Thanks for sharing ❤️

  • @prossvay8744
    @prossvay8744 3 месяца назад +2

    ABCD is rectangle
    so AB=CD ; AD=BC=15
    AB=CD=540/15=36cm
    Let OB=x
    OA=OC=OE=R
    R+x=AB=36
    so x=36-R
    Connect O to C
    In ∆OBC
    OB^2+BC^2=OC^2
    x^2+15^2=R^2
    R^2-x^2=225
    R^2-(36-R)^2=225
    So R=169/8cm=21.125cm.❤❤❤ Best regards.

    • @PreMath
      @PreMath  3 месяца назад

      Excellent job!
      Thanks for sharing ❤️

  • @wackojacko3962
    @wackojacko3962 3 месяца назад +1

    So @ 10:12
    CA² = AB × AE
    CE² = EB × AE
    CB² = AB × EB
    🙂

    • @PreMath
      @PreMath  3 месяца назад +1

      Excellent!
      Thanks for the feedback ❤️

  • @PREPARE_GURU
    @PREPARE_GURU 3 месяца назад +1

    Sir, I want to know that from where you get these type of questions

    • @PreMath
      @PreMath  3 месяца назад +1

      Hello dear, we don't have such magic book!
      It takes enormous amount of time for research and planning to prepare quality videos! Thanks for asking❤️

  • @AmirgabYT2185
    @AmirgabYT2185 3 месяца назад +2

    r=21,125

    • @PreMath
      @PreMath  3 месяца назад +1

      Excellent!
      Thanks for sharing ❤️

  • @soli9mana-soli4953
    @soli9mana-soli4953 3 месяца назад

    R^2 = 15^2 + (36 - R)^2
    15*15 = 36*X

    • @conconstantin7454
      @conconstantin7454 3 месяца назад

      Exactly what I did as well

    • @PreMath
      @PreMath  3 месяца назад

      Excellent!
      Thanks for sharing ❤️

  • @grantschiff7544
    @grantschiff7544 3 месяца назад

    I know 15, so make a quick measuring tape out of paper and done.

    • @PreMath
      @PreMath  3 месяца назад +1

      Thanks for the feedback ❤️

  • @kalavenkataraman4445
    @kalavenkataraman4445 3 месяца назад +1

    21.125

    • @PreMath
      @PreMath  3 месяца назад

      Excellent!
      Thanks for sharing ❤️

  • @himadrikhanra7463
    @himadrikhanra7463 3 месяца назад

    21..?

  • @jimwhalen5675
    @jimwhalen5675 3 месяца назад

    First present the problem clearly and cogentlynwith proper.description

    • @PreMath
      @PreMath  3 месяца назад

      Thanks for the feedback ❤️

  • @riteshkumar8152
    @riteshkumar8152 3 месяца назад +1

    First comment

    • @PreMath
      @PreMath  3 месяца назад

      Excellent!
      Thanks dear❤️

  • @joiceroosita5317
    @joiceroosita5317 3 месяца назад

    Its too easy..

    • @PreMath
      @PreMath  3 месяца назад

      Thanks ❤️

  • @devondevon4366
    @devondevon4366 3 месяца назад +1

    21.125

    • @PreMath
      @PreMath  3 месяца назад

      Excellent!
      Thanks for sharing ❤️