Can you the Perimeter a+b+c? | (Semicircle) |

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  • Опубликовано: 24 дек 2024

Комментарии • 20

  • @jamestalbott4499
    @jamestalbott4499 10 часов назад +2

    Thank you!

    • @PreMath
      @PreMath  7 часов назад

      You are very welcome! 😊🙏

  • @imetroangola17
    @imetroangola17 11 часов назад +2

    *_Nota de Aula:_*
    c² = a²+ b² = (a+b)² - 2ab
    (a+b)² = c² + 2ab. Ora,
    c²=175 e ab= 2×[ABC] = 2×35=70. Daí,
    (a+b)² = 175 + 140 = 315 = 9×35. Assim,
    a + b = 3√35. Portanto,
    *a + b + c = 3√35 + 5√7*

  • @himo3485
    @himo3485 15 часов назад +1

    7 : 28 = 1 : 4 = 1² : 2²
    AD=k CD=2k DB=4k AB=AD+DB=5k
    5k*2k*1/2=7+28 5k²=35 k2=7 k=√7
    a²=(√7)²+(2√7)²=35 a=√35 b²=(2√7)²+(4√7)²=140 b=2√35 c=AB=5√7
    a + b + c = 3√35 + 5√7 (cm)

  • @AnonimityAssured
    @AnonimityAssured 13 часов назад +2

    The title needs correcting.

  • @alexniklas8777
    @alexniklas8777 7 часов назад

    I solved this problem with your method.
    Thanks sir!

  • @Abdelfattah-hr8tt
    @Abdelfattah-hr8tt 11 часов назад

    Im hear following you dear

  • @alexundre8745
    @alexundre8745 14 часов назад +1

    Bom dia Mestre
    Acertei
    Graças ao Sr estou aprendendo Geometria

  • @quigonkenny
    @quigonkenny 14 часов назад

    As ∆ADC = 7cm² and ∆CDB = 28cm², which is a ratio of 1:4, and the two triangles share the same height (CD), then their bases (AD and DB) must share the same ratio. Thus AD = c/5 and DB = 4c/5.
    As ∠BCA = 90° by Thales' Theorem, then as ∠ADC = ∠CDB = 90° and ∠CAB = ∠CAD = ∠BCD = 90°-∠B, then ∆BCA, ∆ADC, and ∆CDB are similar triangles.
    AD/CD = CD/DB
    (c/5)/CD = CD/(4c/5)
    CD² = (c/5)(4c/5) = 4c²/25
    CD = √(4c²/25) = 2c/5
    Triangle ∆CDB:
    A = bh/2 = DB(CD)/2
    28 = (4c/5)(2c/5)/2
    28 = (8c²/25)/2 = 4c²/25
    c² = 28(25/4) = 7(25) = 175
    c = √175 = 5√7
    CA/DC = BC/DB
    a/2√7 = b/4√7
    2√7b = 4√7a
    b = 4√7a/2√7 = 2a
    a² + b² = c²
    a² + (2a)² = (5√7)²
    a² + 4a² = 175
    5a² = 175
    a² = 175/5 = 35
    a = √35
    b = 2a = 2√35
    a + b + c = √35 + 2√35 + 5√7
    [ a + b + c = 3√35 + 5√7 ≈ 30.977cm ]

  • @marcelowanderleycorreia8876
    @marcelowanderleycorreia8876 5 часов назад

    A questão é mais difícil do que parece. Muito inteligente!! Parabéns mestre!! 👍

  • @marcgriselhubert3915
    @marcgriselhubert3915 15 часов назад

    Without thinking outside the box:
    Be h = CD. We have AD.h = 2.7 = 14 and BD.h = 2.28 = 56, so BD = 4.AD, and AD as AD +BD = c, we get that AD= (1/5).c and BD = (4/5).c
    In triangle ADC: h^2 = a^2 -(1/25).(c^2), and in triangle BDC : h^2 = b^2 - (16/25).(c^2), so b^2 - a^2 = (15/25).(c^2) = (3/5).(c^2)
    As a^2 + b^2 = c^2 in triangle ABC, we get then that b^2 = (4/5).(c^2) and a^2 = (1/5).(c^2), giving b = (2.sqrt(5))/5).c and a = (sqrt(5)/5).c
    We had h^2 = a^2 - (1/25).(c^2), so h^2 = (1/5).(c^2) - (1/25).(c^2) = (4/25).(c^2), giving that h = (2/5).c
    We had AD.h = 14, so ((1/5).c).((2/5).c) = 14, and so c^2) = (14.25)/2 = 7.25, giving that c = 5.sqrt(7)
    Now we get a = (sqrt(5)/5).c = sqrt(35) and b = ((2.sqrt(5))/5).c = 2.sqrt(35),
    and finally a + b + c = sqrt(35) + 2.sqrt(35) + 5.sqrt(7) = (sqrt(7)).(5 + 3.sqrt(5)).

  • @davidseed2939
    @davidseed2939 4 часа назад

    my method
    since the left and right triangles have an area ratio of 4 and because those triangles are similar.
    (a property of rt angled triangles divided by the perpendicular)
    then corresponding lengths in the triangles will have a ratio of √4 =2
    hence b=2a.
    now the area of the whole triangle is 35 and this is calculated as ab/2
    35=a(2a)/2 a^2=35, b^2=140 and by Pythagoras c^2 =a²+b²=175
    it follows that a=√35, b=2√35 c=5√7
    so a+b+c=3√35+5√7
    Incidentally, this diagram can be used to prove pythagoras.
    First using angle sums, the two constituent triangles and the whole triangle are all similar.
    For any similar triangles, the area of each triangle is in proportion to the square of corresponding lengths in the triangle. The chosen corresponding lengths are the hypotenuses of each triangle (a,b,c)
    ie area of ΔACD =ka² ΔDCB=kb² ΔABC=kc²
    now the two component triangles have a total area equal to the area of the large triangle.
    ie. ka²+kb²=kc²
    hence
    a² +b ² =c² QED
    😊

  • @PrithwirajSen-nj6qq
    @PrithwirajSen-nj6qq 23 минуты назад

    ABC & ACD are similar triangles.
    Hence a^2/c^2 =7/35=1/5
    > a/c =1/√5--(1)
    ABC & BCD are similar triangles.
    Hence b^/c^2=28/35=4/5
    > b/c =2/√5 --(2)
    From (1) & (2) we get
    a ঃ b ঃ c = 1ঃ 2ঃ√5=kঃ2k ঃ √5 k
    The area of the 🔺 ABC
    = 1/2*ab =35
    1/2*k*2k =35
    >k =√35
    Hence a =√35
    b =2√35
    c =5√7
    Hence Perimeter
    =(√35+2√35+5√7) units
    =(3√35 +5√7) units

  • @wasimahmad-t6c
    @wasimahmad-t6c 12 часов назад

    A=5.353 b=13.076696 c=14

  • @sergioaiex3966
    @sergioaiex3966 13 часов назад

    Solution:
    Triangles with the same height will have areas whose ratios is the same as the ratio of their bases
    In other words:
    ∆ ADC Area/∆ BDC Area
    7/28 = 1/4
    Therefore If AD = k, BD = 4k
    Applying Chords Theorem
    h² = k . 4k
    h² = 4k²
    h = 2k
    ∆ ADC
    A = ½ base × heigth
    7 = ½ k × 2k
    14 = 2k²
    k = √7
    =====
    Applying Pythagorean Theorem in ∆ ADC
    (k)² + (2k)² = a²
    k² + 4k² = a²
    a² = 5k²
    a = k √5
    a = √7 . √5
    a = √35
    ======
    c = k + 4k
    c = 5k
    c = 5√7
    ======
    Applying Thales's Theorem in ∆ ABC and, at the same time, Pythagorean Theorem, to calculate "b"
    a² + b² = c²
    (√35)² + b² = (5√7)²
    35 + b² = 175
    b² = 140
    b = 2√35
    =======
    Final Step
    a + b + c = √35 + 2√35 + 5√7
    a + b + c = 3√35 + 5√7 cm ✅
    a + b + c = 30.9769 cm ✅

  • @PrithwirajSen-nj6qq
    @PrithwirajSen-nj6qq 8 часов назад

    BDC /ADC =
    (1/2*BD*CD)/(1/2*AD*CD)= 28/7
    >BD /AD = 4/1=4x/x
    In respect of the semicircle
    CD is the geometric mean of AD and BD
    Then CD ^2=AD*BD=x*4x
    > CD =2x
    rt 🔺 ADC
    =1/2 *AD *CD =1/2*x*2x=x^2
    Now x ^2 =7
    x =√7
    AD =√7
    BD =4x =4√7
    c =AD +BD =5√7--(1)
    In rt 🔺 ADC
    a =√(AD^2 + CD ^2)
    = √(7+4*7)=√35 --(2)
    In rt 🔺 BDC
    b =√(BD^2+CD^2)
    =√(16*7+4*7)
    =2√35 ---(3)
    From (1),(2) & (3)
    a +b+c
    =√35 +2√35+ 5√7
    =(3√35 +5√8) units

  • @unknownidentity2846
    @unknownidentity2846 15 часов назад

    Let's face this Christmas challenge:
    .
    ..
    ...
    ....
    .....
    The triangle ABC can be divided into the right triangles ACD and BCD. So we can conclude:
    A(ABC) = A(ACD) + A(BCD) = (1/2)*AD*CD + (1/2)*BD*CD = (1/2)*(AD + BD)*CD
    A(ACD) = (1/2)*AD*CD ∧ A(BCD) = (1/2)*BD*CD ⇒ BD/AD = A(BCD)/A(ACD) = (28cm²/7cm²) = 4 ⇒ BD = 4*AD
    According to the theorem of Thales the triangle ABC is a right triangle. So we can apply the right triangle altitude theorem:
    CD² = AD*BD
    CD² = AD*(4*AD)
    CD² = 4*AD²
    ⇒ CD = 2*AD
    A(ACD) + A(BCD) = (1/2)*(AD + BD)*CD
    7cm² + 28cm² = (1/2)*(AD + 4*AD)*(2*AD)
    35cm² = 5*AD²
    7cm² = AD²
    ⇒ AD = √(7cm²) = (√7)cm
    CD = 2*AD = (2√7)cm
    BD = 4*AD = (4√7)cm
    Now we apply the Pythagorean theorem to the right triangles ACD and BCD:
    AC² = AD² + CD² = (√7)²cm² + (2√7)²cm² = 7cm² + 28cm² = 35cm² ⇒ AC = √( 35cm)² = (√35)cm
    BC² = BD² + CD² = (4√7)²cm² + (2√7)²cm² = 112cm² + 28cm² = 140cm² ⇒ BC = √(140cm)² = (2√35)cm
    Now we are able to calculate the perimeter of the triangle ABC:
    P(ABC) = AB + AC + BC = (AD + BD) + AC + BC = (√7)cm + (4√7)cm + (√35)cm + (2√35)cm = (5√7)cm + (3√35)cm ≈ 30.98cm
    And now (as a Christmas gift) let's check the side lengths by applying Heron's formula:
    s = (AB + AC + BC)/2 = [(5√7 + 3√35)/2]cm
    s − AB = [(5√7 + 3√35)/2 − 5√7]cm = [(3√35 − 5√7)/2]cm
    s − AC = [(5√7 + 3√35)/2 − √35]cm = [(5√7 + √35)/2]cm
    s − BC = [(5√7 + 3√35)/2 − 2√35]cm = [(5√7 − √35)/2]cm
    s * (s − AB) * (s − AC) * (s − BC)
    = [(5√7 + 3√35)/2]*[(3√35 − 5√7)/2]*[(5√7 + √35)/2]*[(5√7 − √35)/2]cm⁴
    = [(9*35 − 25*7)*(25*7 − 35)/16]cm⁴
    = [(315 − 175)*(175 − 35)/16]cm⁴
    = [140*140/16]cm⁴
    = 35²cm⁴
    ⇒ A(ABC) = √[s * (s − AB) * (s − AC) * (s − BC)] = 35cm² ✅
    Best regards from Germany

  • @Thampuran-o9o
    @Thampuran-o9o 11 часов назад

    👍👍👍