Can you find area of the Trapezoid? | (Trapezium) |

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  • เผยแพร่เมื่อ 11 ธ.ค. 2024

ความคิดเห็น • 30

  • @wackojacko3962
    @wackojacko3962 12 วันที่ผ่านมา +2

    Love the algebra reinforcement! It's got a lot of rhythm and soul...😊

    • @PreMath
      @PreMath  12 วันที่ผ่านมา +1

      Excellent!
      Glad to hear that!
      Thanks for the feedback ❤️🙏

  • @Dr_piFrog
    @Dr_piFrog 12 วันที่ผ่านมา +1

    Nailed it.

  • @ChuzzleFriends
    @ChuzzleFriends 12 วันที่ผ่านมา

    Draw a segment thru point C and a point E on base AB such that it is perpendicular to the bases of right trapezoid ABCD. This is the height CE of the trapezoid.
    This segment forms a square ADCE and a right △BEC (We know ADCE is a square because AD = CD).
    Label the side length of the square as s.
    Then, BE = AB - AE = (1 + √3) - s.
    Use the Pythagorean Theorem on △BEC.
    a² + b² = c²
    s² + [(1 + √3) - s]² = 2²
    s² + [(1 + √3)² - 2s(1 + √3) + s²] = 4
    s² + [(1 + 2√3 + 3) - 2s - 2s√3 + s²] = 4
    s² + 4 + 2√3 - 2s - 2s√3 + s² = 4
    2s² + 2√3 - 2s - 2s√3 = 0
    2s² - 2s√3 - 2s + 2√3 = 0 (We can try to group)
    2s(s - √3) - 2(s - √3) = 0
    (2s - 2)(s - √3) = 0
    2s - 2 = 0 or s - √3 = 0
    2s = 2 s = √3
    s = 1
    Both lengths are reasonable. BE = (1 + √3) - 1 = √3 or (1 + √3) - √3 = 1.
    Find the areas of the square and the triangle. It will be easier to find the area of the trapezoid this way.
    First, substitute s = 1.
    A = s²
    = 1²
    = 1
    A = (bh)/2
    = (√3 * 1)/2
    = (√3)/2
    So, one answer for the area of the trapezoid is 1 + (√3)/2 square units, a. w. a. (2 + √3)/2 square units (exact), or about 1.87 square units (approximation).
    Next, substitute s = √3.
    A = (√3)²
    = 3
    A = (bh)/3
    = (1 * √3)/2
    = (√3)/2
    So, another answer for the area of the trapezoid is 3 + (√3)/2 square units, a. w. a. (6 + √3)/2 square units (exact), or about 3.87 square units (approximation).
    So, the possible areas of the trapezoid are as follows:
    A = 1 + (√3)/2 u² = (2 + √3)/2 u² ≈ 1.87 u²
    A = 3 + (√3)/2 u² = (6 + √3)/2 u² ≈ 3.87 u²

  • @alster724
    @alster724 12 วันที่ผ่านมา

    I did the standard trapezoid area formula and it worked.
    Also, I did factoring by grouping to solve for x

  • @StuartSimon
    @StuartSimon 11 วันที่ผ่านมา

    After you solved for x, you could just use the trapezoid area formula in each case: A = (x^2 + x + x*sqrt3)/2

  • @jamestalbott4499
    @jamestalbott4499 12 วันที่ผ่านมา

    Thank you!

  • @unknownidentity2846
    @unknownidentity2846 12 วันที่ผ่านมา +2

    Let's find the area:
    .
    ..
    ...
    ....
    .....
    First of all we add point E on AB such that BCE is a right triangle. Since AD=CD, in this case ADCE is a square. Now we apply the Pythagorean theorem to the right triangle BCE:
    BC² = BE² + CE² = (AB − AE)² + CE² = (AB − CE)² + CE²
    2² = [(√3 + 1) − CE]² + CE²
    4 = (√3 + 1)² − 2(√3 + 1)*CE + CE² + CE²
    4 = 3 + 2√3 + 1 − 2(√3 + 1)*CE + 2*CE²
    0 = 2*CE² − 2(√3 + 1)*CE + 2√3
    0 = CE² − (√3 + 1)*CE + √3
    ⇒ Δ = [(√3 + 1)/2]² − √3 = (3 + 2√3 + 1)/4 − 4√3/4 = = (3 − 2√3 + 1)/4 = [(√3 − 1)/2]²
    ⇒ CE = (√3 + 1)/2 ± √Δ = (√3 + 1)/2 ± (√3 − 1)/2
    First solution:
    CE = (√3 + 1)/2 + (√3 − 1)/2 = (√3 + 1 + √3 − 1)/2 = √3
    A(ABCD) = (1/2)*(AB + CD)*AD = (1/2)*(AB + CE)*CE = (1/2)*(√3 + 1 + √3)*√3 = (√3 + 6)/2
    Second solution:
    CE = (√3 + 1)/2 − (√3 − 1)/2 = (√3 + 1 − √3 + 1)/2 = 1
    A(ABCD) = (1/2)*(AB + CD)*AD = (1/2)*(AB + CE)*CE = (1/2)*(√3 + 1 + 1)*1 = (√3 + 2)/2
    Best regards from Germany

  • @santiagoarosam430
    @santiagoarosam430 12 วันที่ผ่านมา

    ABCº=60º---> AD=2√3/2=√3---> AB=√3+(2/2)---> ABCD =(√3)²+(1*√3/2) =3+(√3/2)
    Gracias y saludos

  • @Ibrahimfamilyvlog2097l
    @Ibrahimfamilyvlog2097l 12 วันที่ผ่านมา

    Thanks so much sir ❤❤

    • @PreMath
      @PreMath  12 วันที่ผ่านมา

      You're very welcome! ❤️

  • @alexundre8745
    @alexundre8745 12 วันที่ผ่านมา +1

    Bom dia Mestre

  • @PrithwirajSen-nj6qq
    @PrithwirajSen-nj6qq 12 วันที่ผ่านมา +1

    Complete 🔺 ABC
    (√3+1)/sin ACB=2/sin BAC=2/sin45=2√2
    >sin ACB =(√3 +1)/2√2=sin 75=sin 105
    Then ACB = 75 or 105
    Hence BDE =75-45=30 or
    105 - 45=60
    Hence 🔺 BDE is a 30-60-90 triangle whose angle BDE =30 or 60
    When it is 30, BE =1and side of square is √3
    When it is 60, BE =√3.side of square is 1

  • @marcgriselhubert3915
    @marcgriselhubert3915 11 วันที่ผ่านมา

    Let's name c = AD = DC. In triangle EBC we have (c^2) + (1 +sqrt(3) - c)^2 = 2^2, then (c^2) -(1 + sqrt(3)).c +sqrt(3) = 0, and the c = 1 or c = sqrt(3).
    The area of the trapezoïd is A = (1/2).(DC + AB).AD = (1/2).(c^2).(c + 1 + sqrt(3).
    If c = 1, then A = (2 + sqrt(3))/2; if c = sqrt(3) then A = (6 + sqrt(3))/2.

  • @quigonkenny
    @quigonkenny 12 วันที่ผ่านมา

    Let AD = DC = x. Drop a perpendicular from C to E on AB. As ∠ADC = ∠DCE = ∠CEA = ∠EAD = 90° and adjacent sides AD and DC equal x, then ADCE is a square with side length x. BE = (1+√3)-x.
    Let ∠CBA = θ.
    sinθ = EC/CB = x/2
    x = 2sinθ
    cosθ = BE/CB = ((1+√3)-x)/2
    2cosθ = (1+√3) - 2sinθ
    2sinθ + 2cosθ = 1 + √3
    2(sinθ+cosθ) = 1 + √3
    4(sin²θ+2sinθcosθ+cos²θ) = 1 + 2√3 + 3
    4(1+sin2θ) = 4 + 2√3
    4 + 4sin2θ = 4 + 2√3
    4sin2θ = 2√3
    sin2θ = (2√3)/4 = √3/2
    2θ = 60° | 2θ = 120° --- 0° < θ < 90°
    θ = 30° | θ = 60°
    Trapezoid ABCD:
    A₁ = x(x+1+√3)/2
    A₁ = 2sin30°((2sin30°)+1+√3)/2
    A₁ = 2(1/2)(2(1/2)+1+√3)/2
    A₁ = (1+1+√3)/2 = (2+√3)/2
    [ A₁ = 1 + √3/2 ≈ 1.866 sq units ]
    A₂ = x(x+1+√3)/2
    A₂ = 2sin60°((2sin60°)+1+√3)/2
    A₂ = 2(√3/2)(2(√3/2)+1+√3)/2
    A₂ = (√3)(1+2√3)/2 = (6+√3)/2
    [ A₂ = 3 + √3/2 ≈ 3.866 sq units ]

  • @misterenter-iz7rz
    @misterenter-iz7rz 12 วันที่ผ่านมา

    (1+sqrt(3)-s)^2=4-s^2, 2-s^2=2+sqrt(3)-(1+sqrt(3))s, s^2-(1+sqrt(3))s+2=0,........😅

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 12 วันที่ผ่านมา

    (2)^2=4 {1x+1x ➖}+{3x+3x ➖ }={4+2x^2+6x^2}=12x^4 {90°A+90°B+45°C+45°D}=180°ABCD/12x^4=15x^4ABCD 3^5x^4 3^2^3x^2^2 1^1^3x^1^2 3x^2 (ABCDx ➖ 3ABCDx+2).

  • @wasimahmad-t6c
    @wasimahmad-t6c 12 วันที่ผ่านมา

    1.866

  • @PrithwirajSen-nj6qq
    @PrithwirajSen-nj6qq 12 วันที่ผ่านมา

    Sir, may I offer a conceptual solution of the problem keeping ur construction?
    The side of square will be 1or √3
    If side is 1
    BE= √3 and based on 🔺 BCE we may say CE =2^2 -√3^2=1
    Hence assumption for taking the sides as1 unit is right.
    Again if the side is √3 then BE will be 1unit and based on 🔺 BCE we see CE =2^2-1=√3
    Hence the assumption for taking the side √3 unit is correct.
    Hence parallel sides of trapezium is
    1 and 1+√3(case 1)
    Or √3 and 1+√3(case 2)
    For case 1
    Area of trapezium is 1/2(1+1+√3) sq units
    For case 2
    Area of trapezium is
    1/2(√3+√3+1) sq units

  • @wasimahmad-t6c
    @wasimahmad-t6c 12 วันที่ผ่านมา

    1+1.732+1=3.732÷2×1=1.866

  • @LuisdeBritoCamacho
    @LuisdeBritoCamacho 12 วันที่ผ่านมา +1

    STEP-BY-STEP RESOLUTION PROPOSAL :
    01) AD = DC = X
    02) BC = 2
    03) AB = (1 + sqrt(3))
    04) Drop a Vertical Line from Point C until AB. Find Point E.
    05) AE = X and EB = (1 + sqrt(3) - X)
    06) BC^2 = CE^2 + EB^2
    07) 4 = X^2 + (1 + sqrt(3) - X)^2
    08) Two Solutions : X = 1 and X = sqrt(3)
    09) Solution #1 :
    a) AT = [1 + (1 + sqrt(3))] / 2
    b) AT = (2 + sqrt(3)) / 2
    c) AT = 1 + sqrt(3) / 2
    d) AT ~ 1,87
    10) Solution #2 :
    a) AT = [sqrt(3) + (1 + sqrt(3))] * sqrt(3) / 2
    b) AT = (1 + 2sqrt(3)) * sqrt(3) / 2
    c) AT = (sqrt(3) + (2 * 3)) / 2
    d) AT = 3 + sqrt(3)/2
    e) AT ~ 3,87
    Therefore,
    OUR BEST ANSWER :
    Trapezoid Area equal to approx. 1,87 Square Units or approx. equal to 3,87 Square Units.

    • @PreMath
      @PreMath  12 วันที่ผ่านมา +1

      Excellent!
      Thanks for sharing ❤️

  • @cyruschang1904
    @cyruschang1904 12 วันที่ผ่านมา

    X^2 + (1 + √ 3 - x)^2 = 2^2 = 4
    2X^2 - 2(1 + √ 3)x + (1 + √ 3)^2 = 4
    2X^2 - 2(1 + √ 3)x + 2√ 3= 0
    X^2 - (1 + √ 3)x + √ 3= 0
    X = 1 or √ 3
    area = (1 + √ 3 + 1)/2 or (1 + √ 3 + √ 3)(√ 3)/2 = 1 + (√ 3)/2 or 3 + (√ 3)/2

  • @Birol731
    @Birol731 12 วันที่ผ่านมา

    My way of solution ▶
    [AE]= [EC]= [CD]= [DA] = x
    E ∈ [AB] and [AE] ⊥ [EC]

    [AE]= x
    [EB]= 1+ √3 -x
    By considering the right triangle ΔEBC :
    [CE]= x
    [EB]= 1+ √3 -x
    [BC]= 2
    According to the Pythagorean theorem, we can write:
    x² + (1+ √3 -x)²= 2²
    x² + 1+ √3 - x + √3 + 3- x√3 - x - x√3 + x²= 4
    2x²+4+2√3 - 2x -2√3 x = 4
    boh sides divided by 2
    x²-x-√3x +2+√3 = 2
    x(x-1) - √3(x-1)= 0
    (x-1)[x-√3]= 0

    x₁= 1 ✅
    x₂= √3 ✅
    Both solutions are possible !
    For x= 1
    A(AECD)= x²
    A(AECD)= 1
    A(ΔEBC)= (1+ √3 -1)*1/2
    A(ΔEBC)= √3/2 square units
    A(ABCD)= A(AECD) + A(ΔEBC)
    A(ABCD)= 1+ √3/2
    A(ABCD) ≈ 1,866 square units ✅
    For x= √3
    A(AECD)= (√3)²
    A(AECD)= 3
    A(ΔEBC)= (1+ √3 -√3)*√3/2
    A(ΔEBC)= √3/2 square units
    A(ABCD)= A(AECD) + A(ΔEBC)
    A(ABCD)= 3+ √3/2
    A(ABCD) ≈ 3,866 square units ✅
    Therefore we have two solutions !

  • @PrithwirajSen-nj6qq
    @PrithwirajSen-nj6qq 12 วันที่ผ่านมา

    1)Join AC
    Ang BAC = 45 degrees
    2)
    Triangle BAC
    sin (Ang ACB) /(1+√3)
    = [sin (ang BAC)]/2
    > sin (ang ACB) /(1+√3)
    =(sin 45)/2=1/2√2
    Then sin(ACB)= (1+√3)/2√2=sin75 or sin 105
    Hence ACB =75 or 105 degrees
    Then ang ABC is 180-45-105=30
    Or 180-45-75=60
    3)
    Drop perpendicular CE on AB
    Now triangle is rt triangle whose hypotenuse is 2 units
    Hence side opposite to 30 degree will be 1 unit and that of 60 degree will be √3 unit
    Hence CE =1 or √3
    BE =√3 or 1
    4)
    Hence the parallel sides of trapezium
    √3 and 1+√3
    Or
    1 and 1+√3
    Area of trapezium
    =1/2(1+1+√3)
    Or 1/2(√3+√3+1) sq units

  • @PrithwirajSen-nj6qq
    @PrithwirajSen-nj6qq 12 วันที่ผ่านมา

    Method 2
    Previous method part 1
    Add AC =x √2
    Previous method part 2 keep as it is
    The part 3 now be replaced by the bellow
    AC /sin ABC =2/sin45=2√2
    >x √2/sin30=2√2
    x = 1
    Or AC /sin ABC =2/sin45
    > x √2/sin60=2√2
    x =√3
    The 4th will be as it is

  • @gelbkehlchen
    @gelbkehlchen 9 วันที่ผ่านมา

    Solution:
    E = point exactly below C.
    AD = DC = CE = AE = x
    Pythagoras for the right-angled triangle EBC:
    (1+√3-x)²+x² = 2² ⟹
    (1+√3)²-2*(1+√3)*x+2x² = 4 ⟹
    2x²-2*(1+√3)*x+4+2*√3 = 4 |/2 ⟹
    x²-(1+√3)*x+2+√3 = 2 |-2 ⟹
    x²-(1+√3)*x+√3 = 0 |p-q formula ⟹
    x1/2 = (1+√3)/2±√[(1+√3)²/4-√3] = (1+√3)/2±√[(1+2*√3+3)/4-√3]
    = (1+√3)/2±√[1+√3/2-√3] = (1+√3)/2±√(1-√3/2) ⟹
    x1 = (1+√3)/2+√(1-√3/2) = 1.7320508075688773 and
    x2 = (1+√3)/2-√(1-√3/2) = 1 ⟹
    Checking the result for x1 = (1+√3)/2+√(1-√3/2) = 1.7320508075688773:
    left Side:
    (1+√3-1.7320508075688773)²+1.7320508075688773² = 1²+1.7320508075688773² = 4
    right side: 2² = 4 everything OK
    Area of ​​the trapezoid = 1²+1*√3/2 = 1+√3/2 ≈ 1.8660
    Checking the result for x2 = 1:
    left side:
    (1+√3-1)²+1² = 3+1 = 4
    right side: 2² = 4 everything OK
    Area of ​​the trapezoid = (√3)²+1*√3/2 = 3+√3/2 ≈ 3.8660

  • @RupaliDhasal
    @RupaliDhasal 12 วันที่ผ่านมา

    I'm second

  • @Brotherman7
    @Brotherman7 12 วันที่ผ่านมา

    I’m here to be first.

  • @lasalleman6792
    @lasalleman6792 12 วันที่ผ่านมา

    I set up a triangle ABC. Angle at A is 45 degrees, Line AB is 2.73, Line CB is 2. Use the sine rule. Angle C becomes 74.84. Angle B becomes 60.16 degrees. Height of the triangle CE is 1.73. Area of the set up triangle ABC is thus 2.368. Add area of remaining figure, triangle ADC, which is 1.496. Total area of figure is 3.864. More or less equal to the narrator's suggested solution.