Can you find area of the total Purple shaded region? | (Square) |

Поділитися
Вставка
  • Опубліковано 15 лис 2024

КОМЕНТАРІ • 25

  • @marcgriselhubert3915
    @marcgriselhubert3915 2 дні тому +2

    Very good.

    • @PreMath
      @PreMath  2 дні тому

      Glad to hear that!
      You are very welcome!
      Thanks for the feedback ❤️

  • @michaeldoerr5810
    @michaeldoerr5810 2 дні тому +1

    The answer is 5.1 units. This is easier than it looks. And I am glad that I have reviewed how to apply AA similarity for deducing the area which is a form of subtraction. I shall use that for practice!!!

  • @cyruschang1904
    @cyruschang1904 День тому

    The white triangle base = 2 cm
    the upper right triangle and lower right triangle are similar with a linear ratio of 3 : 2
    the side of the square = x
    x^2 + (1 - 2x/3)^2 = 3^2
    x^2 + x^2/9 = 3^2
    10x^2 = 9(3^2) = 81
    x^2 = 81/10
    shaded area = (81/10) cm^2 - 3 cm^2 = (51/10) cm^2

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 17 годин тому

    (3)^2 (3)^2={9+9}=18 360°ABCD/18=2ABCD (ABCD ➖ 2ABCD+1).

  • @phungpham1725
    @phungpham1725 2 дні тому

    1/EF=2
    2/Angle BAF=angle EFC (having sides perpendicular) --> the two right triangles BAF and CFE are similar and the relevant ratio is 3/2 ( the ratio of the relevant hypotenuses)
    --> CF/a= 2/3--> CF=2a/3-> AF = a/3
    Using Pythagorean theorem
    sqa+sq a/9 = 9 --> sqa=8.1
    Area of the purple region = 5.1 sq cm😅😅o

  • @santiagoarosam430
    @santiagoarosam430 2 дні тому

    EF=2*3/3=2--->. Si AB=c--->FC=2c/3---> BF=c/3---> c²+(c/3)²=3²---> c²=81/10---> Área sombreada =(81/10)-3=51/10.
    Gracias y saludos

  • @jamestalbott4499
    @jamestalbott4499 2 дні тому +2

    Thank you!

    • @PreMath
      @PreMath  2 дні тому

      You are very welcome!
      Thanks for the feedback ❤️

  • @prossvay8744
    @prossvay8744 2 дні тому +1

    Let x is the side of the square .
    Area of the AEF=3 cm^2
    1/2(AF)(EF)=3
    1/2(3)(EF)=3
    So EF=2 cm
    ∆ABF~∆FCE
    So x/3=CF/EF
    BF=√9-x^2
    CF=x-√9-x^2
    x/3=(x-√9-x^2)/2
    2x=3x-3√9-x^2
    3√9-x^2=x
    9(9-x^2)=x^2
    81-9x^2=x^2
    10x^2=81
    So x^2=81/10 cm^2=8.1 cm^2 is area of the square.
    So Purple shaded area=8.1-3=5.1 cm^2
    ❤❤❤

  • @quigonkenny
    @quigonkenny 2 дні тому

    Triangle ∆EFA:
    A = bh/2 = FA(EF)/2
    3 = 3(EF)/2
    EF/2 = 1
    EF = 2
    Let s be the side length of square ABCD. As ∠EFA = 90°, then ∠AFB and ∠CFE are complementary angles and sum to 90°. As ∆ECF and ∆FBA are right triangles, then each contains both of the complementary angles and thus ∆ECF and ∆FBA are similar.
    CF/BA = FE/AF
    CF/s = 2/3
    CF = 2s/3
    CB = CF + FB
    s = 2s/3 + FB
    FB = s/3
    Triangle ∆FBA:
    FB² + BA² = AF²
    (s/3)² + s² = 3²
    s²/9 + s² = 9
    10s²/9 = 9
    10s² = 81
    s² = 81/10
    Shaded area:
    A = s² - 3
    A = 81/10 - 3 = 51/10
    [ A = 5.1 cm² ]

  • @sergioaiex3966
    @sergioaiex3966 День тому

    Solution:
    A ∆AEF = ½ b h
    3 = ½ b 3
    3b/2 = 3
    b = 3 . 2/3
    b = 2 cm
    Considering "x" as the side of the square and knowing that triangles ABF and CEF are similar ("α" and "β" are complementary angles), we have proportions:
    AB = x
    AF = 3
    EF = 2
    CF = y
    x/3 = y/2
    y = 2x/3
    BF = x - y
    BF = x - 2x/3
    BF = x/3
    Applying Pythagorean Theorem in ∆ ABF, we have:
    x² + (x/3)² = 3²
    x² + x²/9 = 9 (×9)
    9x² + x² = 81
    10x² = 81
    x² = 8,1 cm²
    Purple Area = Square ABCD Area - Triangle AEF Area
    Purple Area = 8,1 - 3
    Purple Area = 5,1 cm² ✅

  • @unknownidentity2846
    @unknownidentity2846 2 дні тому

    Let's find the area:
    .
    ..
    ...
    ....
    .....
    From the given values for the right triangle AEF we can conclude:
    A(AEF) = (1/2)*AF*h(AF) = (1/2)*AF*EF ⇒ EF = 2*A(AEF)/AF = 2*(3cm²)/(3cm) = 2cm
    The purple triangles ABF and CEF are similar right triangles:
    ∠ABF = ∠ECF = 90° ∧ ∠BAF = ∠CFE ⇒ ∠AFB = ∠CEF
    Therefore we can conclude:
    CF/AB = EF/AF = (2cm)/(3cm) = 2/3 ⇒ CF = 2*AB/3
    Now we apply the Pythagorean theorem to the right triangle ABF. With s being the side length of the square ABCD we obtain:
    AF² = AB² + BF²
    AF² = AB² + (BC − CF)²
    AF² = AB² + (BC − 2*AB/3)²
    AF² = s² + (s − 2*s/3)² = s² + (s/3)² = s² + s²/9 = 10*s²/9
    ⇒ A(ABCD) = s² = 9*AF²/10 = 9*(3cm)²/10 = 8.1cm²
    Now we are able to calculate the area of the purple region:
    A(purple) = A(ABCD) − A(AEF) = 8.1cm² − 3cm² = 5.1cm²
    Best regards from Germany

  • @Emerson_Brasil
    @Emerson_Brasil 2 дні тому

    *Solução:*
    A[APF] = 3×EF/2 = 3 →EF=2.
    Seja ∠BAF=α. Assim ,
    ∠BFA=90° - α → ∠EFC=α. Daí,
    AB = 3 cos α, BF = 3 sen α e CE= 2 cos α.
    Como ABCD é um quadrado, então
    AB = BF + FC
    3 cos α = 3 sen α + 2 cos α
    cos α = 3 sen α. Ora,
    sen² α + cos² α = 1
    (cos α/3)² + cos² α = 1
    cos² α/9 + cos² α = 1
    *_cos² α = 9/10._*
    A [ABCD]=AB²= (3 cos α)²=
    9 cos² α = 81/10.
    Portanto,
    A[painted] = 81/10 - 3 = (81 - 30)/10
    *A[painted] = 51/10 = 5,1 cm²*

  • @himo3485
    @himo3485 2 дні тому

    3*FE/2=3 FE=2
    ABF∞FCE EC=2x FC=2y FB=3x AB=3y
    3y=2y+3x y=3x AB=9x
    (3x)²+(9x)²=3² 90x²=9 x²=1/10
    Purple shaded area : 9x*9x - 3 = 81x² - 3 = 81/10 - 3 = 51/10 = 5.1cm²

  • @Ibrahimfamilyvlog2097l
    @Ibrahimfamilyvlog2097l 2 дні тому +1

    Thanks sir ❤❤

    • @PreMath
      @PreMath  2 дні тому

      You're very welcome! ❤️

  • @alexundre8745
    @alexundre8745 2 дні тому +1

    Bom dia Mestre
    Forte Abraço do Rio de Janeiro
    Deus lhe Abençoe

    • @PreMath
      @PreMath  2 дні тому +1

      Obrigada querido❤️
      Amor e orações dos EUA! 😀
      Deus te abençoe❤️

  • @MrPaulc222
    @MrPaulc222 День тому

    My method differed slightly. I wrote some unnecessary info which I deleted before posting.
    AEF has sides 2, 3, sqrt(13).
    However, using AFE's right angle as a guide, it appears that ABF and CFE are similar.
    FCE's side lengths are 2/3 of ABF's due to EF and FA corresponding sides being 2 and 3 respectively.
    FC = (2/3)x.
    Area ABF = (x*(1/3)x)/2 = (1/6)x^2
    Due to the 2/3 fraction, ECF is 4/9 of that so (4/54)x^2 = (2/27)x^2
    ABF + ECF = (1/6)x^2 + (2/27)x^2 = 9/54 + 4/54 = (13/54)x^2
    EC is (1/3)*(2/3) = 2/9
    Now ADE: (x*(7/9)x)/2 = (7/18)x^2
    The three purple triangles are (13/54)x^2 + (7/18)x^2 = ((13 + 21)/54)x^2
    = (34/54)x^2 = (17/27)x^2
    This means that the remaining(10/27)x^2 = 3
    If (10/27)x^2 = 3, then (17/27)x^2 = 3*(17/10) = 51/10 = 5.1cm^2

  • @wasimahmad-t6c
    @wasimahmad-t6c 2 дні тому

    3×3=9-1=8squrooth × squrooth=8

  • @nenetstree914
    @nenetstree914 2 дні тому

    5.1

  • @michaelgarrow3239
    @michaelgarrow3239 День тому

    Metric- is for people that count on their fingers…

  • @misterenter-iz7rz
    @misterenter-iz7rz День тому

    very difficult. NO idea at all.😂😂😂😂😂😂😂