Can you find area of the Yellow shaded Triangle? | (Trapezoid) |

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  • čas přidán 4. 07. 2024
  • Learn how to find the area of the Yellow shaded Triangle. Important Geometry and algebra skills are also explained: Trapezoid; Trapezium; Trapezoid area formula; Triangle area formula; Pythagorean Theorem. Step-by-step tutorial by PreMath.com
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Komentáře • 94

  • @anatoliy3323
    @anatoliy3323 Před 28 dny +3

    It's so gladly to hear your voice and listen to your math lesson

    • @PreMath
      @PreMath  Před 28 dny +1

      Thanks dear, good to hear from you as well😀
      I' glad you are safe and sound!
      Love and prayers from the USA! ❤️

  • @RAG981
    @RAG981 Před 29 dny +15

    I don't follow using the trapezium, since once you have AD = 25/4, and height ADC is 5, surely the area of ADC is 1/2x25/4x5 = 125/8.

    • @LucareonVee
      @LucareonVee Před 28 dny +1

      I did the same thing. Didn't bother with the trapezoid once I had AD and AB

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      Thanks for the feedback ❤️

  • @thewolfdoctor761
    @thewolfdoctor761 Před 29 dny +7

    Like the professor I (1) determined the lengths of missing sides of triangle ABC to be 5 and 10 and (2) determined that ACD is an isosceles triangle. I dropped a perpendicular to AC from D at E on AC. Resulting triangle DEC is similar to ABC. I used proportions of the two similar triangles to determine length of DE, i.e. height of triangle ADC, to be (5*SQRT(5))/4. So Area of yellow = 1/2 (5*SQRT(5)) * (5*SQRT(5)/4) = 125/8.

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      Thanks for sharing ❤️

    • @rabotaakk-nw9nm
      @rabotaakk-nw9nm Před 26 dny

      AD/AE=AC/BC => AD=25/4
      [ACD]=[ABD]=½AD•AB=125/8 😁

  • @MrPaulc222
    @MrPaulc222 Před 28 dny +1

    For the blue triangle, a^2 + b^2 = 125, so the most reasonable assumption is that a and b are 5 and 10.
    tan(-1)(1/2) is 26.6 deg.
    Draw a line vertically up frpm C. then across to D. The top right vertex is E and AEC is 25cm^2.
    Yellow area is AEC - DEC.
    Angles at C total 53.2 deg, so

    • @PreMath
      @PreMath  Před 28 dny +1

      No worries!
      Thanks for sharing ❤️

  • @himo3485
    @himo3485 Před 29 dny +3

    1 : 2 : √5
    AB=x BC=2x
    x^2+(2x)^2=(5√5)^2 5x^2=125 x=5
    AD=DC=y
    (10-y)^2+5^2=y^2
    100-20y+y^2+25=y^2 20y=125 y=25/4
    Yellow shaded area
    = 25/4*5*1/2=125/8=15.625cm^2

    • @davidseed2939
      @davidseed2939 Před 28 dny

      Call AB=y, BC=x
      From area xy=50
      from hypotenuse x^2+y^2=125. =100+25
      Guess y=5,x=10.
      So use the 2:1 radio in the yellow triangle
      Vert height =b/4, area = b^2/8
      But b^2=125
      So area=125/8

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      Thanks for sharing ❤️

  • @batavuskoga
    @batavuskoga Před 28 dny +1

    In the last step, there no need to calculate the trapezium's area.
    You know the base and height of the triangle : base=25/4, height=5
    area = base * height / 2 = 25/4 * 5 / 2 = 125/8
    But it is a nice exercise

    • @PreMath
      @PreMath  Před 28 dny

      Correct!
      Glad to hear that!
      Thanks for the feedback ❤️

  • @jimlocke9320
    @jimlocke9320 Před 29 dny +1

    Another way to compute a (length BC) and b (length AB): Construct 4 triangles congruent to ABC and assemble them into a square with sides of length AC or 5√5. The square's area is (5√5)(5√5) = 125. Each triangle has area 25, given, so 4 have area 100 and the area of the small square in the middle is 25. Therefore, its sides have length √(25) = 5. Note that the figure implies that a is greater than b. So, a - b = 5. However, from area formula for a triangle, (1/2)ab = 25 or ab = 50. Substitute b = a - 5 and solve: a(a - 5) = 50, finding a = 10 and, from b = a - 5, b = 5. Skip ahead to about 7:20. At about 10:46, we find x (length AD) = 25/4. Let AD be the base of yellow triangle ACD, then its height is 5 and area = (1/2)(25/4)(5) = 125/8, as PreMath found at 12:16.

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!👍🌹
      Thanks for sharing ❤️

  • @mvrpatnaik9085
    @mvrpatnaik9085 Před 26 dny

    The way this rigorous problem is solved is impressive. Thanks!

  • @marcgriselhubert3915
    @marcgriselhubert3915 Před 28 dny +1

    Another solution when BA = 5 and BC = 10 are known: The triangle ADC is isosceles (easy), so its area is (1/2).AC.DP with P the orthogonal projection of D on (AC)
    DP = AP.tan(t) = (1/2).AC.tan(t), so the area of the yellow triangle is (1/4).(AC^2).tan(t)
    tan(t) = 5/10 = 1/2 (in triangle ABC)
    The yellow area is(1/4).((5.sqrt(5))^2).(1/2)
    = (1/4).(125).(1/2) = 125/8.

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      Thanks for sharing ❤️

  • @phungpham1725
    @phungpham1725 Před 29 dny +1

    Alternative solution:
    1/Drop the height BH to AC, we have: 1/2 BHxAC= 25-> 1/2 BHx5sqrt5=25-> BH=2sqrt5
    2/ From D drop AM perpendicular to AC intersecting BC at D’. We have ADCD’ is a diamond so AD=DC=CD’=D’A
    and BM=AM=CM
    Focus on the triangle BHM
    sq HM=sqBM-sqBH=sq(5sqrt5/2)-sq(2sqrt5)
    --> HM=3sqrt5/2
    --> CH=HM+CM=4sqrt 5
    By using triangle similarity:
    MD’/BH=CM/CH-- > MD’=5sqrt5/4
    Because DM=MD’
    So area of the yellow triangle= 1/2 x 5sqrt5/4 x 5 sqrt5= 125/8 sq cm 😅

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      Thanks for sharing ❤️

  • @xualain3129
    @xualain3129 Před 29 dny +1

    My version with trigonometry.
    AB=5*sqrt(5)*sin(theta)
    BC=5*sqrt(5)*cos(theta)
    1/2*AB*BC=1/2*125*sin(theta)*cos(theta)=25 -->sin(2*theta)=4/5 -->cos(2*theta)=3/5
    cos(2*theta)=2*cos(theta)^2-1 -->cos(theta)^2=4/5 -->sec(theta)^2=5/4
    tan(theta)^2=1/4 -->tan(theta)=1/2
    Let the height of the isosceles angle ADC be h and the base is AC=5*sqrt(5)
    Then h=1/2*AC*tan(theta)=5/4*sqrt(5)
    Area(ACD)=1/2*AC*h=125/8

    • @PreMath
      @PreMath  Před 28 dny +1

      Excellent!
      Thanks for sharing ❤️

  • @josedavis4242
    @josedavis4242 Před 29 dny +3

    Different method using trigonometry:
    1)Follow steps as yours till finding out sides AB and BC
    2) extend AD to point E to make a rectangle ABCE, hence AC is the diagonal and area of triangle ACE= triangle ABC= 25 cm^2.
    3) draw a perpendicular from point D to line BC, and let the meeting point be P.
    4) AB/ BC = tan Theta = 5/10 = 1/2
    5) Consider triangle DPC,
    DP/PC = tan 2 Theta = 2 tan Theta/(1- tan^2 theta)
    = 2*( 1/2) / 1- (1/4)
    = 4/3
    Hence DP/PC = 4/3; DP= AB =5
    ie 5/PC = 4/3
    therefore PC = 15/4
    6)Area of triangle DPC = 1/2 * DP *PC
    = 0.5* 5*15/4
    = 75/8 cm^2
    7) area of triangle DPC = area of triangle DCE ( congruent triangles) 75/8 cm^2
    8) area of yellow shaded region (ADC) = (AREA of triangle ACE - AREA of triangle DCE) = 25- (75/8)
    = 15.625 cm^2
    Thank you 😊

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      You are very welcome!
      Thanks for sharing ❤️

  • @santiagoarosam430
    @santiagoarosam430 Před 29 dny +2

    Los ángulos DAC y DCA son iguales→ Si "E" es La proyección ortogonal de D sobre AC→ AE=EC=5√5/2.
    AB*BC=a*b=2*25=50→ a=50/b→ a²+b²=(5√5)²=(50/b)²+b²→ b=10→a=5 → Pendiente de AC=p=a/b=1/2→ DE=p*AE=(5√5/2)/2=5√5/4 → Área ACD=AC*DE/2 =(5√5)(5√5/4)/2=125/8 cm².
    Gracias y un saludo cordial.

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      Thanks for sharing ❤️

  • @allanflippin2453
    @allanflippin2453 Před 29 dny +1

    Once you found the "x" value (the two equal sides of the isosceles triangle), it's actually easier to find that triangle's area by ignoring the trapezoid area. A triangle area is h*b/2. In this case, h = 5 and b = x (which you found was 25/4). 5 * (25/4) / 2 = 125/8 or 15.625.

    • @PreMath
      @PreMath  Před 28 dny

      Correct!👍
      Thanks for sharing ❤️

  • @professorrogeriocesar
    @professorrogeriocesar Před 26 dny

    Excelente!!

  • @alexniklas8777
    @alexniklas8777 Před 28 dny +1

    From the system of equations:
    2S(∆ABC)= a×b;
    AC^2= AB^2+BC^2,
    AB=5; BC=10.
    Cos(

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      You are very welcome dear!
      Thanks for sharing ❤️

  • @hongningsuen1348
    @hongningsuen1348 Před 29 dny +1

    Equation (4) has 2 forms: (a - b) = 5 or (a - b) = - 5. (a is BC, b is AC as in your solution)
    This leads to 2 sets of a, b values.1st set is a = 10, b = 5 (as in your solution). 2nd set is a = 5, b = 10. The second set is rejected by 2 arctan 10/5 = 126.87 which is an obtuse angle. With that problem settled, tan(theta) = b/a = 1/2.
    Triangle ADC is isosceles triangle as angle CAD = theta (corresponding angle with angle BCA) = angle ACD.
    Height of isosceles triangle ADC bisects AC. Hence height = [(5sqrt5)/2] tan(theta) = [5(sqrt5)]/4.
    Area of yellow triangle = (1/2)(5sqrt5)[5(sqrt5)](1/4) = 125/8

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      Thanks for sharing ❤️

  • @guyhoghton399
    @guyhoghton399 Před 28 dny +1

    We get _|AB| = 5, |BC| = 10_ as shown.
    In _ΔABC:_
    _sinθ = 1/√5, cosθ = 2/√5_
    Let _x = |AD|_
    _∠ADC = 180° - 2θ_
    Using the sine rule in _ΔADC:_
    _sinθ/x = sin(180° - 2θ)/(5√5)_
    ⇒ _1/(x√5) = sin(2θ)/(5√5)_
    ⇒ _x = 5/(2sinθcosθ) = 5/[2(1/√5)(2/√5)] = 25/4_
    Area of _ΔADC = ½|AC|.|DC|sinθ = ½(5√5)(x)sinθ = ½(5√5)(25/4)(1/√5) = _*_125/8_*

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      Thanks for sharing ❤️

  • @phungpham1725
    @phungpham1725 Před 29 dny +2

    1/ axb= 50
    sq(a+b)=sqa+sqb+2ab=125+100=225->a+b=15
    By Vieta theorem we have the equation:
    Sqx-15x+50=0😊
    -> x=5 and x’=10
    Assume b b=5 and a=10
    2/The triangle ADC is an isosceles one.
    Drop the height DH to the base AC, we have DH is the perpendicular bisector of AC . Moreover the triangle DHA and ABC are similar so, DH/AH=AB/AC=5/10= 1/2
    -> DH= 1/2AH= 1/4 AC=5sqrt5/4
    Area of the yellow triangle =1/2 x 5sqrt5/4 x 5sqrt5=125/8 sq cm😅

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      Thanks for sharing ❤️

  • @jamestalbott4499
    @jamestalbott4499 Před 29 dny +3

    Thank you, enjoyed the application of the algebraic identities!😀

    • @PreMath
      @PreMath  Před 28 dny

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

  • @abdessamadsafri8791
    @abdessamadsafri8791 Před 29 dny +5

    Really amazing

    • @PreMath
      @PreMath  Před 28 dny

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

  • @denisrenaldo3506
    @denisrenaldo3506 Před 16 dny

    No need to compute AD. You can draw the height DH perpendicular to AC and prove that triangle DHA (or DHC) is similar to ABC. Then DH/AH=AB/BC=1/2. So, DH=AH/2=AC/4 and then S = DHxAC/2= AC^2/8= 125/8

  • @prollysine
    @prollysine Před 28 dny +1

    T=a*b/2 , T=50/2 , T=25 cm^2 , sides are integers , a^2+b^2=(5sqr5)^2 , a^2+b^2=125 , 10^2+5^2=100+25 , a=10 , b=5 ,
    ß=arctg(b/a) , arctg(5/10)=26.5651° , ß=26.5651° , 2ß=53.1301° , triangle : b , x , 2ß , , b=5 , ß=53.1301° ,
    x=5/tg(53.1301°) , x=3.75 ,
    triangle T"=5*3.75 /2 , T"=9.375 , yellow area: 25-9.375=15.625 cm^2 ,

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      Thanks for sharing ❤️

    • @prollysine
      @prollysine Před 27 dny

      @@PreMath Thanks...

  • @sergeyvinns931
    @sergeyvinns931 Před 29 dny +2

    AB=5, BC=10, angle CAD=Theta, AD=CD, let`s draw teigtDF on AC, DF/CF=AB/BC, DF/CF=1/2, CF=5\/5/2, areaof the Yellow shaded Triangle =5\/5*5\/5/8=125/8=15,625.

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      Thanks for sharing ❤️

  • @gaylespencer6188
    @gaylespencer6188 Před 28 dny +1

    If b*h=50, then b=50/h. Now you have a quadratic, (50/h)^2+H^2-125=0. h is then 5, and take it from there by getting the arctan of theta = 26.565.... which doubled is 53.130... So 5/tan(53.130...) = 3.75. So base of yellow triangle is 6.25 (10-3.75) times height of 5 divided by 2 is 15.625.

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      Thanks for sharing ❤️

  • @yalchingedikgedik8007
    @yalchingedikgedik8007 Před 29 dny +2

    Very good
    Thanks Sir
    Simply solution to this difficult case .
    That’s nice
    With my respects
    ❤❤❤❤

    • @PreMath
      @PreMath  Před 28 dny

      So nice of you dear😀
      Glad to hear that!
      Thanks for the feedback ❤️

  • @zeroone7500
    @zeroone7500 Před 28 dny

    5:53 √[(a - b)^2] = |a - b|
    |a - b| = a - b if a > b
    |a - b| = b - a if a < b

    • @PreMath
      @PreMath  Před 28 dny

      Thanks for sharing ❤️

  • @Birol731
    @Birol731 Před 29 dny +3

    My way of solution ▶
    AB= b
    BC= a
    A(ΔABC)= 25 cm²

    a*b/2= 25
    ab= 50
    according to the Pythagorean theorem:
    a²+b²= c²
    c= 5√5
    c²= 25*5
    c²= 125

    (a+b)²= a²+b²+2ab
    (a+b)²= 125+2*50
    (a+b)= √225
    a+b= 15
    a+b= 15
    ab= 50
    b= 50/a

    a+(50/a)= 15
    a²+50= 15a
    a²-15a+50=0
    Δ= 15²-4*1*50
    Δ= 25
    √Δ= 5
    a= (15+5)/2
    a= 10 cm

    b= 5 cm
    tan(Θ)= AB/BC
    tan(Θ)= 5/10
    tan(Θ)= 0,5
    Θ= arctan(0,5)
    Θ= 26,565°
    2Θ= 53,13°
    sin(2Θ)= AB/DC
    0,8= 5/DC
    DC= 25/4 cm
    Ayellow= A(ΔACD)
    A(ΔACD)= (1/2) sin(Θ)*AC*CD
    A(ΔACD)= (1/2)*sin(26,565°)*5√5*25/4
    A(ΔACD)= 15,625 cm²
    A(ΔACD)= 125/8 cm²

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      Thanks for sharing ❤️

  • @laxmikantbondre338
    @laxmikantbondre338 Před 28 dny +1

    Once we find x in above solution i.e. AD = 25/4 we can directly find area of yellow traingle by 1/2 AD x AB

    • @PreMath
      @PreMath  Před 28 dny

      Correct!
      Thanks for the feedback ❤️

  • @misterenter-iz7rz
    @misterenter-iz7rz Před 29 dny +1

    ab=50, a^2+b^2=125, a+b=sqrt(125+2×50)=15, (a,b)=(5,10) or (10,5), in view of the figure, we say a=5, b=10, ADC is an isosceles triangle, then the area is 5sqrt(5)/2×5sqrt(5)/4=125/8.😊

    • @PreMath
      @PreMath  Před 28 dny +1

      😀
      Thanks for sharing ❤️

  • @waheisel
    @waheisel Před 28 dny

    At 6:05 (a-b)^2=25 has two roots; a-b=5 or -5. In other words a=5 and b=10 is another solution. That results in angle BCD being an obtuse angle. That is not excluded by the problem. So a second solution is that the area of triangle ACD=125/2 or 62.5 by my calculations.

    • @PreMath
      @PreMath  Před 28 dny

      Thanks for sharing ❤️

  • @giuseppemalaguti435
    @giuseppemalaguti435 Před 29 dny +2

    5√5cosθ5√5sinθ=50...sin2θ=4/5...θ=arcsin(4/5)/2..Ash=5√5*(5√5tgθ/2)/2=(125/4)tgθ=(125/4)√(1-coθ)/(1+cosθ)=(125/4)/√((1-3/5)/(1+3/5))=125/8

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      Thanks for sharing ❤️

  • @unknownidentity2846
    @unknownidentity2846 Před 29 dny +3

    Let's find the area:
    .
    ..
    ...
    ....
    .....
    The blue triangle is a right triangle. Therefore we can conclude:
    AB² + BC² = AC²
    A(ABC) = (1/2)*AB*BC
    (AB ± BC)² = AB² ± 2*AB*BC + BC² = AC² ± 2*AB*BC = AC² ± 4*(1/2)*AB*BC = AC² ± 4*A(ABC)
    ⇒ AB ± BC = √[AC² ± 4*A(ABC)]
    AB + BC = √[AC² + 4*A(ABC)] = √[(5√5cm)² + 4*25cm²] = √(125cm² + 100cm²) = √(225cm²) = 15cm
    AB − BC = √[AC² − 4*A(ABC)] = √[(5√5cm)² − 4*25cm²] = √(125cm² − 100cm²) = √(25cm²) = 5cm
    These two equations lead to AB=10cm and BC=5cm. Of course AB=5cm and BC=10cm is also a valid solution for the original set of equations. Now let's handle these two solutions. For that purpose we add point E on BC such that ABED is a rectangle. In this case the triangle CDE is a right triangle and we obtain:
    DE/CE
    = tan(2θ)
    = 2tan(θ)/[1 − tan²(θ)]
    = 2(AB/BC)/[1 − (AB/BC)²]
    Case 1 (AB=10cm and BC=5cm): DE/CE = 2*2/(1 − 2²) = −4/3
    Case 2 (AB=5cm and BC=10cm): DE/CE = 2*(1/2)/(1 − (1/2)²) = 4/3
    In the first case we have 2θ>90°. So this is not a useful solution and we can focus on the second case (AB=5cm and BC=10cm):
    DE/CE = 4/3
    AB/CE = 4/3
    ⇒ CE = (3/4)*AB = (3/4)*(5cm) = (15/4)cm
    ⇒ AD = BE = BC − CE = 10cm − (15/4)cm = (40/4)cm − (15/4)cm = (25/4)cm
    Now we are able to calculate the area of the yellow triangle:
    A(ACD) = (1/2)*AD*h(AD) = (1/2)*AD*DE = (1/2)*AD*AB = (1/2)*[(25/4)cm]*(5cm) = (125/8)cm²
    Best regards from Germany

    • @PreMath
      @PreMath  Před 28 dny +1

      Excellent!❤️
      Thanks for sharing ❤️

  • @ludosmets2018
    @ludosmets2018 Před 29 dny +2

    In the last step, no need to calculate the area of the trapezoid. The area of the yellow triangle is simply 1/2 (25/4)(5)..

    • @PreMath
      @PreMath  Před 28 dny

      Correct!👍
      Thanks for the feedback ❤️

  • @LuisdeBritoCamacho
    @LuisdeBritoCamacho Před 29 dny +2

    STEP-BY-STEP RESOLUTION PROPOSAL :
    01) BC = X
    02) AB = Y
    03) AC = Z = 5*sqrt(5)
    04) Z^2 = 125
    05) X * Y = 50
    06) X^2 + Y^2 = 125
    07) Solutions : X = 10 and X = 5 or X = 5 and Y = 10. Let's choose BC = X = 10 and AB = Y = 5.
    08) tan(teta) = 5/10 ; tan(teta) = 1/2.
    09) Define a Point between Point B and Point C, wich the Vertical Line (Orthogonal Projection) passing through Point D. Define Point E.
    10) tan(2*teta) = 2*tan(teta) / (1 - tan^2(teta))
    11) tan(2*teta) = 1 / (1 - 1/4) ; tan(2*teta) = 1 / (3/4) ; tan(2*teta) = 4/3
    12) tan(2*teta) = DE / CE
    13) 4 / 3 = 5 / CE
    14) CE = 15 / 4 ; CE = 3,75
    15) Yellow Area = (AD * AB) / 2
    16) AD = BE = 10 - 3,75 = 6,25
    17) Yellow Area = (AD * AB) / 2 ; Yellow Area = (6,25 * 5) / 2 ; Yellow Area = 31,25 / 2 ; Yellow Area = 15,625 sq un
    So, our ANSWER is : The Yellow Shaded Area equals 15,625 Square Units.
    Best Regards from the Department of Ancient Greek Arabic and Persian Mathematics and Geometry (Tri+Gonio+Metry) in the Universal Islamic Institute for Study of Ancient Thinking, Knowledge and Wisdom. Cordoba Caliphate.

    • @PreMath
      @PreMath  Před 28 dny +1

      Excellent!👍
      Thanks for sharing ❤️

  • @sureshmehetre2882
    @sureshmehetre2882 Před 27 dny

    Super, sir I studied in SSS 1976

  • @DraCat1993
    @DraCat1993 Před 29 dny +1

    Great, in my decision I did not come to the conclusion that the yellow triangle is isosceles, and I was looking for the length of the upper base of the trapezoid through the sine of the double angle

    • @PreMath
      @PreMath  Před 28 dny

      No worries!
      Thanks for sharing ❤️

  • @aljawad
    @aljawad Před 26 dny

    I solved for the base and hight of the triangle and then used trigonometry to reach the same conclusion.

  • @wackojacko3962
    @wackojacko3962 Před 28 dny +1

    Yom! ...that's Yiddish for yah mon. @ 5:11 , forward and backward math! 🙂

    • @PreMath
      @PreMath  Před 28 dny

      Thanks for the feedback ❤️
      Cheers!

  • @michaelstahl1515
    @michaelstahl1515 Před 28 dny +1

    I enjoyed your Video one more time. I used sin ( teta ) and tan ( teta ) in your triangle and found the same solution .

    • @PreMath
      @PreMath  Před 28 dny

      Great 👍
      Glad to hear that!
      Thanks for sharing ❤️

  • @duantran-bk4mv
    @duantran-bk4mv Před 29 dny +2

    S(yellow)= 15,625

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      Thanks for sharing ❤️

  • @kalavenkataraman4445
    @kalavenkataraman4445 Před 29 dny +3

    15.625 Sq units

    • @PreMath
      @PreMath  Před 28 dny

      Excellent!
      Thanks for sharing ❤️

  • @nenetstree914
    @nenetstree914 Před 28 dny

    125/8 ??

    • @PreMath
      @PreMath  Před 28 dny

      Thanks for sharing ❤️

  • @DavidPalen-zw4rt
    @DavidPalen-zw4rt Před 29 dny +1

    The diagram does not specify that AD is parallel to BC. You call it a trapezoid but this is not indicated in the diagram.

    • @brettgbarnes
      @brettgbarnes Před 29 dny +2

      Angles ABC and BAD are indicated as right angles on the diagram.

    • @tshepomotau4354
      @tshepomotau4354 Před 27 dny

      Co interior angles add up to 180° and 90°+90° add up to 180°. Therefore the lines are still parallel.