Can you find area of the Yellow shaded Triangle? | (Trapezoid) |
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- Опубліковано 4 лип 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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It's so gladly to hear your voice and listen to your math lesson
Thanks dear, good to hear from you as well😀
I' glad you are safe and sound!
Love and prayers from the USA! ❤️
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.
I did the same thing. Didn't bother with the trapezoid once I had AD and AB
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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.
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AD/AE=AC/BC => AD=25/4
[ACD]=[ABD]=½AD•AB=125/8 😁
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
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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
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
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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
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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
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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.
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Really amazing
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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 😅
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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
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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.
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The way this rigorous problem is solved is impressive. Thanks!
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
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Excelente!!
Very good
Thanks Sir
Simply solution to this difficult case .
That’s nice
With my respects
❤❤❤❤
So nice of you dear😀
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Thank you, enjoyed the application of the algebraic identities!😀
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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.
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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.
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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
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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
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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
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From the system of equations:
2S(∆ABC)= a×b;
AC^2= AB^2+BC^2,
AB=5; BC=10.
Cos(
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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😅
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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_*
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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)..
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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.
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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 ,
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@@PreMath Thanks...
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
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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.
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5:53 √[(a - b)^2] = |a - b|
|a - b| = a - b if a > b
|a - b| = b - a if a < b
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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.
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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
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.😊
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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.
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Super, sir I studied in SSS 1976
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
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S(yellow)= 15,625
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I enjoyed your Video one more time. I used sin ( teta ) and tan ( teta ) in your triangle and found the same solution .
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Yom! ...that's Yiddish for yah mon. @ 5:11 , forward and backward math! 🙂
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Cheers!
I solved for the base and hight of the triangle and then used trigonometry to reach the same conclusion.
15.625 Sq units
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125/8 ??
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The diagram does not specify that AD is parallel to BC. You call it a trapezoid but this is not indicated in the diagram.
Angles ABC and BAD are indicated as right angles on the diagram.
Co interior angles add up to 180° and 90°+90° add up to 180°. Therefore the lines are still parallel.