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.
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😅
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
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 😊
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.
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
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
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.
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
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 😅
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
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.😊
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.
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.
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.
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
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
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.
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
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.
Solution: ∡BAC = 90°-θ ⟹ ∡CAD = 90°-∡BAC = 90°-(90°-θ) = 90°-90°+θ = θ ⟹ ∆ACD is an isosceles triangle. I drop the perpendicular to AC and get the point E. ∆ECD is similar to ∆ABC because both triangles are right-angled and both triangles have the angle θ. For ∆ABC the following applies: (1) AB*BC/2 = 25 |*2/BC ⟹ (1a) AB = 50/BC |in (2) ⟹ (2) AB²+BC² = (5*√5)² = 125 (2a) 50²/BC²+BC² = 125 |*BC² ⟹ (2b) 2500+BC^4 = 125*BC² |-125*BC² ⟹ (2c) BC^4-125*BC²+2500 = 0 |with x = BC² ⟹ (2d) x²-125*x+2500 = 0 |p-q formula ⟹ (2e) x1/2 = 125/2±√(125²/4-2500) = 125/2±37.5 ⟹ (2f) x1 = 125/2+37.5 = 100 and (2g) x2 = 125/2-37.5 = 25 ⟹ 1st case: BC1² = x1 = 100 |√() ⟹ BC1 = 10 |in (1a) ⟹ (1b) AB1 = 50/BC1 = 50/10 = 5 2nd case: BC2² = x2 = 25 |√() ⟹ BC2 = 5 |in (1a) ⟹ (1c) AB2 = 50/BC2 = 50/5 = 10 Both pairs of values satisfy both equations, but since BC is the longer side, BC = 10. Now the following applies: Area ∆ECD/Area ∆ABC = (AC/2)²/BC² ⟹ Area ∆ECD/25 = (5*√5/2)²/10² |*25 ⟹ Area ∆ECD = (5*√5/2)²/10²*25 = 7.8125 ⟹ Yellow area = 2*Area ∆ECD = 2*7.8125 = 15.625[cm²]
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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 😁
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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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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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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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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From the system of equations:
2S(∆ABC)= a×b;
AC^2= AB^2+BC^2,
AB=5; BC=10.
Cos(
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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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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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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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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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The way this rigorous problem is solved is impressive. Thanks!
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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Really amazing
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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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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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Thank you, enjoyed the application of the algebraic identities!😀
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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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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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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...
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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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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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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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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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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Very good
Thanks Sir
Simply solution to this difficult case .
That’s nice
With my respects
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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
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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Excelente!!
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:
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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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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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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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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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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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Super, sir I studied in SSS 1976
Yom! ...that's Yiddish for yah mon. @ 5:11 , forward and backward math! 🙂
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Cheers!
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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I solved for the base and hight of the triangle and then used trigonometry to reach the same conclusion.
S(yellow)= 15,625
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15.625 Sq units
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Solution:
∡BAC = 90°-θ ⟹ ∡CAD = 90°-∡BAC = 90°-(90°-θ) = 90°-90°+θ = θ ⟹ ∆ACD is an isosceles triangle. I drop the perpendicular to AC and get the point E. ∆ECD is similar to ∆ABC because both triangles are right-angled and both triangles have the angle θ.
For ∆ABC the following applies:
(1) AB*BC/2 = 25 |*2/BC ⟹ (1a) AB = 50/BC |in (2) ⟹
(2) AB²+BC² = (5*√5)² = 125
(2a) 50²/BC²+BC² = 125 |*BC² ⟹
(2b) 2500+BC^4 = 125*BC² |-125*BC² ⟹
(2c) BC^4-125*BC²+2500 = 0 |with x = BC² ⟹
(2d) x²-125*x+2500 = 0 |p-q formula ⟹
(2e) x1/2 = 125/2±√(125²/4-2500) = 125/2±37.5 ⟹
(2f) x1 = 125/2+37.5 = 100 and (2g) x2 = 125/2-37.5 = 25 ⟹
1st case: BC1² = x1 = 100 |√() ⟹ BC1 = 10 |in (1a) ⟹ (1b) AB1 = 50/BC1 = 50/10 = 5
2nd case: BC2² = x2 = 25 |√() ⟹ BC2 = 5 |in (1a) ⟹ (1c) AB2 = 50/BC2 = 50/5 = 10
Both pairs of values satisfy both equations, but since BC is the longer side, BC = 10.
Now the following applies: Area ∆ECD/Area ∆ABC = (AC/2)²/BC² ⟹
Area ∆ECD/25 = (5*√5/2)²/10² |*25 ⟹
Area ∆ECD = (5*√5/2)²/10²*25 = 7.8125 ⟹
Yellow area = 2*Area ∆ECD = 2*7.8125 = 15.625[cm²]
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.