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No Calculators! | Can you find Area of the Yellow triangle? | #math #maths #geometry

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[jimlocke]

Solution using the tangent double angle formula: tan(α) = BE/AB = 4/8 = 1/2. tan(2α) = BC /AB = BC/8. The tangent double angle formula is tan(2α) = (2tan(α))/(1 - tan²(α)) = (2)(1/2)/(1 - (1/2)²) = 1/(1 - 1/4) = 1/(3/4) = 4/3. So, BC/8 = 4/3 and BC = 32/3. Let BC be the base and AB be the height of ΔABC. Area of ΔABC = (1/2)bh = (1/2)(32/3)(8) = 128/3 sq. units, as PreMath also found.

[alokverma]

Sir,
By angle bisector theorem

AB/AC=BE/EC
AC = 8*EC/4=2EC eq(1)

& AC^2=8^2+(4+EC)^2 eq(2)

From eq 1 &2

4*EC^2=8^2 + (4+EC)^2

SSolving above quadratic eq we get EC = -4 (inadmissible) & 20/3

Thus area = 1/2*(4+20/3)*8

42.66 sq unit

[prossvay]

In right triangle ABE
Let Alfa=x
Tan(x)=4/8=1/2
In right triangle ABC
Tan(2x)=(BC)/8
Tan(2x)=2tan(x)/1-tan(x)^2
Tan(2x)=2(1/2)/1-(1/2)^2=4/3
So BC/8=4/3
BC=32/3
So yellow triangle area =1/2(8)(32/3)=128/3 square units.❤❤❤ Thanks teacher. Best regards

[ybodoN]

💡 in a 3:4:5 right triangle,
• if we bisect the angle formed by the minor cathetus and the hypotenuse, we get a 1:2:√5 right triangle.
• if we bisect the angle formed by the major cathetus and the hypotenuse, we get a 1:3:√10 right triangle.

[ybodoN]

Since ⊿ABE is a 1:2:√5 right triangle and ∠BAE = ∠EAC, then ABC is a 3:4:5 (Pythagorean) special right triangle.
Therefore, the area of the yellow triangle is 8 / 3 × 4 × 8 / 2 = (8 × 8 × 4) / (2 × 3) = 256/6 = 128/3 square units 🤓
Thank you PreMath 🙏

[quigonkenny]

Let β = 90°- α. As ∠EAB and ∠DAE both equal α and ∠ABE and ∠AED both equal 90°, then ∠BEA and ∠EDA both must equal β, and ∆ANE and ∆AED are similar.

Triangle ∆ABE:
BE² + AB² = EA²
EA² = 4² + 8² = 16 + 64
EA = √80 = 4√5

Triangle ∆AED:
DA/AE = EA/AB
DA/4√5 = 4√5/8
DA = 16(5)/8 = 10

ED/AE = BE/AB
ED/4√5 = 4/8 = 1/2
ED = (4√5)(1/2) = 2√5

Drop a perpendicular from D to EC at F. As ∠BEA = β and ∠AED = 90°, ∠DEF = α. As ∠EFD = 90°, then ∠FDE = β, and ∆EFD is similar to ∆ABE and ∆AED.

Triangle ∆EFD:
FD/DE = BE/EA
FD/2√5 = 4/4√5 = 1/√5
FD = 2√5/√5 = 2

EF/FD = AB/BE
EF/2 = 8/4 = 2
EF = 2(2) = 4

Let FC = x. As ∆DFC = 90° and ∠C is a shared internal angle, ∆ABC and ∆DFC are similar triangles.

Triangle ∆DFC:
FC/DF = BC/AB
x/2 = (8+x)/8
8x = 2(8+x) = 16 + 2x
6x = 16
x = 16/6 = 8/3

Triangle ∆ABC:
A = bh/2 = (8+8/3)(8)/2
A = (32/3)4 = 128/3 ≈ 42.667

[ROCCOANDROXY]

No need for ED and AE being perpendicular. Just need angle bisector AE and AB = 8 and BE = 4.

Note: tan(alpha) = 1/2 and tan(2 * alpha) = (4 + EC)/8 = 4/3 implies EC = 20/3 implies BC = 32/3 implies Area = 128/3.

[wsmv]

Another solution without pythagorean theorem or trig:

⊿ABE and ⊿AFE are congruent, they have the same area. Together, their combined area is 32.

⊿ECF is similar to ⊿ABC (they both have a right angle and share the same angle BCA = ECF.)
Length of EF = 4 = half of AB. Therefore the area of ⊿ECF is 1/4 of the area of ⊿ABC.

So: Area x = 32 + 1/4 x
3/4 x = 32
x = 128/3

[sandanadurair]

Let us use m-n-theorem

Where
alpha=BAE, beta=CAE,
theta=AEC.
m=BE, n=CE

In this problem
theta=90+alpha,
alpha =alpha
beta = alpha
m = 4, n to be found.
Tan(alpha) = 4÷8 = 1/2,
Cot(alpha) = 8÷4 = 2
cot(90+alpha) = -tan(apha) = -(1/2)

By m-n theorem



(4+n)cot(90+alpha) = 4.cot(alpha)-n.cot(alpha)

-(4+n)tan(alpha) = 4.cot(alpha)-n.cot(alpha)
Substituting respective values
We get
-(4+n).(1/2)=4*2-2n

n = 20/3
BC = 4+20/3=32/3
Area = (1/2)(32/3)(8) = 42.67

[marcgriselhubert]

In triangle ABC: AB/BC = tan(angleBCA) = tan (90° -2.alpha) = cotan(2.alpha)
Or in triangle ABE: tan(alpha) = BE/AB = 4/8 = 1/2, so tan(2.alpha) = (2.tan(alpha))/ (1 - (tan(alpha)^2) = 1/(3/4) = 4/3, so cotan(2.alpha) = 3/4
We then have: AB/BC = 3/4, so BC = (4/3).AB = (4/3).8 = 32/3. Finally the area of ABC is: (1/2).AB.BC = (1/2).8.(32/3) =128/3

[hongningsuen]

A wonder use of 2 congruence tests and 2 similarity tests.
A shorter method is to use angle bisector theorem: dd = bc - mn where d is the angle bisector and the angle bisector divides side a into m and n segments, b and c are the other two sides.
dd = 8^2 + 4^2 = 80 (Pythagoras theorem)
Let EC be n and AC be x
There are 2 unknowns n and x hence 2 equations are needed for solution.
8x - 4n = 80 (angle bisector theorem)
x^2 = 8^2 + (4 + n)^2 (Pythagoras theorem) Hence x = sqrt (n^2 + 8n + 80)
Substitute n for x,
8 x sqrt(n^2 + 8n + 80) - 4n = 80
sqrt(n^2 + 8n + 80) = (4n - 80)/8
n^2 + 8n + 80 = (n^2/4) -10n + 100
3n^2 - 8n - 80 = 0
(3n - 20)(n + 4) = 0
n = 20/3 (n = -4 rejected)
Area of triangle = 1/2 x 8 x (4 + 20/3) = 128/3

[spafon]

Now that I know angle bisector theorem, I always love to see two triangles next to each other with the same angle.

This is a good problem to use that theorem. By angle bisector theorem, EC/AC=BE/AB. Therefore EC/AC=1/2. If we label EC as x, then AC=2x. So now applying pythagorean to ABC we get 8^2+ (4+x)^2= (2x)^2. This gives EC=20/3, and thus BC=32/3 and the triangle area is (1/2)*8*(32/3)=128/3.

[Birol]

My way of solution is ▶
for the ΔABE
tan(α) = 4/8
tan(α)= 1/2

Let's say EC=x
tan(2α)= 2tan(α)/[1-tan²(α)]
tan(2α)= (4+x)/8
⇒
(4+x)/8= 2*(1/2)/[1-(1/2)²]
(4+x)/8= 1/(3/4)
(4+x)/8= 4/3
32= 12+3x
3x= 20
x= 20/3 length units

ΔABC= BC*AB/2
BC= 8
AB= 4+20/3
AB= 32/3
⇒
ΔABC= 8*(32/3)*(1/2)
ΔABC= 128/3
ΔABC ≈ 42, 7 square units

[santiagoarosam]

F es la proyección ortogonal de D sobre EC→ Ángulo FED=α
AE=√4²+8²=4√5→ ED=AE/2=2√5→ EF=4→ DF=2 → Razón de semejanza entre DFC y ABC =s=2/8=1/4→ Razón entre áreas =s²=1/16
Áreas: ABE+AED+EFD=16+20+4=40 → ABFD=40=15ABC/16→ ABC=40*16/15 =128/3
Gracias y saludos.

[bijaykumarrath]

There is a simple solution.AE is angle bisector. So AB/AC=BE/EC. Let EC=x, 8/Ac=4/x
ThusAC=2x. InABC rtangle triangle AB*2+BC*2=Ac*2.
(8)2+(4+x)2=(2x)2 _x=20/3

Area 1/2 *8*32/3=128/3

[Ensign_Cthulhu]

Lesson learned: I am too slow to draw in additional lines to make the problem easier. It's a problem I've had since high school. I seem to have a mental block against doing it.

[toninhorosa]

Great solution teacher!!!
👏👏👏👏👏
I solved a little different.
∆ABE applying Pythagoras
AE^2 = 8^2 + 4^2
AE^2 = 64+16 = 80
AE = √80
∆ABE~∆ADE 3 angles equals
and alpha + Beta = 90°
AB/BE = AE/ED
8/4 = √80/ED
8ED = 4√80
ED = (4√80)/8
ED = √80/2
∆ABD applying Pythagoras
AD^2 = AE^2 + DE^2
AD^2= (√80)^2 + (√80/2)^2
AD^2 = 80 + 20
AD^2 = 100
AD = 10
∆ADE~∆DEP 3 angles equals
AD/ED = ED/DP
10/(√80/2) = (√80/2)/DP
10DP = 20
DP = 2
∆DEP Applying Pythagoras
ED^2 = DP^2 + EP^2
(√80/2)^2 = 2^2 + EP^2
20 = 4 - EP^2
EP^2 = 20 - 4 = 16
EP = 4
From point D, we draw a line parallel to line BC until we reach line AB and mark point G, forming straight line DG and ∆ADG.
∆ADG ~ AABC
DG = BP = 4+4 = 8
AG = 8 - DP = 8-2= 6
AG/DG = AB/BC
6/8 = 8/BC
6BC = 64 = BC = 64/6
BC = 32/3
Yellow ∆area =( 1/2)*8*(32/3)
Yellow ∆ area = 128/3
Yellow ∆ area =
42, 666... units^2.

[wackojacko]

Drilling for Congruence is like drilling for Black Gold/Texas Tea. Become a millionaire using SSS(Side Side Side) SAS(Side Angle Side) ASA(Angle Side Angle) AAS(Angle Angle Side) RHS(90°Angle Hypotenuse Side) Theorems. 🙂

[DB-lgsq]

شكرا لكم على المجهودات
يمكن استعمال
tanBAE=1/2, tan2BAE=4/3
EC=20/3
S(ABC)=128/3

[mhcheung]

Great work
I would like to modify your work.
Once EF is drawn, triangle ABC and triangle CEF are similar. Take CE as x and CF as y, a system of 2 linear equations would be set up by ratio of sides of the two similar triangles.