Calculate area of the Green shaded Triangle | Fun Geometry | Important Geometry skills explained

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Calculate area of the Green shaded Triangle | Fun Geometry | Important Geometry skills explained

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[lindafromcalifornia]

Thank you for sharing your way of tackling this problem.

[normanc]

I drew a perpendicular line from E to line AB, lets call the line as EF (which is the height of the triangle). Now angles EAB & ADE are equal, angles DAE & FEA are equal. Triangles AED & EFA are similar trianges. Therefore, the ratios of AD/8 = 8/EF, then (AE)(EF) = (8)(8) => area of green triangle being 64/2 = 32. It looks complicated at first sight, but I could solve it by mental calculation in less than a minute.

[hoaihuong]

Thank you worked out the answer. Honestly, I did not know where to start.

[johankotze]

Take the perpendicular from E to AB, and call it F. EF is then the height of the green triangle.
In triangle ADE, call angle ADE = a and angle DAE = b. a+b = 90 (complementary angles)
Therefore angle EAF = b and AEF = a
Triangles ADE and AEF are similar (AAA)
Take the side of the square as x
Now proportions:
h/8 = 8/x this gives h = 64/x
Area = 1/2*(height)*(base)
Area = 1/2 *(64/x)*(x) = 32

[jjcadman]

Excellent! I always struggle with these solutions that require using an auxiliary line, but they're great. Thanks for the excellent video!

[davidellis]

Here's another solution that I think may be more direct and intuitive. Construct the altitude EF of the green triangle. Angles ADE and EAF are both complementary to angle DAE, so they are equal, making right triangles ADE and EAF similar. Then AD/AE = AE/EF, so AD*EF = AE^2 = 64. The area of the green triangle is (AB*EF)/2 = (AD*EF)/2 = 64/2 = 32.

[ybodoN]

Very nice puzzle and great solution.
Amazing special case is when E = F.
More special case is when E = F = D.

[MSPC]

There is even more creative Solution. With the available data we can also assume that the Traingle AED is isosceles right angle traingle. Using this we can easily prove the Traingle AEF is also isosceles right angle traingle with side as 8.

[alexniklas]

From point E we lower the perpendicular to AD, point K. ∆AEK~∆ADE.
AE/AD=AK/AE; AD×AK=AE×AE=8×8=64. Area ∆AEB
S=AB×AK/2=64/2=32
Thanks sir! 😊

[tommyb]

Interesting.
I solved in another way which ia think is simpler and using just the needed elements to solve it.
Key component. Draw a perpendicular line from E to AB. (which is the height of the triangle)
In the triangle ADE we name the angle DAE alpha and angle ADE beta. Sum of alpha and beta is 90.
We then find that angle BAE is 90-alpha = beta
We draw a perpendicular line from E to the AB line (which is the height of the ABE triangle)
We obtain a new triangle AEP (P is the point on the AB line) which is a straight triangle. Angle EAP is beta thus angle AEP is beta.
So this new triangle is proportional to the AED triangle.
The side of the square is length X.
The EP line (which is the height of the green triangle is named Y.
...
Thus 8/x = y/8 = z
We need to find y.
y=8*z
z=8/x
y = 64/x
Area of triangle is (x*y)/2 = (x*64/x)/2 = 64/2 = 32.

[Okkk]

Observe that the green area is given by (1/2)*AE*AB*(sin <BAE) and angle <BAE is equal to angle <ADE (Circle alternate angle theorem by constructing a circle with a diameter on AD). Now AB=AE/sin(<BAE) substituting that in the area formula we have the green area=(AE^2)/2=(8^2)/2=32.

[HappyFamilyOnline]

Great explanation👍
Thanks for sharing😊

[montynorth]

Unfortunately I did not spot that clever solution.
The position of point E is not defined, other than the fact that it can be anywhere on the semicircle with line AD as diameter.
Thus point E can be positioned in the exact middle of the square where the diagonals would cross.
There are then 4 congruent right angled triangles with adjacent side lengths of 8 units either side of the 90 degrees, thus giving an area of 1/2 x 8 x 8 for each of them including the green one.

[timeonly]

Angle(ADE) ≅ angle(BAE), since AD⟂AB and AE⟂DE, (2 angles with sides ⟂ to each other are themselves congruent).
Let the measure of angle(ADE) = angle(BAE) = θ .
So, in right triangle ∆(ADE), we have sin(θ) = 8/s (def'n of sine),
and, Area(BAE) = 1/2 (8)(s) sin(θ ) = 4s (8/s) = 32. Done!

[AnonimityAssured]

Oh, goody, goody, a minimal-information problem – my favourite kind! This can be solved in a few seconds, just by making use of the paucity of information provided.

Spoiler alert.

Without loss of generality, we can imagine the figure modified a bit so that DEB become a straight line segment. That makes the calculation simple:
A = 8² / 2;
= 64 / 2;
= 32.

[vitragsinghi]

Many people are having trouble accepting this solution. Think of it like this, for a square, lines drawn from adjacent vertices can never form a right angle, unless both the lines are diagonals. So the figure is not possible as point E and F are the same points. Thus the seeming paradox.

[gordonclarke]

I like to work on a strategy for solving your problems, based on the given facts in the description. Unfortunately, in this case, the key fact that the large rectangle is a square is not provided until your verbal description in the video itself, so I was looking at a problem that had too little information to solve the problem until after the video started.

[gibbogle]

Let s = side of square. Height of green triangle = 8*cos (DAE) = 8*8/s = 64/s.
A = (64/s)*s/2 = 32. No need to complicate a simple solution.

[santiagoarosam]

La horizontal por E corta a AD en el punto F 》La razón de semejanza entre los triángulos AED y AFE es s=8/AD 》AF=(AE)s=8s=8(8/AD)=64/AD 》Área verde =AB×AF/2 = AD(64/AD)/2=64AD/2AD=32
Un saludo cordial.

[bigm]

Wow, you certainly simplified that problem ❤😂