Calculate area of the Green shaded Triangle in a square | Important Geometry skills explained

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Calculate area of the Green shaded Triangle in a square | Important Geometry skills explained

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[procash]

Amazing

Complex Problem, Simple Solution

Premath Guru Ji rocks

[solimana-soli]

I've found this fast solution: AF = h, AD = b, for the first Euclid theorem we have
9² = h*b
h = 81/b ( if we want distinguish each side )
Area = 81/b*b*1/2 = 81/2
...yes, I am a fan of Euclid

[EnnioPiovesan]

ABPF is the rectangle with sides AF=the projection of EA on hypotenuse AD of right triangle EAD, and AB=AD= the hyypotenuse so, by the first Euclid's theorem it is equivalent to the square with side AE, whose area is 81. So the green area is 81/2.

[Dalesmanable]

I just went: h = 9 cos(alpha) and 9 = b cos(alpha), so that A = 0.5 x 9 x cos(alpha) x 9/cos(alpha) = 40.5. No need to worry about similar triangles.

[phungpham]

Consider the right triangle AED.
EF is the height to the hypotenuse so sq EF= AExFD ( the right triangle altitude theorem) = h.(b - h) = h.b - sqh
sq EF+ sq h = 81= h.b Area of the green triangle= 81/2= 40.5 sq units

[alster]

The similarity technique was a bit tricky but the problem became easier after obtaining bh. From there, I solved it on my own and fast forwarded to the end to double check if it matches which turned out it did.

[HerrPedroapan]

How about assuming ED to be 9 as well? The the area of FBP will be (b*b)/4 or (81+81)/4=40, 5?

[spiderjump]

Let AD be a and AF be b
and EF be e.
Triangles AEF and EDF are similar.( AA)
Hence e/b=a—b/e
e^2=b(a—b)
9^2 —b ^2 =ab—b^2
ab=81
Area of green triangle = 1/2ab=40.5 units squared

[montynorth]

Also, ED could be 9, dividing the square into 2 equal rectangles. Then AD = Sq.root of 162.
Then area of square = 162.
4 equal triangles = 162 /4 = 40.5.

[vidyadharjoshi]

Let AF = A, FD = B, EF = H. Grean Triangle Area = (1/2)*A*(A + B). EFsq = 81 - Asq = EDsq - Bsq.
81 - EDsq = Asq - Bsq. 81 + WDsq = (A + B)sq . 2*81 = Asq - Bsq + Asq +2AB + Bsq = 2*Asq + 2AB = 4*(Green Triangle Area)
Grean Triangle Area = 81/2 = 40.5

[geoninja]

I did it Olympiad style - assuming the triangle off the side of the square was isoceles, then the side of the square is 9*sqrt(2), then the triangle in question is 1/4 of the square area.- :)

[peterkrauliz]

Since the length of ED can be assumed as any length, the solution is easy, say ED is the same as AE, meaning 9. Thus alpha and beta will both be 45 deg. No problems and 40.5 is at hand!

[theoyanto]

Well, I think my brain had been switched to sleep mode while I wasn't paying attention, and I did read your notes.
I'm almost ashamed to admit I missed it completely. I almost vanished up Pythagoras alley... But on the positive side, I did find a way not to accomplish it.🤓👍🏻

[ybodoN]

Quick evaluation: the smallest possible square ABCD is obtained when E = D = F...😉

[wackojacko]

Complementary angles, yes! Complimentary angles, No! Those would be angles we get for free....🙂

[Aligakore]

Found it using the Pythagorean theorem only. I have labelled AF=a, FD=b, DE=c and EF=d. The area of the green triangle is therefore (a+b).a/2.
I used three equations :
1. a^2 + d^2 = 81
2. 9^2 + c^2 = (a+b)^2
3. b^2 + d^2 = c^2
After replacing c^2 by b^2 + d^2 in equation 2 and developing (a+b)^2, equation 2 can be written as 9^2 + b^2 + d^2 = a^2 + 2ab + b^2.
I have subtracted equation 2 to equation 1 and the result is a^2 - 9^2 -b^2 = 81 - a^2 - 2ab - b^2. The -b^2 can be simplified and after grouping I got 2a^2 + 2ab = 2*81. Simplify by two : a^2 + ab = 81 => a(a+b)=81 which is twice the area we are looking for so A=81/2=40.5 square units 😊

[santiagoarosam]

Simplificando, podemos decir que el cateto de longitud 9 un tiene una pendiente del 100% 》Lado del cuadrado =c =2[9/(sqrt2)] 》Área verde =(1/2)c×c/2 =9^2/2 =81/2 =40, 5
Gracias y un saludo.

[misterenter-izrz]

Rather strange figure, clearly we have to find ab/2, where a is the side of lower square and b is the height of the upper rectangle. Then EF^2=81-b^2, so 81-b^2+a^2=ED^2, thus (a+b)^2=a^2+b^2+2ab=81+81-b^2+a^2, 2b^2+2ab=162, or b^2+ab=81, I fail😅
I see I misinterpret the figure. 🙄

[kckr]

Opposite side length to 90 degrees in triangle aed is b+h. But u put as b. Can u pls confirm

[kituluipitu]

Can one conclude that triangle AED is a 3-4-5 triangle from the information given ? Then by way of similar triangles one can easily get the base and height of triangle BPF.