Find the blue shaded area | Geometry Problem | 2 Different Methods | Very Important Geometry Skills

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Find the blue shaded area | Geometry Problem | 2 Different Methods | Very Important Geometry Skills

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If you rotate this triangle, you can apply the ladder theorem which relates the areas:
1/(AEF+AFB) + 1/(ADB+AFB) = 1/ABC + 1/AFB.
So 1/(33+44) + 1/(22+44) = 1/T + 1/44.
T is the total area of the triangle, and we find T=184.8. The area of CDFE= T-(22+33+44)= 184.8-99=85.8

brandonk

If you know Mass Points, you can solve as follows. Let G be the point where AD and BE meet. Since AG:GF=44:22=2:1, assign weights A=n, D=2n. Similarly assign B=3m, E=4m. The weight of C is 2n-3m along BDC and 4m-n along AEC. Then 2n-3m=4m-n, so 3n=7m. If we set n=7, m=3 then the weights of A and C are 7 and 5. Thus the area of BEC is (7/5)*77, or 107.8. Now subtract 22 to yield the blue area of 85.8.

davidellis

It can be very easily solved by using area lemma and menalaus theorem
Let required area is x
In triangle ABE =>AF :FE= 4:3.
In traingle ABD and DAC=> BD:DC=66:33+x
In triangle ABE and BEC => CE:EA=22+x:77
Using menalaus theorem in triangle BEC, 66/33+x * 99+x/77 * 3/4=1
Which gives x as 85.8 💁👀

SumitVerma-lgqh

Hello, I'm from Brazil and I'm really enjoying your channel. I would like you to prove the following:

(a - 1)^a < a^(a - 1), where "a" is a positive real number greater than 1,

gabrielfreitas

you could just use Menelaus in the ACD triangle

mice

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