Find the blue shaded area | Math Olympiad Geometry Problem | Important Geometry and Algebra Skills

Find the blue shaded area | Math Olympiad Geometry Problem | Important Geometry and Algebra Skills

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[sumithpeiris]

Simple Geometry Solution

Obviously, BDCE is a cyclic quadrilateral with diameter BC = a and center F
Hence < BDC = 90 and so < ACD = 30, hence < DFE = 2 X 30 = 60
So Triangle DFE is equilateral with DE = a/2, AD = b/2 and AE = c/2 since Triangles ADE and ABC are similar

Let S = S(ABC) = S so that S(ADE) = S/4
S(BCD) = 2 X 8 = 16, S(BCE) = 30
We can write 3 equations
S = (V3/4)bc
S - 16 =
S - 30 =

(2) X (3) (S-16)(S-30) = (3/64)b^2.c^2 = (3/64)(16S^2 /3) from (1)

Simplfying

3S^2 - 184S + 1920 = 0
(3S - 40)(S-48) = 0
S = 40/3 is not possible since S > 8+15

So S = 48 and the Blue Shaded Area = 48 - 8 -15 = 25

Sumith Peiris
Moratuwa
Sri Lanka

[manuelgabrielrojasleon]

There is a mistake: (min 18:39) the BDF area is [BDF]=(Xcos(P))(Xsin(P))=8, because Xcos(P) is the altitude and 2Xsin(P) is the base

[isaacrichter]

Great work. Your way of explaining the root you take is so clear and charming to watch. Well done!! and all the best to you!!

[suchaisuteparuk]

Geometry only

Follow the video thru 8:15 min
triangle ABC is similar to triangle AED
AB/AE = AC/AD = BC/ED; but BC = 2ED
then AB = 2AE, AC = 2AD
let AE = x, AD = y

Draw CD
We get AB = 2x, DB = 2x - y, and area of DBC = 16
Since area of ABC : area of DBC = AB : DB (Both triangles have the same height)
area of ABC:16 = 2x:(2x-y)
area of ABC = 32x/(2x-y) **1
Draw BE
We get AC = 2y, EC = 2y-x, and area of ECB = 30
Since area of ABC : area of EBC = AC : EC
area of ABC = 60y/(2y-x) **2

**1 = **2;
32x/(2x-y) = 60y/(2y-x)
8x/(2x-y) = 15y/(2y-x)
re-arrange
16xy - 8(x^2) = 30xy - 15(y^2)
8(x^2) + 14xy - 15(y^2) = 0
(4x - 3y)(2x + 5y) = 0
x = 3y/4, -5y/2 --> neglect Negative value
x = 3y/4

Substitute x in **1
area of ABC = 32*(3y/4)/(3y/2 -y) = 24y/(y/2) = 48 ****

Area of ADFE (blue) = area of ABC - area of BDF - area of CEF = 48 - 8 - 15 = 25

Geometry only no Trigonometry
Even you are in grade 9, you can do it.

[hans]

Ich hab mit den Mittelpunkt F einen Thaleskreis um die Punkte B, D, E und C gezeichnet und fand mich dabei genial. Leider kam ich damit aber kein bisschen weiter.

[Istaphobic]

I'm curious, because triangles ABC and ADE are similar, why is the ratio of their areas equal to the division of the squares of their bases?

[arulbiswas]

But how to do it in short way?
I can't solve any of your problems, is it normal?
Am a class 10th student from cbse in India.

[markkusalmelainen]

I fell off to close my eyes..boring. There is far easier way to go.