Find The Yellow Shaded Area | Math Olympiad | A nice Geometry Problem

Find The Yellow Shaded Area | Math Olympiad | A nice Geometry Problem

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[harikatragadda]

If the base of Green is b, then the base of Blue is 4b, as they are Similar triangles with same height h.
By Similarity, b/h = h/4b
So h² = 4b² and h = 2b
Green Area= 1 = ½*h*b = b²
b= 1
Yellow Area = (5b)²-5 = 20

[EddieDraaisma]

A geometrical solution: triangles ABD and ACD are similar, area ACD = 4 * area ABD, then AC = 2 * AB. Rectangle with sides AB and AC has area 2 * AB * AB = 10 (2 * area of triangle ABC), so square with sides AB must have area 5. Repeat the two triangle structure 3 times on the other sides of the square, in the middle a square with sides AC - AB = AB and thus area 5 remains. So total area of square = 4 * (1 + 4) + 5 = 25; yellow area = 25 - 5 = 20.

[santiagoarosam]

Relación entre áreas =1/4》Razón de semejanza s=1/2》Si cateto largo verde =h =cateto corto azul》cateto corto verde =h/2 》cateto largo azul =2h》2h×h=2×4=8》h=2》Lado del cuadrado =h/2 +2h =2/2 +2×2 =5》Área amarilla =5×5 - (1+4) =20
Gracias y un saludo cordial.

[yurenchu]

The green right triangle and blue right triangle are similar. The area of the blue triangle is 4 times the area of the green triangle, hence the hypotenuse of the blue triangle is √4 = 2 times the hypotenuse of the green triangle. However, the hypotenuses of the green triangle and the blue triangle are the right sides of a larger "green+blue" right triangle that is also similar to the green triangle and the blue triangle. Therefore, the ratio of the right sides of the "green+blue" right triangle is 2:1 .

Since the blue triangle is similar to the sum triangle, its width is twice its height. The width and height of the blue triangle define a rectangle with an area that's twice the area of the blue triangle, so a rectangle with area 4+4 = 8. With the ratio width : height = 2:1, this means width = 4 and height = 2 .

Likewise, we can derive that the height of the green triangle is 2, and the width of the green triangle is 1.

So the bottom of the square has length (1+4) = 5. The area of the square is 5*5 = 25. And therefore the yellow area is
(area of square) - (green area + blue area) = 25 - (1+4) = 20 .

[AndreasPfizenmaier-yw]

Without calculating: two triangles with the same height have the same ratio in areas and base legs. So square side= 1+4 =5 and yellow area = 25-5

[Dimitar_Stoyanov_]

Let:
BC = a (square's side)
BD = b (△ABD's shorter leg)
DC = c (△ADC's longer leg)
AD = h
S▫ – square's area
S₁ – △ABD's area
S₄ – △ADC's area ;

△ABD ~ △ADC, therefore
⇒ S₁ / S₄ = 1 / 4
⇒ b / h = 1 / 2 ⇒ h = 2b
⇒ h / c = 1 / 2 ⇒ c = 2h

For △ABD (green)
S₁ = b·h / 2 = 2b² / 2 = b² = 1
⇒ b = 1

For △ADC (blue)
S₄ = c·h / 2 = 2h² / 2 = h² = 4
⇒ h = 2, but because we need c
⇒ c = 2h = 4

For the square
a = b + c = 1 + 4 = 5
⇒ S▫ = a² = 5² = 25

And finally
Yellow Area = S▫ – S₁ – S₄
⇒ Yellow Area = 25 – 1 – 4 = 20

[RahulKumar-idcq]

I request to share two questions daily one based on General Analysis and one based on excellent logic..

[vkr]

BD/DC=1/4 .AD/BD=DC/AD AD*AD=DC*BD, BD=x AD=2x BD*AD/2=1, BD=1, DC=4, BC=5, dzeltenās daļas laukums=25-5=20.