Math Olympiad | Can you find area of the Green Square? | Quarter circle | #math #maths #geometry

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Math Olympiad | Can you find area of the Green Square? | Quarter circle | #math #maths #geometry

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[hongningsuen]

It is very kind of you to leave the easy solution for us to find such that we look smart.
Here is my solution in exam answer format:
1. Let the side of square be 2X.
2. Draw perpendicular bisector from centre O to chord CD. Let the point of bisection be M.
Hence DM = CM = X
3. OM also bisects right angle AOB into two 45 degrees angles (property of perpendicular bisector of chord).
4. OM must bisect AB perpendicularly as well as it cuts the square into 2 equal rectangles. Let the point of bisection of AB be N. AN = BN = X
5. As triangle AON is a right-angled triangle, angle OAN = 180 - 90 - 45 = 45. Hence triangle AON is an isosceles triangle with ON = AN = X (equal sides of isosceles triangles).
6. OM = AN + AD = X + 2X = 3X.
7. Draw OD to form triangle ODM. OD = 15 (radius of circle), DM = X, OM = 3X.
8. By Pythagoras theorem, OD^2 = DM^2 + OM^2. Hence 15^2 = X^2 + (3X)^2. Hence X^2 = 225/10.
9. Area of green square = 4X^2 = 90.

[santiagoarosam]

AB=a ; M=Punto medio de DC → En el triángulo DMO: 15²=(a/2)²+(a +a/2)²=(10/4)a² → a²=4*15²/10=90 =Área ABCD.
Gracias por sus vídeos. Un saludo cordial.

[misterenter-izrz]

Let s×s be the square, then OE perpendicular to AB, =s/2, by Pythagoras theorem, s^2, s^2=15^2, s^2, the answer, is 15^2×2/5=90.😊

[jimlocke]

At 2:15, label the intersection of the axis of symmetry with AB as point E and with CD as point F. If x is the length of the side of the square, AE = DF = x/2 because E and F are midpoints. ΔAEO is an isosceles right triangle, so EO = AE = x/2. Consider ΔDFO. OD = hypotenuse = radius of circle. DF = x/2. OF = OE + EF = x/2 + x. Applying the Pythagorean theorem, r² = (x/2)² + (3x/2)² = x²/4 + (9/4)x² = 10x²/4, so x² = (4/10)r². We are given r = 15, so r² = 225 and x² = (4/10)(225) = 90 square units, as PreMath also found.

[smartsbs]

Let the midpoint of A and B be E, and the midpoint of C and D be F and AB = x
OE = x/2, EF = x -> OF = 3x/2


DF = x/2, OD = 15
In right triangle ODF, OD^2 = OF^2 + DF^2
15^2 = (3x/2)^2 +(x/2)^2
225 = 9x^2/4 + x^2/4
225 =10x^2/4
x^2 = 225 * 4/10 = 90

[DB-lgsq]

شكرا لكم على المجهودات
يمكن استعمال Hهو المسقط العمودي لDعلى(OP)
OH=x(جذر2)
DH=x(جذر2)/2
OD^2=OH^2+DH^2

x^2=90

[ChuzzleFriends]

Think outside the box!
Create three quadrants identical to the given one such that together the quadrants form a full circle.
Label BO = x.
The diagonals of two of the green squares form a chord. Use the Intersecting Chords Theorem.
(s√2)(s√2) = (x + 15)(15 - x)
2s² = 225 - x²
The sides of the four green squares make a square towards the center of the circle with side lengths identical to the green ones. BO is part of the diagonal. So, the diagonal of this new square is 2x units long.
s = (d√2)/2
= [(2x)√2]/2
= x√2
Substitute.
2(x√2)² = 225 - x²
2(2x²) = 225 - x²
4x² = 225 - x²
5x² = 225
x² = 45
x = 3√5
Substitute again.
s = (3√5)√2
= √45 * √2
= √90
= 3√10
A = s²
= (3√10)²
= 90
So, the area of the green square is 90 square units.

[marcgriselhubert]

As often, let's use an adapted orthonormal, center O, vertical first axis. Then A(a; -a) B(a;a) C(3.a;a) D(3.a;-a) and the length of the square is 2.a
The equation of the circle is x^2 + y^2 = 15^2 = 225 and D is on the circle, so (3.a)^2 + a^2 = 225, so a^2 = 225/10 = 45/2.
Now the area of the square is (2.a)^2 = 4.(a^2), so it is 4.(45/2) = 90. Very quick!

[AndreyDanilkin]

you can complete the remaining sectors of the circle with the same squares. Then: x^2+(3x)^2=30^2 ->10x^2=900 -> x^2=90

[unknownidentity]

Let's do some math:
.
..
...
....


May s be the side length of the green square, may r=15 be the radius of the quarter circle and may R and S be the midpoints of AB and CD, respectively. Then the right isosceles triangle OAB can be divided into the two congruent right 45°-45°-90° triangles OAR an OBR. So we can conclude:

AR = BR = OR = s/2

The triangle ODS is also a right triangle, so we can apply the Pythagorean theorem and finally calculate the area of the green square:

OD² = DS² + OS²
OD² = DS² + (OR + RS)²
r² = (s/2)² + (s/2 + s)²
r² = (s/2)² + (3s/2)²
r² = s²/4 + 9s²/4 = 10s²/4 = 5s²/2

⇒ A(ABCD) = s² = 2*r²/5 = 2*15²/5 = 90

[kennethstevenson]

You could have found OB to equal X/ Root 2. Then the triangle OBD would have 15^2 = (x/ root 2}^2 + (x root 2)^2). Simplified you would have 225 =(5x^2)/2 ; 450 = 5x^2 ; x^2 = 90 ; Area = 90 Sq. Un.

[ybodoN]

OP = OQ = OD = OC = 15 since they are all radii of the quarter circle.
Draw the diagonals AC and BD. Note that AC = BD is twice OA = OB.
The diagonals of a square are _s√2_ where _s_ is the side of the square.
Then (s√2)² + (½ s√2)² = 15² and the area of the green square is 90.

[massimiliano__]

As the distance between O and AB is AB/2, then the distance between O and CD is 3*AB/2; then 15^2= (3AB/2)^2+(AB/2)^2; that means (AB)^2= 90

[alokranjan]

Extremely beautiful problem. How beautifully Pithagorus formula has been used twice to find the result ❤❤

[LuisdeBritoCamacho]

Answer: The Green Square Area = 90 Square Units. Why?
First, the Area of a Quarter of our Semicircle is equal to (225Pi) / 4 ~ 176, 7 Square Units.
Let's call OA = OB = "a" and the Side Length of the Green Square "x"
Now : As triangle [OAB] is a Right Triangle; x^2 = 2a^2 (is very important to keep this equality in mind!)
Angle DBO = 90º 'cause angle OBA = 45º.
Triangle DBO is a Right Triangle with Side Lengths equal to : OD = 15 ; OB = a ; BD = x*sqrt(2)
Now, 15^2 = a^2 + [x*sqrt(2)]^2 ; 225 = a^2 + 2x^2 ; 225 = a^2 + 2*(2a^2) ; 225 = a^2 + 4a^2 ; 225 = 5a^2 ; (225 / 5) = a^2 ; a^2 = 45 ; a = sqrt(45).
x^2 = 2a^2 ; x^2 = 2*(sqrt(45))^2 ; x^2 = 2 * 45 = 90 ; x^2 = 90 sq un. QED.

[adept]

OK ⟂ CD. Pythagorean theorem in ▲ODK: (x/2)² + (x + x/2)² = R² = 225. x²(1/4 + 9/4) = 225. x² = 90.

[andreadevescovi]

I use another way but the same reasoning. Nice

[quigonkenny]

Let the side length of the square be s. Draw radius OE so that it bisects DC at F and intersects BA at G. As it also bisects BA, due to being perpendicular to CD (as any radius bisecting a chord), DCBA is centered in the quarter circle and the construction is symmetrical about OE.

∆BOA is an isosceles right triangle, as BO = OA. As OG bisects BA perpendicularly and ∠OAG and ∠GBO are each 45°, ∆OGB and ∆AGO are also isosceles right triangles. Therefore OG = BG = GA = s/2.

Triangle ∆ODF:
DF² + FO² = OD²
(s/2)² + (3s/2)² = 15²
s²/4 + 9s²/4 = 225
10s²/4 = 225
s² = 225(2/5) = 45(2) = 90

[sundareshvenugopal]

The method I used is this,

Area of quadrant = Area of triangle OAB + Area of square ABCD + twice area of flanking section/collar BCQ + area of arc DC below chord DC.

In general, if r is the radius, y is distance OB = OA, x is the side length of square, then

x = √2*y.

Area of triangle OAB = 1/2*(y^2)

Area of square ABCD = 2*(y^2)

Area of collar BCQ =
MIN(a = 2*π*(y^2)/8, b =(π*((r-y)^2))/8) + ABS(a-b)/2

Area below chord DC, above arc DC =
(π*(r^2))/8 - (y*√(2*(r^2) - (y^2)))/4

Here r = 15.

Apologies, for any mistakes in the derivation.

[skoijlg]

Let's build a circle with a radius of 15, divide it into 4 quarters. Let's put a square in each quarter. Let's apply the chord theorem. For one of their chords, we take the diameter, the continuation of the radius Oq. Point B divides it into two segments: (15+0.5x√2) and (15-0.5X√2). The second chord is a continuation of the segment BC, which is divided by the point B into two segments x and 2x. We make up the equation (15+0.5x√2) *(15-0.5X√2)=2x*x. x^2=90.