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.
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.
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.
I basically drew both those constructions yet didn't at all notice that I could get a valid equation from them. Once you mentioned I can, I immediately noticed both the x and the extra x/2 of the height of the DFO triangle, TYSM for this, much appreciated
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.
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!
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.
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
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.
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
I have been doing these maths questions for two years and I still cannot work them out. Not even close. Is it really possible to have the fore site to see the solutions? Is there anyone out there can that can? :-/ .
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
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.
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.
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.
Какое слабое преподавание математики - проводим две диагонали в квадрате , из центра четверти круга к одной из вершин , находящихся на окружности , проводим радиус . Из получившегося прямоугольного треугольника диагональ квадрата d=\|Х*2+Х*2=X\|2 , меньший катет = 1/2d=(X\|2)/2 . Согласно теоремы Пифагора R*2=(X\|2)*2+((X\|2)/2)*2 , 2R*2=5X*2 , Х*2=(2х15*2)/5= 90 .
Very nice sir ❤😊😊😊😊
Glad to hear that!
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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.
Excellent!
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This was my solution as well. It's simplified greatly to note that ON=X
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.
Excellent!
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Y yo)
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.
Excellent!
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@@PreMath Thank you as well!
I basically drew both those constructions yet didn't at all notice that I could get a valid equation from them. Once you mentioned I can, I immediately noticed both the x and the extra x/2 of the height of the DFO triangle, TYSM for this, much appreciated
Excellent..... I salute you
Excellent!
Thanks dear ❤️🌹
Thank you for sharing 😊
Thanks for watching!❤️🌹
Extremely beautiful problem. How beautifully Pithagorus formula has been used twice to find the result ❤❤
Square's sides are x.
From O to M (midpoint of AB) is the square root of
(x/sqrt(2))^2 - ((1/2)x)^2
(x^2)/2 - (1/4)x^2 = (OM^2)
(1/2)x^2 - (1/4)x^2 = (OM)^2
sqrt(1/4)x^2 = (1/2)x
OD = 15
((3/2)x)^2 + ((1/2}x)^2 = 225
(9/4)x^2 + (1/4)x^2 = 225
(10/4)x^2 = 225
10x^2 = 900
x^2 = 90 un^2
x - (1/4)x^2 = (OM)^2
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.
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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!
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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
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Thank you!
You are very welcome!
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Why didn't you think that the hypotenuse of the triangle above the rectangle was ½x√2 ?
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.
Excellent!
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@@GordonSimpson-hr4yfthe diagonals of the green square divide it into four triangles all congruent with OAB.
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
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
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We came up with basically the same solution and posted at about the same time!
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
Excellent!
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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.
Excellent!
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Harmonious!🙂
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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
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I have been doing these maths questions for two years and I still cannot work them out. Not even close. Is it really possible to have the fore site to see the solutions? Is there anyone out there can that can? :-/ .
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
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OK ⟂ CD. Pythagorean theorem in ▲ODK: (x/2)² + (x + x/2)² = R² = 225. x²(1/4 + 9/4) = 225. x² = 90.
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r^2=(l/2)^2+(l+l/2)^2=(10/4)l^2=(5/2)l°2=225...l^2=225*2/5=90
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I use another way but the same reasoning. Nice
Cool, thanks!❤️
Where do you get 5x squared from?
We combined the like terms!
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Sir , OD= x/√2 and BD =√2x . OD²= OB²+ BD². Putting the values and solving we get x²- 90 .
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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.
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Area of the green square= 90 square units.❤❤❤ Thanks sir
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You are very welcome!
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That was easy
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.
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5:10 a is half of b. So a squared is a quarter of b squared.
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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.
Yes sir it's area will be 112.5
(15)^2=225 (15)^2=225 (225+225)=450° (450°-360°)=090° 3^√30 3^√5^√6 √3^1^3^2 √1^√13^2 (x+2x-3)
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I'll guess 35.77 with angle DOC = 45°
AAAAA!!!! It was to hard!
OA=OB=x AB=√2x BD=2x
OD=15
x^2+(2x)^2=15^2 5x^2=225 x^2=45
Green Square area : √2x＊√2x=90
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Какое слабое преподавание математики - проводим две диагонали в квадрате , из центра четверти круга к одной из вершин , находящихся на окружности , проводим радиус . Из получившегося прямоугольного треугольника диагональ квадрата d=\|Х*2+Х*2=X\|2 , меньший катет = 1/2d=(X\|2)/2 . Согласно теоремы Пифагора R*2=(X\|2)*2+((X\|2)/2)*2 , 2R*2=5X*2 , Х*2=(2х15*2)/5= 90 .
Excellent!
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