Can you find the area of the Green Square? | Circle | (Step-by-step explanation) |

[@bigm383]

Thanks Professor, great solution!❤

[@PreMath]

Glad to hear that! You are very welcome! Thanks ❤️

[@jimlocke9320]

Let the side of the square have length s. At 3:00, construct AO. Since the area of the circle is π, its radius is 1, as found in the video. So, OA = OB = OE = OF = 1. OP = PE - OE. PE is equal to the side of the square, or s, and OE = 1, so OP = s - 1. AP is half the length of the side of the square, or s/2. Apply the Pythagorean theorem to ΔAPO: 1² = (s - 1)² + (s/2)², 1 = s² -2s + 1 + s²/4, 0 = 5s²/4 - 2s, which has two roots, s = 0 and 5s/4 = 2, or s = 8/5. We discard s = 0, so s = 8/5. The area of the square is s² = (8/5)² = 64/25 = 2.56 sq. units, as PreMath also found.

[@PreMath]

Excellent! Thanks ❤️

[@jimlocke9320]

@@PreMath , thanks for the compliment and thank you for posting these challenging geometry problems, along with your solutions! You are enhancing our interest in geometry two ways. First, you stimulate us to try to solve the problem on our own before watching your solution, and, when we find that our methods are alternatives to yours, post our solutions. Secondly, when we watch your solution, we may see alternative ways to solve the problem. Take pride that we are carefully following and studying your solution. Sometimes, we will see a way to simplify your solution and you should not feel embarrassed that you overlooked the simplification! Other times, we are just finding different methods to reach the same end result. Keep up the good work!

[@PreMath]

@@jimlocke9320 Thanks dear ❤

[@ybodoN]

APE and BPE are 1:2:√5 right triangles, so ∠AEP and ∠BEP are tan⁻¹ 1/2 ⇒ ∠AEB is tan⁻¹ 4/3 (by trig. identities). Therefore, the chord AB is the major cathetus of a 3:4:5 right triangle whose hypotenuse is a diameter of the circle. Since the area of the circle is π, its diameter is 2 so AB is 2·4/5 = 8/5 and the area of the square is 64/25 = 2.56 u².

[@PreMath]

Excellent! Thanks ❤️

[@thomast.2060]

thank you for this video I used the triangle OPA : OA = r = 1 , OP = x , AP = 1/2 AB = ( 1 + x )/2 Pythagoras => x² + (( 1 + x )/2 )² = 1² x² + 1/4 ( 1 + 2x + x² ) = 1 5/4 x² + 1/2 x - 3/4 = 0 x² + 2/5 x - 3/5 = 0 => x1,2 = - 1/5 +/- sqrt( 1/25 + 15/25 ) with x = 3/5 we get AB = 8/5 and the Area of the square is : A = ( 8/5 )² = 64/25

[@PreMath]

Excellent! You are very welcome! Thanks ❤️

[@thomast.2060]

thank you@@PreMath

[@MrPaulc222]

Pretty close to how I did it (I invented F rather than P though). Using your labelling rather than mine, it gave me a right triangle of sides (1/2)x, (x-1), and 1, from which x can be calculated by Pythagoras.

[@wes9627]

Use intersecting chord theorem. Radius of circle is 1 and diameter is 2. Let AB represent the side length of the square. Extend a horizontal line from E through O to F on the opposite side of the circle. Mark G where this line intersects AB. Then AG*BG=EG*FG yields (AB/2)^2=AB(2-AB) or 5AB^2-8AB=0; AB=8/5 units. Green area=AB^2=64/25 square units.

[@Ramkabharosa]

Let AD meet the circle at G & draw a perpendicular from G to meet EF at H. Since AD is parallel to EF, ∠GOE=∠ AOF. So we get |DG| = |EH| = |PF| = 1. By the tangent-secant theorem from the point D, we get |DE|²=|DG|.|DA|. ∴ [(2-x)/2]²=x(2-x). Since 2-x≠0, (2-x)/4=x. ∴ 5x-2=0. So x=2/5. ∴ area(ABCD)= (2 -2/5)²=(8/5)² = 64/25.

[@wackojacko3962]

Basic Concept Reviews are very nice! And like labeling everything, before proceeding too solve I include writing out Basic Concepts that I feel may help in solving any given problem by inspection. 🙂

[@PreMath]

Glad to hear that! Thanks ❤️

[@luigipirandello5919]

Very nice solution. Easy to understand. Thank you.

[@phungpham1725]

1/ Let a be the side of the square and F the intersecting point of DA and the circle. We have r=1 and BF is the diameter So OP= EP-EO=a-1 -->FA=2OP= 2a-2--> DF=DA-FA=2-a By using tangent theorem Sq ED=DF x DA--> sqa/4=a.(2-a)--> 5sqa-8a=0 a=8/5 Area=64/25=2.56 sq units

[@PreMath]

Excellent! Thanks ❤️

[@marcgriselhubert3915]

Let's use an adapted orthonormal: E(0;0) A(k;k/2) B(k, -k/2) where k is the length of the square. The equation of the circle is x^2 + y^2 + ax + by +c = 0 where a, b and c are unknown. A is on the circle so c = 0; B is on the circle so k^2 + (k^2)/4 + ak + (bk)/2 = 0; B is on the circle so k^2 + (k^2)/4 +ak - (bk)/2 = 0. Its easy to obtain that c =0; a = -(5/4).k; b =0. So the equation of the circle is x^2 + y^2 -(5/4).k = 0, or (x - (5/8).k)^2 + y^2 = ((5/8).k)^2 So we have O((5/8).k; 0) and the radius R of the circle is (5/8).k Now we know that R = 1 (as the area of the circle is Pi), so (5/8).k = 1 and then k = 8/5 is the length of the square. Finally the area of the square is (8/5)^2 = 64/25.

[@AdemolaAderibigbe-j8s]

Let's label the length of the side of the square ABCD as"Y". (Y must be positive to be a viable solution) Set up the product of the component lengths of the intersecting chords APB and the diameter EOF for the circle of radius 1 (the two chords intersect at point "P" ): (AP)(PB) = (EP)(PF) or (Y/2)(Y/2)= Y(2-Y). Hence Y = 8/5 and area of square ABCD = Y^2 = 64/25

[@MarieAnne.]

Let s = side length of square. Then in △AOP, we get: ∠APO = 90° (diameter EF perpendicular to tangent DC and AB parallel to DC, so EF perpendicular to AB) AP = AB/2 = s/2 (diameter that is perpendicular to chord bisects the chord) OP = EP − OE = s−r = s−1 OA = r = 1 Using Pythagorean theorem, we get (s/2)² + (s−1)² = 1² s²/4 + s² − 2s + 1 = 1 5s²/4 − 2s = 0 s/4 (5s − 8) = 0 s = 0 or 8/5 Since s is sidelength of square and is > r, then s = 8/5 = 1.6 Area of square = s² = 64/25 = 2.56

[@murdock5537]

∎ABCD → AB = BC = CD = AD = 2a; r = 1 → OP = 2a - r = 2a - 1 → ∆ AOP → (2a - 1)^2 + a^2 = 1 → a = 4/5 → (2a)^2 = 64/25; sin⁡(ϑ) = AP/AE = √5/5 or: r = 1; sin⁡(ϑ) = √5/5 → cos⁡(ϑ) = 2√5/5 → sin⁡(2ϑ) = 2sin⁡(ϑ)cos⁡(ϑ) = 4/5 = AP/AO = a/r → a = 4/5 → (2a)^2 = 64/25

[@PreMath]

Excellent! Thanks ❤️

[@MrPaulc222]

This could be trickier than it looks at first. r=1. Call the square's sides s. Make a midpoint on AB and call it F. AOF is a right triangle with sides (1/2)s, s-1, and 1. ((1/2)s)^2 + (s-1)^2 = 1^2 (1/4)s^2 + (s-1)^2 = 1^2 (1/4)s^2 + s^2 - 2s + 1 = 1 (5/4)s^2 - 2s = 0 5s^2 - 8s = 0 s = (8/5) so square is (64/25) sq units. 2.56 sq units. I did assimilate additional information which I later discarded and deleted from this answer. EDIT: No intersecting chords needed - just straightforward Pythagoras for this one. Now to watch how you did it :) I see you used another method and, unusually, mine appears simpler :) Thank you once again.

[@adept7474]

▲АВЕ -Inscribed. AB = x, AE = BE = (x√5)/2. S(ABE) = (AB × AE × BE)/4R = 5x³/16 = EP × BP = x²/2. 5x/8 = 1. x = 8/5, S(square) = 64/25.

[@PreMath]

Excellent! Thanks ❤️

[@tombufford136]

At a quick glance: The chord theorem, proposition 35 in Euclid's geometry is useful. The products of the two segments of two intersecting chords of a circle are equal.

[@marcelowanderleycorreia8876]

Very good!! Thank you Sir!

[@PreMath]

You are very welcome! Thanks ❤️

[@soli9mana-soli4953]

In this problem applying the intersecting chords theorem on AB and diameter passing on EO, or applying Euclid’s theorem on right triangle whose hypotenuse is the diameter and its height AH, or applying the tangent secant theorem on points C,E,B we get always the same result: S^2 = 2S*(2 - 2S). Being 2S the side of the square 😊

[@PreMath]

Thanks ❤️

[@santiagoarosam430]

Area del círculo =π → Radio del círculo =r =1 Los lados DA y CB del cuadrado definen las cuerdas FA y GB; ambas tienen una flecha de longitud "f" → La cuerda AB tiene una flecha de longitud 2f=DF → Potencia del punto D respecto a la circunferencia: DE²=DF*DA → (r-f)²=(2f)(2r-2f)→ 5f²-6f+1=0→ f=1/5→ AB=2r-2f =2-(2/5) =8/5 → Área ABCD =64/25 =2.56 Interesante problema. Gracias y un saludo cordial.

[@PreMath]

You are very welcome! Thanks ❤️

[@misterenter-iz7rz]

Let 2s×2s be the square, 2s=1+sqrt(1-s^2), (2s-1)^2=1-s^2, 5s^2-4s=0, s=4/5, as s >< 0, then the answer is (8/5)^2=64/25=2.56.😊

[@PreMath]

Excellent! Thanks ❤️

[@christopherlinder7618]

I didn't know the chord multiplication theorem. I immediately drew in sin M and cos M (angle at midpoint M, I don't write O as it looks like zero) and came up with 2 sin M as the vertical side and 1 + cos M as the horizontal side, and since it's a square you get the equation 2x = 1 + sqrt(1 - x^2) if you sub x = sin M and figure cos from the trig Pythagoras. This gives you (2x-1)^2 = 1 - x^2 or finally x(5x - 4) = 0. Since x is obviously not 0 from the sketch, it must be 4/5 = 0.8 , making the cos M = 0.6 and yielding the famous 3, 4, 5 shape. But since 0.8 was sin M and 2sinM is 1.6 was the side of the square, you get 2.56 for the area. I like the chord theorem solution, though, as it avoids trigonometry.

[@unknownidentity2846]

Let's find the size of the green area: . .. ... .... ..... May s be the side length of the square and may r be the radius of the circle. From the given area of the circle we can conclude: A(circle) = πr² = π ⇒ r = 1 The line through the points E and O may intersect AB in point F. Then we get two congruent right triangles OFA and OFB, so we can apply the Pythagorean theorem: OA² = FA² + OF² r² = (s/2)² + (s − r)² r² = s²/4 + s² − 2sr + r² 0 = 5s²/4 − 2sr 0 = 5s²/8 − sr 0 = s*(5s/8 − r) Since s≠0, we can follow: 5s/8 = r ⇒ s = (8/5)r = 8/5 ⇒ A(square) = s² = 64/25 = 2.56 Best regards from Germany

[@PreMath]

Great! Thanks ❤️

[@contillojakexander2003]

Thank you so much

[@papilgee4evaeva]

When we got to the step where 4x(2 - x) = (2 - x)(2 - x), I divided both sides by (2 - x). The resulting work led me to the acceptable value of x. Not sure if that works every time, but it worked here. 🙂

[@PreMath]

It'd work as long as x is less than 2. Thanks ❤️

[@AmirgabYT2185]

S=2,56

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! Thanks ❤️

[@raya.pawley3563]

Thank you

[@mathbynisharsir5586]

Fantastic video sir ❤❤❤❤❤❤

[@PreMath]

So nice of you dear. Thanks ❤️

[@weird8599]

I have solved it in my own way . btw great solution premath sir :)

[@PreMath]

Great job! Glad to hear that! 😀

[@LuisdeBritoCamacho]

Being the Area of the Circle Pi square units, the only solution is the solution below. The Area of the Square is equal to 1,6^2 square units = 2,56 square units. I solved this Problem by Geometrical Optimization.

[@prossvay8744]

Green square area= (1.6)^2=2.56 suqare units. ❤❤❤ Thanks

[@PreMath]

Excellent! You are very welcome! Thanks ❤️

[@الثورة-ص7ق]

Let x^2 be the green square area.lets take triangle rectangle PAE at P, cos(alpha/2)=2x/x_/5 =2/_/5 》cos(alpha)=3/5.now lets take triangle rectangle PAO, cos(alpha)=×-1, now we get x-1=3/5》x=8/5 then x^2=2,56 square units

[@PreMath]

Thanks ❤️

[@GaryBricaultLive]

A much simpler way of resolving the ab=cd and only leaves one solution instead of two. The steps are as follows; 4x(2-x) = (2-x)(2-x) starting following the first simplification step 4x = 2-x divide both sides by (2-x) 4x + x = 2 add 'x' to both sides 5x = 2 combine 'x' terms x = 2/5 divide both sides by x

[@PreMath]

Thanks

[@jan-willemreens9010]

... Good day, Radius circle is R = 1 ... assume side length green square X ... I DA I = I EO I + I OP I ... X = 1 + I OP I ... we can find I OP I with Pythagoras in for instance right triangle (AOP) ... I OA I = R = 1 , I AP I = X/2 , so I AP I^2 + I OP I^2 = I OA I^2 ... so after doing the algebraic steps we obtain I OP I = SQRT(4 - X^2)/2 ... recalling X = 1 + I OP I ... X - 1 = SQRT(4 - X^2)/2 ... after a few algebraic steps solving for X we obtain ... 5X^2 - 8X = 0 ... X(5X - 8) = 0 ... X = 0 (rejected) v X = 8/5 .... finally Area green square = X^2 = ( 8/5 )^2 = 64/25 u^2 .... thank you for your clear alternative strategy ... best regards, Jan-W

[@PreMath]

Excellent! You are very welcome! Thanks ❤️

[@giuseppemalaguti435]

r=1...l=r+√(r^2-(l/2)^2)=1+√(1-l^2/4)...(l-1)^2=1-l^2/4...5l^2/4-2l=0...5l/4=2...l=8/5

[@PreMath]

Thanks ❤️

[@sergioaiex3966]

Solution: A = π r² π = π r² r² = 1 *r = 1* *Square Side = 2x* Let's assume a point F, that lies on the square side, in a streight line, such a way, OF = 2x - r *OF = 2x - 1* Applying The Pythagorean Theorem, we gonna have: OA² = OF² + FA² 1² = (2x - 1)² + x² 1 = 4x² - 4x + 1 + x² 5x² - 4x = 0 x (5x - 4) = 0 x = 0 Rejected *x = 4/5* Accepted L = 2x L = 2 . 4/5 *L = 8/5* A = L² = (8/5)² *A = 64/25 Square Units* *A = 2,56 Square Units*

[@PreMath]

Excellent! Thanks ❤️

[@xianartman]

Maybe I jumped to conclusions, but if: 4x(2-x)=(2-x)(2-x) Then can you not divide by (2-x)? Thus 4x=(2-x) 4x+x=2 5x=2 Therefore: X=2/5

[@sirchedr]

Scanavi - Сканави

[@hermannschachner977]

1+cos = 2 sin; 1,6 = 2 x 0,8 . . . . 1.6^2 = 2,56

[@PreMath]

Thanks ❤️

[@jarikosonen4079]

When x=2, the 'square' is just a dot at the point E.

[@PreMath]

True! Thanks ❤️

[@JSSTyger]

A = 2.56

[@PreMath]

Excellent! Thanks ❤️

[@weird8599]

i did it by pythagoras

[@weird8599]

I became your new subscriber

[@PreMath]

Excellent! Welcome aboard🌹 Thanks ❤️