Can you find the area of the Yellow semicircle? | (Fun Geometry Problem) |

Learn how to find the area of the Yellow semicircle. Important Geometry and Algebra skills are also explained: Pythagorean theorem; Thales' theorem; area of the circle formula; similar triangles. Step-by-step tutorial by PreMath.com
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Can you find the area of the Yellow semicircle? | (Fun Geometry Problem) | #math #maths | #geometry
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[@Ashi-cq7tj]

Hi Me first 😂😂

[@PreMath]

Yes you are!❤️

[@rabotaakk-nw9nm]

ΔAOP: b²=r²+(vʼ5)² => r²=b²-5 Crossing chords: (r+vʼ5)(r-vʼ5)=b(8-b) r²-5=8b-b²; b²-5-5-8b+b²=0 b²-4b-5=0; b=-1 X; b=5; r²=5²-5=20 A=½πr²=½π20=10π≈31.416 😁

[@PreMath]

Excellent! Thanks for sharing ❤️

[@phungpham1725]

The quadrilateral PCBO is cyclic so, APxAC=AOxAB--> 8b=2sq r -> b=sqr/ 4 (1) And b= sqrt(sqr+5) (2) From (1) and (2) we have: sq r/4 = sqrt (5+sqr) Or sqr.sqr/16= 5+sqr Just label sqr=X We have: sqX-16X-80=0 Or (sqX-2.8X +64) -64-80=0 sq(X-8)=144-> X-8=+/- 12 X=sqr= 20 Area of the semicircle=1/2. Pi.20= 10pi😅

[@PreMath]

Excellent!👍 Thanks for sharing ❤️

[@facegamer8379]

Dear Premath, Thank you for your clear and concise math videos. They make learning so much easier!

[@PreMath]

Glad you like them! You are very welcome! Thanks for the feedback ❤️

[@josedavis4242]

Let AP be "x" PC = 8-x Extend OP to touch the circle and let that point be "Q" Radius being "r" PQ = r - √5 Consider the semicircle to be full circle and extend PO to the point on other side of circle. Let it be "M" PM = r + √5 With the intersecting theorem, AP * PC = PQ *PM x * (8-x) = (r-√5)*(r+√5) Solving this, r^2 = 5+ 8x - x^2 ---------------(1) Pythagoras theorem in triangle APO, r^2 = x^2 - 5 -------------(2) Considering (1) &(2) 10 + 8x -2x^2 = 0 Simplfying, x^2 - 4x -5 = 0 Hence, x = 5 Substituting, r^2 = 20 Hence π r^2 / 2 = 31.4

[@PreMath]

Excellent! Thanks for sharing ❤️

[@IOSARBX]

PreMath, I love this video!

[@PreMath]

Excellent! Glad to hear that! Thanks dear❤️

[@johnbrennan3372]

|CB| squared = 4r^2 - 64. |PC|= 8- sqroot(r^2 +5). In the triangle CPB:(8- sqroot of (r^2 + 5 ))^2)+ 4r^2-64)= r^2 +5. That gives r^2=20 etc .

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marcgriselhubert3915]

Let's use an orthonormal center O, first axis (OB) We have C(Rcos(t); R.sin(t)) with R the radius of the circle and t = angleBOC, and A(-R; 0) Then VectorAC(R.(1 +cos(t); R.sint) = 2.R.cos(t/2).VectorU with U(cos(t/2); sin(t/2)). And so AC = 2.R.cos(t/2) = 8, and R.cos(t/2) = 4 (eq. 1) angleOAP =t/2, and in triangle OAP: OP = AO.tan(t/2), so we have: R.tan(t/2) = sqrt(5) (eq. 2) From eq. 1: (cos(t/2))^2 = 16/(R^2) From eq. 2: (tan(t/2))^2 = 5/(R^2) Knowing that 1 + (tan(t/2))^2 = 1/((cos(t/2))^2), we get: 1 + (5/R^2) = (r^2)/16 or R^4 -16.(

[@PreMath]

Excellent! Thanks for sharing ❤️

[@innovativeeducation5814]

no need to find BC. Take the third ratio AP/AB = OA/CA AP = sq rt ( r^2 + 5 ) get a quadratic equation in r^4 - 16 r^2 -80 = 0 solve for 'r'

[@misterenter-iz7rz]

s/8=sqrt(5)/r, s=8sqrt(5)/r, 320/r^2+64=4r^2, r^4-16r^2-80=(r^2-2)(r+4)=0, r^2=20, therefore the answer is 10pi 😊

[@PreMath]

Excellent!👍 Thanks for sharing ❤️

[@MrPaulc222]

I will try to wing this and hope that I stumble across a viable method. Call the portion of the 8 chord towards the top right, x. This gives a right triangle of (8-x)^2 - (sqrt(5))^2 = r^2 64 - 16x + x^2 - 5 = r^2 r^2 = x^2 - 16x + 59 Triangle ACB is similar, with a hypotenuse of 2r and the longer of the other sides being 8. Maybe I don't need to square sides after all: (2r)/8 = (8-x)/r gives 2r^2 = 64 - 8x when cross-multiplying. Two equations: (1) r^2 = x^2 - 16x + 59 (2) 2r^2 = 64 - 8x so r^2 = 32 - 4x Conclusion: x^2 - 16x + 59 = 32 - 4x Find x: x^2 - 12x + 27 = 0 (12+or-sqrt(144 - 4*27))/2 = x (12+or-sqrt(36))/2 = x (12+or-6)/2 = x The value of x must be less than 8, so discard the positive result. x = 3 This makes length AP 8 - 3 = 5 Therefore, 5^2 - (sqrt(5))^2= r^2 Go longhand: 25 - 5 = r^2 so r^2 = 20 (20pi)/2 = 10 pi, so area is 31.42 sq un (rounded).

[@PreMath]

Excellent!🌹👍 Thanks for sharing ❤️

[@ravikrpranavam]

As usual you have explained the solution very well.

[@PreMath]

Many thanks, my dear friend🌹 Glad to hear that! Thanks for the feedback ❤️

[@jimlocke9320]

At 3:50, PreMath has equated the ratio of short side to long side for each of the similar triangles, ΔABC and ΔAOP. I chose to equate the ratios of hypotenuse to long side instead, arriving at the equation AB/AC = AP/AO. Apply the Pythagorean theorem to ΔAOP: AP = √(r² + 5). The equation AB/AC = AP/AO becomes 2r/8 = √(r² + 5)/r. Cross multiply: 2r² = 8√(r² + 5). Replace r² by x: 2x = 8√(x + 5). Divide both sides by 2: x = 4√(x + 5). Square both sides: x² = 16(x + 5) and x² -16x -80 = 0. Skip ahead to 7:30, where PreMath has found the same equation, solves it for r² = x = 20 and then computes the area of the semicircle as (1/2)πr² = (1/2)π(20) = 10π square units.

[@PreMath]

Excellent!👍 Thanks for sharing ❤️

[@devondevon4366]

32.416 Made an arithmetic error, but this was by approach Draw a perpendicular line from C to B (Thales Theorem) to form a right triangle ABC. Let's focus on the triangle on the left the short leg = sqrt 5, and the long leg = radius of the semi-circle. For triangle ABC, the long leg is 8, and the short leg is unknown. Let it be n. Since both triangles are similar, then sqrt 5 n ------- = ------- r 8 8 ( sqrt 5) = rn (cross multiply 8(sqrt 5) _____ = n (divide both sides by n) r So, now we have three sides of triangle ABC in terms of r 8, 8 ( sqrt 5)/ r, and 2 r (2 r is the diameter of triangle ABC as the diameter = twice the radius) Notice that the two legs of ABC are 8 and 8(sqrt 5)/n and the hypotenuse = 2r Let's employ the Pythagorean Theorem. (8)^2 + { 8(sqrt 5)/r]^2 =( 2r)^2 64 + 320/r^2 = 4r ^2 64 r^2 + 320 = 4 r^4 (multiply both sides by r^2) 16 r^2 + 80 = r^4 (divide both sides by 4) r^4 -16 r^2 -80 = 0 let n = r^2 , then n^2 - 16 n -80 =0 (n-20)(n+4) = 0 n = 20 and n= -4 So n = 20, not the negative 4 Hence r^2 = 20 since n= r^2 Hence r = sqrt 20 or 4.4721359 Hence area of circle = pi r^2 = pi * 4.4721259^2 = pi *20 = 62.8313 Hence, the semi-circle area is half or 32.416

[@PreMath]

No worries. We are all lifelong learners. Thanks for sharing ❤️🙏

[@prossvay8744]

Let CP=x and Radius=R AP=8-x OA=R OA^2+OP^2=AP^2 R^2+(√5)^2=(8-x)^2 R^2+5=64-16x+x^2 R^2-x^2+16x=59 (1) x(8-x)=(R-√5)(R+√5) 8x-x^2=R^2-5 R^2=8x-x^2+5 (2) (1) 8x-x^2+5-x^2+16x-59=0 2x^2-24x+54=0 x^2-12x+27=0 x=3 ; x=9 rejected (2) R^2=24-9+5=20 So semicircle area=1/2π(20)=10π=31.42 square units.❤❤❤

[@PreMath]

Excellent!👍 Thanks for sharing ❤️

[@santiagoarosam430]

AP=a=√(5+r²)→ Potencia de P respecto a la circunferencia =a(8-a)=(r-√5)(r+√5)→ 8a-a²=r²-5=8(√(5+r²))-5-r²→ (r²)²-16r²-80=0→ r=2√5→ Área semicírculo amarillo =10π . Gracias y saludos.

[@PreMath]

Excellent!🌹 Thanks for sharing ❤️🇺🇸

[@calototube]

It was not necessary to find the square root of r2.

[@PreMath]

Thanks for the feedback ❤️

[@devondevon4366]

Exactly, since the area of the circle = pi r^2, but perhaps pre-math wanted to show step-by-step

[@yvesdelombaerde5909]

Yes indeed, if you have got r squared, keep it short and multiply with pi/2

[@Waldlaeufer70]

ABC and APO are similar triangles due to the common angle at A and the right angle at C. r : √5 = 8 : BC Let a = BC ar = 8√5 a = 8√5/r a² = 64*5/r² = 320/r² Pythagoras a² = (2r)² - 8² a² = 4r² - 64 a² = a² 320/r² = 4r² - 64 320 = 4r⁴ - 64r² r² = x 4x² - 64x = 320 x² - 16x = 80 x² - 16x + 8² = 80 + 8² (x - 8)² = 144 x - 8 = +- 12 x1 = 8 + 12 = 20 x2 = 8 - 12 = -4 (rejected) x = 20 r² = x r² = 20 A(semicircle) = 1/2 * r² π = 1/2 * 20 π = 10 π

[@PreMath]

Excellent! Thanks for sharing ❤️

[@marcelowanderleycorreia8876]

Very good!!!

[@PreMath]

Glad you think so! Thanks for the feedback ❤️

[@zdrastvutye]

there are 2 triangles with the same proportions: 10 vdu5:for a=0 to 15:gcola:print a;:next a:vdu0:gcol8 20 print "premath-can you find the area of the yellow semicircle":nu=50 30 dim x(1,2),y(1,2):l1=8:l2=sqr(5):sw=l1^2/(l2+l1)/10:goto 50 40 l3=sqr(abs(4*r*r-l1^2)):dgu1=l2/r:dgu2=l3/l1:dg=dgu1-dgu2:return 50 r=sw:gosub 40 60 dg1=dg:r1=r:r=r+sw:r2=r:gosub 40:if dg1*dg>0 then 60 70 r=(r1+r2)/2:gosub 40:if dg1*dg>0 then r1=r else r2=r 80 if abs(dg)>1E-10 then 70 90 print r;"die gesuchte flaeche=";r*r*pi/2 100 la=l1:lb=l3:lc=2*r:lh=(la^2+lc^2-lb^2)/2/lc:h=sqr(la^2-lh^2) 110 x(0,0)=0:y(0,0)=0:x(0,1)=r:y(0,1)=0:x(0,2)=r:y(0,2)=l2 120 x(1,0)=0:y(1,0)=0:x(1,1)=2*r:y(1,1)=0:x(1,2)=lh:y(1,2)=h 130 masx=1200/2/r:masy=850/h:if masx

[@professorrogeriocesar]

❤❤❤

[@PreMath]

Excellent! Glad to hear that! Thanks for the feedback ❤️🌹

[@quigonkenny]

Draw BC. By Thales' Theorem, as A and B are ends of a diameter and C is on the circumference, ∠BCA = 90°. As ∠AOP = 90° as well, and ∠PAO (∠CAB) is common, triangles ∆AOP and ∆BCA are similar. BC/CA = OP/OA BC/8 = √5/r BC = 8√5/r PA/OP = AB/BC PA/√5 = 2r/(8√5/r) = r²/4√5 PA = r²/4 OP² + AO² = PA² (√5)² + r² = (r²/4)² 5 + r² = r⁴/16 r⁴ - 16r² - 80 = 0 (r²+4)(r²-20) = 0 r² = -4 ❌ | r² = 20 ==> r = √20 = 2√5 Area = πr²/2 = π(20)/2 = 10π sq units

[@PreMath]

Excellent! Thanks for sharing ❤️

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... The triangle AOP is a right triangle. According to Thales theorem the triangle ABC is also a right triangle. Now let's have a look at the interior angles of these triangles: ∠AOP = 90° ∠OAP = α ∠APO = β ∠ACP = 90° ∠BAC = α ∠ABC = β Therefore these triangles are similar. So we can conclude: OP/AO = BC/AC With R being the radius of the semicircle we obtain by applying the Pythagorean theorem: OP²/AO² = BC²/AC² OP²/AO² = (AB² − AC²)/AC² (√5)²/R² = [(2*R)² − 8²]/8² 5/R² = (4*R² − 64)/64 320 = 4*R⁴ − 64*R² 80 = R⁴ − 16*R² R⁴ − 16*R² − 80 = 0 R² = 8 ± √(8² + 80) = 8 ± √(64 + 80) = 8 ± √144 = 8 ± 12 Since R²>0, the only useful solution is R²=8+12=20. Now we are able to calculate the area of the semicircle: A(semicircle) = πR²/2 = π*20/2 = 10π Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing ❤️

[@مطبخالشاطرسراج]

Draw OM to perpendicular to AC, therefore AM = MC = (AM + MC) / 2 = 8 / 2 = 4 "Because M is the midpoint of AC, as MO passes through the centre O and perpendicular to AC" Then focus on triangle AOP and AMO, both are right triangles, and we can use the euclidean theorem, but let MP = n, (OP)² = MP × AP "Euclid", but MP = n, and AP = n + 4, and OP = sqrt5. So: (sqrt5)² = n(n + 4), 5 = n² + 4n, n² + 4n - 5 = 0, (n + 5)(n - 1) = 0, n = -5 "refused", n = 1 "approved", therefore AP = n + 4 = 1 + 4 = 5. Then (AO)² = (AP)² - (PO)² "Pythagorean theorem" = 5² - (sqrt5)² = 20, AO = sqrt20 = 2sqrt5 which is the radius. The semicircle area = pi "≈ 3.14" × r² ÷ 2 ≈ 3.14 × (2sqrt5)² ÷ 2 ≈ 31.4.

[@giuseppemalaguti435]

8=2rcosarctg(√5/r)..r^2=20..Ay=π(20/2)=10π

[@PreMath]

Excellent! Thanks for sharing ❤️

[@wackojacko3962]

@ 1:19 You're a funny guy. You say think outside the box, but stay within the circle! 🙂

[@PreMath]

Thanks for the scrutiny 😀

[@JobBouwman]

Let x = r^2 Using Pythagoras (squared) and Thales (squared), we get: x/(x + 5) = 16/x So x^2 - 16x - 80 = 0 Hence x = 20 = r^2, so the area is 10pi

[@tijanimaths6006]

شكرا لك على هذا العمل الرائع

[@PreMath]

You are very welcome! Thanks for the feedback ❤️

[@alster724]

Got it!

[@PreMath]

Excellent! Keep rocking... Thanks for sharing ❤️

[@kassuskassus6263]

We can also draw a perpendicular to AC from the center O which will intersect the segment AC in the middle. We obtain the same result,

[@PreMath]

Thanks for the feedback ❤️

[@Людмила-ч7к4р]

Через точку B провести касательную к окружности. Продолжим AC до пересечения с касательной и обозначим точкой D. Треугольник ACO подобен треугольнику ADP. AB =2xAO. k=2:1. OP=√5, значит, BD=2√5. Продолжим OP до пересечения с окружностью. Точку пересечения обозначим K. Среди ним точки K и P. OKDP квадрат со стороной 2√5.OK=BD=r, r'2=20. Площадь круга πr'2=π20. Площадь полукруга =π10. Россия.

[@kennethkan3252]

8^2-✓5^2=r^2 39=r^2 (39丌)÷2=(r^2×丌)÷2 19.5×丌

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL : 01) sqrt(5) / R = CB / 8. OA = Radius = R and CB = Y 02) Y^2 + 64 = 4R^2 03) Y = ((8*sqrt(5) / R) 04) ((8*sqrt(5) / R)^2 + 64 = 4R^2 05) ((64 * 5) / R^2) + 64 = 4R^2 06) 320 / R^2 + 64 = 4R^2 07) 320 + 64R^2 = 4R^4 08) Variable Change : R^2 = A 09) 320 + 64A = 4A^2 10) Dividing by 4 : 11) 80 + 16A - A^2 = 0 12) A^2 - 16A - 80 = 0 13) - 80 = + 4 * - 20 14) - 20A + 4A = - 16y 15) A^2 + 4A - 20A - 80 = 0 16) A(A + 4) - 20(A + 4) = 0 17) (A - 20)+(A + 4) = 0 18) A = 20 or A = - 4 19) R^2 = 20 ; R = sqrt(20) ; R = 2*sqrt(5) 20) SA = (Pi * R^2) / 2 22) SA = 20Pi / 2 ; SA = 10Pi Sq Un 23) ANSWER : Semicircle Area equal to 10Pi Square Units. Greetings from the Center for Ancient Mathematical Thinking and Wisdom - Cordoba Caliphate Islamic University.

[@PreMath]

Excellent! You are the best🌹 Thanks for sharing ❤️

[@cyruschang1904]

✓(r^2 + 5) ÷ r = 2r ÷ 8 = r ÷ 4 4✓(r^2 + 5) = r^2 16(r^2 + 5) = r^4 r^2 = x x^2 - 16x - 16(5) = 0 x = (16 + ✓(16(16) + 4(16)(5)))/2 = 8 + 12 = 20 = r^2 area = r^2(π/2) = 10π

[@PreMath]

Excellent! Thanks for sharing ❤️

[@cyruschang1904]

@@PreMath Thank you for the fun quiz 🙂

[@MohamedAitki]

Gold.

[@alexniklas8777]

S=10π

[@PreMath]

Excellent! Thanks for sharing ❤️

[@karimMalim-k6e]

Why Root of 20

[@mohanraut890]

I tried to solve it for first time and somehow my answer was 4π+π√21 💀💀