Can you calculate area of the Pink shaded semicircle? | (Olympiad) |

Learn how to find the area of the Pink shaded semicircle. Important Geometry and Algebra skills are also explained: Pythagorean theorem; Circle area formula; perpendicular bisector theorem; circle theorem. Step-by-step tutorial by PreMath.com.
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Can you calculate area of the Pink shaded semicircle? | (Olympiad) | #math #maths | #geometry
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[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! Thanks for the feedback ❤️🙏

[@AzouzNacir]

Let E and F be the points of intersection of the line PQ with the complete circle. We have PE*PF=PC*PD. Since QE²=QA*QB=1*2=2, then QE=√2. From this we find (√2-r)*(√2+r)=r². Therefore, r=1. From this, the shaded area is equal to π*1²/2=π/2.

[@phungpham1725]

Thank you! I solved it almost the same as you!😅😅😅

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@marioalb9726]

R= ½(2+1) = 1,5 cm Pytagorean theorem, twice: R² - r² = r² + (R-1)² 2r² = R² - (R-1)² = 1,5² - 0,5² = 2 r² = 1 A = ½πr² = π/2 cm² ( Solved √ )

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@allanflippin2453]

Interesting problem and I have to admit I had no clue how to solve it before I watched the video. I decided to try solving this in a general case. I let a=1 and b=2. R = (a+b)/2 and OQ = (b-a)/2. Now setting up the two right triangles sharing dimension x (OP). ((a+b)/2)^2 - r^2 = x^2 and x^2 = ((b-a)/2)^2 + r^2. With x^2 in common, set the two equations equal: ((a+b)/2)^2 - r^2 = ((b-a)/2)^2 + r^2. Multiply everything by 4 to remove the fractions: (a+b)^2 - 4r^2 = (b-a)^2 + 4r^2. Move r to right right side and a, b to the left side: (a+b)^2 - (b-a)^2 = 8r^2. Multiply out the two squared equations and cancel out redundant terms. 4ab = 8r^2. Isolate r^2 = ab/2. The semi-circle area is pi * r^2 or pi * ab/4. With the original numbers this comes out to pi * 2 * 1 / 4 or pi/2.

[@soli9mana-soli4953]

Once known that radius =3/2 and OQ=1/2, set OP=x 1) tangent secant theorem from O (X+r)*(x-r)=(1/2)² r² - x² +1/4=0 2) intersecting chords th. (3/2+x)*(3/2-x)=r*r r²+x² - 9/4=0 Summing 1)+2) 2r - 8/4=0 r=1 Area=π/2

[@geraldgiannotti8364]

Very Nice Solution!

[@alanthayer8797]

As usual THANKS for different unique solutions Daily sir!

[@michaelkouzmin281]

Just one more solution: 1. Let us draw 2 auxiliary lines: vertical (perpendicular to AB) pathing through Q, It will intersect with large circle at point F and let us conect O and F. 2. Let y= PF, r= CP, then QF = r+y; 3. AB = 1+2 =3; R=3/2; 4. OQ = R-1 = 3/2 -1 = 1/2; 5. Let us consider right tringle OQF : R^2 = OQ^2+ QF^2 => (3/2)^2 = (1/2)^2 + (r+y)^2 (1) 6. Let us apply intersecting chords theorem (chords CD & FQ... and down to intersection.): (r+y+r)*y = r^2 => (2ry+y^2) = r^2 (2) 7. We have got a system of equations : {(1)and(2)} let us expand (1): 9/4 = 1/4 + r^2 + (2ry + y^2) (3) Let us put (2) into (3) 8/4 = r^2+r^2; 2r^2 = 8/4 r^2 = 1; 8. A(pink) = pi*r^2/2 = pi*1/2 = pi/2 sq units.

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@santiagoarosam430]

La alineación PQ corta a la circunferencia por arriba en E y por abajo en F ; CP=PD=PQ=r ; OA=OB=OF=(1+2)/2=3/2 ; OQ=(3/2)-1=1/2---> QF=√(3/2)²-(1/2)²=√2 ---> Potencia de P respecto a la circunferencia =r²=(√2 +r)*(√2 -r)---> r=1---> Área del semicírculo rosa =π/2. Gracias y un saludo cordial.

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@marioalb9726]

Intersecting chords theorem, over point Q: h² = 2*1. ---> h =√2 cm Intersecting chords theorem, over point P: r² = (h+r)(h-r) r² = h² - 2hr - r² = (√2)² - 2√2r - r² 2r² + 2√2r -2 = 0 r² + √2r - 1 = 0 --> r= 1 cm Area of pink semicircle: A = ½πr² = π/2 cm² ( Solved √)

[@PreMath]

Thanks for the feedback ❤️🙏

[@cyruschang1904]

(1.5)^2 - r^2 - r^2 = (0.5)^2 r^2 = 1 Pink area = π/2

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@cyruschang1904]

@ Thank you 😊

[@SkinnerRobot]

Brilliant!

[@cyruschang1904]

@@SkinnerRobot 🙏

[@marcgriselhubert3915]

Very good. However I propose an analytic solution (which is less elegant, I know): We use an orthonormal center O and first axis (OB). The radius of the big semi circle is 3/2 and its equation is x^2 + y^2 = 9/4. Point P has same abscissa than point Q: -1/2, so P(-1/2; p) with p unknown positive real which is the radius of the small semi circle. The equation of this small semi circle is (x+(1/2))^2 + (y-p)^2 = p^2 or x^2 + y^2 +x -2.p.y +1/4 = 0. At the intersection of the two semi circles we have: 9/4 +x -2.p.y +1/4 = 0 or x = 2.p.y -5/2. We replace x by this value in the equation of the big semi circle to obtain the ordinates of the points C and D: (2.p.y -5/2)^2 + y^2 = 9/4 or (4.(p^2)+1).(y^2) -10.p.y +4 = 0. The half sum of these ordinates is (5.p) / (4.(p^2)+1) and is also is p which is the ordinate of P as P is the middle of [C, D], so we have 5 / 4.((p^2)+1) = 1 which gives p = 1. So the radius of the small circle is 1 and the area we are looking for is Pi/2.

[@PreMath]

Excellent! Thanks for the feedback ❤️🙏

[@NANZH-v6j]

Another solution ，is to use auxiliary lines to solution equal angle, 😊 cd and ab intersect at point h, connect pa, connect pq and po, then ， angle qpo=angle cha---- 1， angle qpo=arctan1/2r---- 2， angle cha=angle paq-angle apc=arctanr - (π/2-arctan1/2r-arctan1/r)----3， Substituting equations 2 and 3 into equation 1, so，arctanr+arctan1/r=π/2, so，r=1, area=π/2

[@alexundre8745]

Bom dia Mestre Respeitosamente desejo-lhe um domingo abençoado Grato pelas aulas.

[@PreMath]

You are very welcome! Thanks for the feedback ❤️🙏

[@johnvriezen4696]

You can clearly see the pink area appears to be a half of a pie. That's why the answer is 1/2π.

[@PreMath]

Thanks for the feedback ❤️🙏

[@waheisel]

I think we need the converse of "if a radius is perpendicular to a chord, then it bisects the chord." In other words we need the theorem that states "if a radius bisects a chord then it is perpendicular to it."

[@JoanRosSendra]

R²+(1/2)²=PO² (3/2)²-R²=PO² Igualando ambas ecuaciones... R²+(1/2)²=(3/2)²-R² Resolviendo: R=1 Área rosa=π/2 u² Saludos

[@AmirgabYT2185]

S=π/2≈1,571≈1,57

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@Christopher-e7o]

x,2z+5=8

[@texitaliano64]

Il raggio della semicirconferenza di diametro AB vale (1+2) quindi il raggio R R=3/2 la distanza QO vale R-1 QO=(3/2)-1 detto r il raggio della semicirconferenza piccola definisco h=OP con pitagora abbiamo che h=sqrt(R^2-r^2) h=sqrt((3/2)^2-r^2)) QO=sqrt(h^2-r^2) QO=QO sqrt(h^2-r^2)=(3/2)-1 sqrt((sqrt((3/2)^2-r^2)))^2-r^2)=(3/2)-1 sqrt((3/2)^2-r^2-r^2)=(3/2)-1 sqrt((9/4)-2*r^2)=(3/2)-1 (9/4)-2*r^2=(3/2)^2+1-2*(3/2)*1 -2*r^2=-2 -r^2=-1 r=sqrt(1) r=1 Ora calcolo la superficie del semicerchio S=pi*r^2/2 S=pi*1^2/2 S=pi/2

[@baljitsingh77]

Why opd is right angle

[@LuisdeBritoCamacho]

MY RESOLUTION PROPOSAL : 01) PO^2 = (3 / 2)^2 - r^2 02) PO^2 = r^2 + (1 / 2)^2 03) (9 / 4) - r^2 = r^2 + (1 / 4) 04) 2r^2 = (9 / 4) - (1 / 4) 05) 2r^2 = 8 / 4 06) 2r^2 = 2 07) r^2 = 1 08) r = 1 09) Pink Semicircle Area (PSA) = (r^2 * Pi) / 2 10) PSA = (Pi / 2) sq un 11) PSA ~ 1,571 sq un MY BEST ANSWER : The Pink Semicircle Area is equal to (Pi/2) Square Units, or approx. equal to 1,571 Square Units.

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... Let R be the radius of the white semicircle and let r be the radius of the pink semicircle. Since Q is the point of tangency, we know that the triangle OPQ is a right triangle. Therefore we can apply the Pythagorean theorem: OQ² + PQ² = OP² OQ² + r² = OP² The triangle OCD is an isosceles triangle (OC=OD=R). Since P is the midpoint of CD, the triangles OPC and OPD are congruent right triangles. So we can apply the Pythagorean theorem again: OC² = OP² + PC² R² = OP² + r² ⇒ OP² = R² − r² By combining these two equations we obtain: OQ² + r² = OP² = R² − r² 2r² = R² − OQ² r² = (R² − OQ²)/2 Now we are able to calculate the area of the pink semicircle: R = AB/2 = (AQ + BQ)/2 = (1 + 2)/2 = 3/2 OQ = OA − AQ = r − AQ = 3/2 − 1 = 1/2 r² = (R² − OQ²)/2 = [(3/2)² − (1/2)²]/2 = (9/4 − 1/4)/2 = (8/4)/2 = 2/2 = 1 ⇒ A = πr²/2 = π*1²/2 = π/2 Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing ❤️🙏

[@louisdsouza6976]

You are too fast , we have are having our .note books.and noting down, nefore we copy you have changed. Please put in methodlogy of teaching, you are wasting youf time and the time of U tube watchers

[@PreMath]

Thanks for the feedback ❤️🙏

[@LuisdeBritoCamacho]

@@louisdsouza6976 🤔