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.
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.
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 r^2 -x^2 +1/4=0 2) intersecting chords th. (3/2+x)*(3/2-x)=r*r r^2+x^2-9/4=0 Summing 1)+2) 2r - 8/4=0 r=1 Area=pi/2
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 id Pi/2.
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.
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
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
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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.
Thank you! I solved it almost the same as you!😅😅😅
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You can clearly see the pink area appears to be a half of a pie. That's why the answer is 1/2π.
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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 √ )
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@ 7:11 , ... when all are one and one is all 😊
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.
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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
r^2 -x^2 +1/4=0
2) intersecting chords th.
(3/2+x)*(3/2-x)=r*r
r^2+x^2-9/4=0
Summing 1)+2)
2r - 8/4=0
r=1
Area=pi/2
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 √)
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(1.5)^2 - r^2 - r^2 = (0.5)^2
r^2 = 1
Pink area = π/2
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@ Thank you 😊
S=π/2≈1,571≈1,57
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Bom dia Mestre
Respeitosamente desejo-lhe um domingo abençoado
Grato pelas aulas.
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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 id Pi/2.
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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.
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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.
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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
Thanks for the feedback ❤️🙏
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
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