Tomando como centro el punto medio de OP, giramos 180º el semicírculo interior y la figura se transforma en dos semicírculos concéntricos que delimitan media corona circular, cuya superficie es igual al área sombreada → El área de una corona circular es el producto (π*c²), siendo "c" la semicuerda tangente al círculo interior → En este caso, c=r=PD→ 23π cm² es el área de la zona sombreada Gracias y un saludo.
There's NO NEED OF ANY CALCULATIONS !!! We already know that area of white semicircle is: A = ½πr² = ⅛πd² , where "d" is diameter of white semicircle, and also is a chord "c" of purple semicircle. A = ⅛ π c² = ½ (¼πc²) Also we already know the formula of a circular ring area, respect to the chord. A = ¼πc² Half circular ring is: A = ½ (¼πc²) Exactly the same formula, same area. Purple shaded area is equal to white semicircle area. A = 23π cm² (Solved without any calculations ) We only need to flip the white semicircle, so that, purple shaded area becomes a half circular ring.
Area of the small semicircle= 23π 1/2πr^2=23π r^2=46 So r=√46 Draw big circle Let R is the Radius of the big circle (√46)(√46)=(R+√46)(R-√46 R^2-46=46 R^2=92 So Area of the Purple shaded area=1/2(π)(R^2)-1/2(π}(r^2)=π/2(92-46)=23π cm^2.❤❤❤
My way of solution ▶ the small semicircle has the radius r [OP]= [PC] = r A= 23π cm² 23π= πr²/2 r²= 46 b) by considering the right triangle ΔOPC, we have: [OP]= r [PC]= r [CO]= R ⇒ by applying the Pythagorean theorem for the right triangle ΔOPC we get: R²= r²+r² R²= 2r² R²= 2*46 R²= 92 c) Apurple= πR²/2 - πr²/2 Apurple= (π/2)*(R²-r²) Apurple= (π/2)*(92-46) Apurple= 23π cm²
This is an example of easier than it looks. Area is 23pi units square. I alao want to comment further that it the perpendicular bisector theorem were not applicable, we would not have symmetric semicircles with regard to difference in area. I could be wrong. I hope that this is a more than sufficient takeaway.
Let r be the radius of semicircle P and R the radius of semicircle O. The area of semicircle P will be πr²/2 and the area of semicircle O will be πR²/2, so their difference, the purple area, will be as follows: Aᴘ = πR²/2 - πr²/2 Aᴘ = (R²-r²)π/2 --- [*] Triangle ∆DOP: OP² + DP² = OD² r² + r² = R² R² = 2r² R = √(2r²) = √2r Aᴘ = (R²-r²)π/2 --- [*] Aᴘ = (2r²-r²)π/2 Aᴘ = πr²/2 Thus the purple shaded area is the same as the area of semicircle P, which is given as 23π cm².
...I justified my answer! It's reasonable enough I guess since we're dealing with non- terminating non-repeating decimals. Which is irrational. It seems to have a negative connotation but in this case not so much. 🙂
R = CO = CP·√2 = r√2 from isosceles right △COD, then the big semicircle area is twice the small semicircle area, and their difference equals to the small semicircle area.
When you obtain that R^2=2r^2 there is no need to compute any length using radical. You simply deduct that the area of the large circle (pi*R^2) is the double of the area of the small one (pi*r^2) and it's the same for the semicircle areas. So, the area of the purple region which is the difference of those two, is equal to the small semicircle area which is 23 pi cm^3. That's it !
Premath has oversimplified the problem. The unshaved section does not need to be a semicircle. It could be a full circle as well, possibly other segments as well. Same answer 23pi.
STEP-BY-STEP RESOLUTION PROPOSAL : 01) OP = CP = CD = r 02) (23 * Pi) = (Pi * r^2) / 2 ; 46 * Pi = Pi * r^2 03) 46 = r^2 ; r = sqrt(46) ; r ~ 6,8 cm 04) OA = OC = OD = OB = R 05) OD^2 = OP^2 + PD^2 06) R^2 = 46 + 46 07) R^2 = 92 ; R = sqrt(92) ; R ~ 9,6 cm 08) Area of Big Semicircle = (92 * Pi) / 2 = 46 * Pi 09) Purple Shade Region Area = (46 * Pi) - (23 * Pi) 10) Purple Shade Region Area = 23PI sq cm Therefore, OUR BEST ANSWER : The Area of Purple Shaded Region is 23Pi Square Centimeters.
Wow very beautiful sharing thank you so much for sharing Sir ❤
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Very nice and enjoyable
Excellent Sir
You are very good explain
Thanks PreMath
❤❤❤❤❤
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White semicircle:
A₁ = 23π = ½πr² = ⅛π(2r)² = ⅛πc²
Chord "c' squared :
c² = 23*8 = 184 cm²
Flip white semicircle, and Purple shaded area becomes a half circular ring:
A = ½(¼πc²) =⅛.π.184
A = 23π cm² ( Solved √ )
How do you know the idea of the problem or how do you analyze the problem to solve it?
Tomando como centro el punto medio de OP, giramos 180º el semicírculo interior y la figura se transforma en dos semicírculos concéntricos que delimitan media corona circular, cuya superficie es igual al área sombreada → El área de una corona circular es el producto (π*c²), siendo "c" la semicuerda tangente al círculo interior → En este caso, c=r=PD→ 23π cm² es el área de la zona sombreada
Gracias y un saludo.
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Thanks for sharing ❤️
There's NO NEED OF ANY CALCULATIONS !!!
We already know that area of white semicircle is:
A = ½πr² = ⅛πd² , where "d" is diameter of white semicircle, and also is a chord "c" of purple semicircle.
A = ⅛ π c² = ½ (¼πc²)
Also we already know the formula of a circular ring area, respect to the chord.
A = ¼πc²
Half circular ring is:
A = ½ (¼πc²)
Exactly the same formula, same area.
Purple shaded area is equal to white semicircle area.
A = 23π cm² (Solved without any calculations )
We only need to flip the white semicircle, so that, purple shaded area becomes a half circular ring.
CP=PD=PO=r
r*r*π/2=23π r=√46
AO=OB=R
√46*√46=(R-√46)(R+√46)
46=R^2-46 R^2=92
Purple Shaded Region :
R^2π/2 - 23π = 46π - 23π = 23π(cm^2)
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Area of the small semicircle= 23π
1/2πr^2=23π
r^2=46
So r=√46
Draw big circle
Let R is the Radius of the big circle
(√46)(√46)=(R+√46)(R-√46
R^2-46=46
R^2=92
So Area of the Purple shaded area=1/2(π)(R^2)-1/2(π}(r^2)=π/2(92-46)=23π cm^2.❤❤❤
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A₁ = ½πr² = 23π --> r=√46 cm
R= r√2 = 2√23 cm
A = A₂ - A₁ = ½πR² - 23π
A = 46π - 23π
A = 23π cm² ( Solved √ )
My way of solution ▶
the small semicircle has the radius r
[OP]= [PC] = r
A= 23π cm²
23π= πr²/2
r²= 46
b) by considering the right triangle ΔOPC, we have:
[OP]= r
[PC]= r
[CO]= R
⇒
by applying the Pythagorean theorem for the right triangle ΔOPC we get:
R²= r²+r²
R²= 2r²
R²= 2*46
R²= 92
c) Apurple= πR²/2 - πr²/2
Apurple= (π/2)*(R²-r²)
Apurple= (π/2)*(92-46)
Apurple= 23π cm²
Excellent!
Thanks for sharing ❤️
This is an example of easier than it looks. Area is 23pi units square. I alao want to comment further that it the perpendicular bisector theorem were not applicable, we would not have symmetric semicircles with regard to difference in area. I could be wrong. I hope that this is a more than sufficient takeaway.
Excellent!
Thanks for the feedback ❤️
I took me a few minutes, but you can usually find a helpful right angle triangle in these problems😊
Let r be the radius of semicircle P and R the radius of semicircle O. The area of semicircle P will be πr²/2 and the area of semicircle O will be πR²/2, so their difference, the purple area, will be as follows:
Aᴘ = πR²/2 - πr²/2
Aᴘ = (R²-r²)π/2 --- [*]
Triangle ∆DOP:
OP² + DP² = OD²
r² + r² = R²
R² = 2r²
R = √(2r²) = √2r
Aᴘ = (R²-r²)π/2 --- [*]
Aᴘ = (2r²-r²)π/2
Aᴘ = πr²/2
Thus the purple shaded area is the same as the area of semicircle P, which is given as 23π cm².
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Very easy. More generally, if the area of the little semi circle is A, then the area of the purple region is also A.
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I got R=sqrt(92) pretty quickly as it is sqrt(46) * sqrt(2).(length OC being the diagonal of an imagined square).
46 pi - 23pi
r=√(46π/π)=√46...R=√2r=√92=2√23...Aviolet=πR^2/2-π23=π46-π23=π23
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...I justified my answer! It's reasonable enough I guess since we're dealing with non- terminating non-repeating decimals. Which is irrational. It seems to have a negative connotation but in this case not so much. 🙂
😀
Thanks for the feedback ❤️
R = CO = CP·√2 = r√2 from isosceles right △COD, then the big semicircle area is twice the small semicircle area, and their difference equals to the small semicircle area.
Thanks for the feedback ❤️
When you obtain that R^2=2r^2 there is no need to compute any length using radical. You simply deduct that the area of the large circle (pi*R^2) is the double of the area of the small one (pi*r^2) and it's the same for the semicircle areas. So, the area of the purple region which is the difference of those two, is equal to the small semicircle area which is 23 pi cm^3. That's it !
Thanks for the feedback ❤️
r=√(46); R= √(r^2+r^2)= √(46+46)=√92=2√23.
S=πR^2/2= π(2√23)^2/2=46π
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Thanks but easy
S=23π≈72,29 cm²
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Premath has oversimplified the problem. The unshaved section does not need to be a semicircle. It could be a full circle as well, possibly other segments as well. Same answer 23pi.
S = 23 pi
Simply mental mputation gives 23pi.
Thanks for the feedback ❤️
Very nice sharing sir❤❤
Thanks for the feedback ❤️
STEP-BY-STEP RESOLUTION PROPOSAL :
01) OP = CP = CD = r
02) (23 * Pi) = (Pi * r^2) / 2 ; 46 * Pi = Pi * r^2
03) 46 = r^2 ; r = sqrt(46) ; r ~ 6,8 cm
04) OA = OC = OD = OB = R
05) OD^2 = OP^2 + PD^2
06) R^2 = 46 + 46
07) R^2 = 92 ; R = sqrt(92) ; R ~ 9,6 cm
08) Area of Big Semicircle = (92 * Pi) / 2 = 46 * Pi
09) Purple Shade Region Area = (46 * Pi) - (23 * Pi)
10) Purple Shade Region Area = 23PI sq cm
Therefore,
OUR BEST ANSWER :
The Area of Purple Shaded Region is 23Pi Square Centimeters.
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