Can you find area of the Purple Semicircle? | (Rectangle) |
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- Опубликовано: 11 сен 2024
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Thanks for the feedback ❤️
R=8
Area of large semicircle would be 32pi.
However, the blue part of it is 15pi, meaning that the two smaller semicircles total 17pi.
As full circles, their areas would total 34pi and their radii would total 8
Knock off the pi for a momemt
R + r = 8
R^2 + r^2 = 34
8 - R = r
R^2 + (8 - R)^2 = 34
R^2 + 64 - 16R + R^2 = 34
2R^2 - 16R + 30 = 0
(16+or-sqrt(256 - 4*2*30))/4 = R
(16+or-sqrt(16))/4 = R
(16+or-4)/4 = R
R = 5 or R = 3
As the purple area's diameter must be >8, R must be >4, so choose 5
Solution: purple area = 12.5pi un^2
Purple area is 39.27 un^2 (rounded)
Excellent!
Thanks for sharing ❤️
Interesting but not too difficult, p+w=8. 1/2(64-w^2-p^2)pi=15pi, p^2+w^2=34, then (p,w)=(5,3) or (3,5), rejected, therefore the answer is 25/2 pi.😊
Let's find the area:
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Let's label the radii for the big, the pink and the white semicircle as b, p and w, respectively. The radius b of the big semicircle is obviously equal to the height of the rectangle: b=8cm. From the sketch we can conclude:
A(blue region) = A(big semicircle) − A(pink semicircle) − A(white semicircle)
(15π)cm² = (π/2)*(b² − p² − w²)
30cm² = b² − p² − w² = (8cm)² − p² − w² = 64cm² − p² − w²
⇒ p² + w² = 34cm²
2*b = 2*p + 2*w
⇒ p + w = b = 8cm
From the combination of these two equations we obtain:
p² + (8cm − p)² = 34cm²
p² + 64cm² − (16cm)*p + p² = 34cm²
2*p² − (16cm)*p + 30cm² = 0
p² − (8cm)*p + 15cm² = 0
(p − 3cm)*(p − 5cm) = 0
So we have two possible solutions:
p = 3cm² ⇒ A(pink semicircle) = πp²/2 = (9π/2)cm²
p = 5cm² ⇒ A(pink semicircle) = πp²/2 = (25π/2)cm²
According to the sketch the pink semicircle should be larger than the white one. That would correspond to the second solution.
Best regards from Germany
I soved like this:
R = 8 (radius of big circle)
a = radius of small circle
b = radius of middle circle
2a + 2b = 16
a + b = 8 => b = a - 8
Area 1/2 big Circle=(π/2)*8^2
Area 1/2 big Circle=(π/2)64
Area 1/2(small+middle Circle)
= (64/2)π - (30/2)π
Area 1/2(small+middle Circle)
= (34/2)π
Area 1/2 small Circle =
(π/2)*a^2
Area 1/2 middle Circle =
(π/2)*b^2
(π/2)*a^2 + (π/2)*b^2 = 34(π/2)
(π/2)(a^2+b^2) = (π/2)*34
a^2 + b^2 = 34
How in First equation:
b = 8 - a
Then:
a^2 + (8-a)^2 = 34
a^2 + 64 - 16a + a^2 = 34
2a^2 - 16a + 30 = 0 (÷2)
a^2 - 8a + 15 = 0
a = (8+-√(64-60))/2*1
a = (8+-2)/2
a1 = (8-2)/2 = 3
a2 = (8+2)/2 = 5
a+b = 8
Then:
a = 3 (radius of small Circle)
b= 5 (radius of middle Circle)
Then
1/2 area of middle Circle =
((5^2)/2)*π = 12,5π cm^2
1/2 area of middle Circle =
= 39,27 cm^2.
Excellent!
Thanks for sharing ❤️
12.5 pi or 39.26691
The width of the rectangle = distance of the largest semi-circle from the point of tangency (circle theorem)
Hence, rectangle length = 2(8) = 16 Hence, the diameter of the largest semi- circle= 16 hence, its radius =8
Hence, it area = pi 8^2/2 = 32 pi
Hence, the blue-shaded region is the area of the largest semi-circle - the area of the other two semi-circles
BS(blue shaded)= largest semi-circle - an area of the two smaller semi-circles
Hence, largest semi-circle - BS = area of the two smaller semi-circles
32 pi - 15 pi= area of the TWO smaller semi-circles
17 pi = the COMBINED area of the Two smaller semi-circles
Let the length of the smallest semi-circle = r and the length of the purple semi-circle = P,
then 2r + 2P= 16
Hence, r + P = 8 (divide both sides by 2)
Hence, P = 8 -r
Hence, its area = pi [(8-r)^2]/2 = pi [64 + r^2 -16r]/2 since it is a semi-circle
and the area of r = [pi r^2]/2 since it is a semi-circle
Hence, pi [ 64 + r^2 -16 r]/2 + [pi r^2]/2 = 17 pi [recall that their combined area = 17 pi ( 32pi -15pi)]
64 + r^2 -16 r + r^2 = 34 multiply both sides by 2 to get rid of 2 in the denominator
2 r^2 -16 r + 30 =0
r^2 - 8r + 15 = 0 divide both sides by 2
r -3)(r-5) =0
r=3 and r= 5
Hence, the radius of the smallest SEMI circle = 3, and the radius of the purple SEMI circle =5
Hence, the area of the purple semi-circle = pi 5^2/2
= 25 pi/2
= 12.5 pi Answer
S=25π/2=12,5π≈39,29
purple + white = (8 x 8)π/2 - 15π = 17π
purple radius = x, white radius = y
(x^2 + y^2)/2 = 17
x^2 + y^2 = 34
x + y = 8 => x = 8 - y
2y^2 - 16y + 30 = 0
y^2 - 8y + 15 = 0
(y - 3)(y - 5) = 0
purple area = (25π/2) cm^2
Let a is Radius of the big semicircle; b is a Radius of purpled semicircle and C is is a small semicircle
So 2b+2c=2a
b+c=a
Radius of big semicircle a=8
So b+C=8 (1)
Area of the purples semicircle add area of the small semicircle=1/2(π)(8^2)-15π=17π
1/2(π)(b^2)+1/2(π)(c^2=17π
so b^2+c^2=34 (2)
(1): C=8-b
(2): b^2+(8-b^2)=34
so b=5 ; b=3 rejected
So Purple semicircle area=1/2(π)(5^2)=25π/2cm^2=39.27 cm^2.❤❤❤
Let O be the center of the largest (cyan) semicircle, P be the center of the middle (purple) semicircle, and Q be the center of the smallest (white) semicircle. Let R be the radius of semicircle P and r be the radius of semicircle Q.
As semicircle O is fully inscribed in a rectangle of height 8, then its radius equals 8, and therefore the width is twice the radius, at 16. As that width is also equal to the diameters of semicircles P and Q:
2R + 2r = 16
R + r = 8
r = 8 - R
Semicircle O:
Aᴏ = π(8)²/2 = 64π/2 = 32π
As the cyan shaded area is equal to 15π but the entire area of semicircle O is equal to 32π, then the sums of the areas of semicircles P and Q are 32π-15π = 17π.
πr²/2 + πR²/2 = 17π
r²/2 + R²/2 = 17
r² + R² = 2(17) = 34
(8-R)² + R² = 34
64 - 16R + R² + R² = 34
2R² - 16R + 30 = 0
R² - 8R + 15 = 0
(R-5)(R-3) = 0
R = 5 | R = 3 ❌ --- 2R > 8
Semicircle P:
Aᴘ = πR²/2 = π(5)²/2 = 25π/2 sq units
จากที่คุณบอกมาว่า 8=r+R จะได้ r=8-R แต่จริงๆแล้วมันก็ได้ R=8-r และ R>r เสมอ
Area of the big semi circle: (Pi/2).8^2 = 32.Pi
Area of the white semi circle + area of the pink semi circle:
32.Pi - 15.Pi = 17.Pi
Let's name R the radius of the pink semi circle, then 8 - R is the radius of the white semi circle, and we have:
17.Pi = (Pi/2).(R^2) + ((Pi/2).((8 - R)^2), or
R^2 + (R^2 -16.R + 64) = 34, or
R^2 -8.R + 15 = 0. Deltaprime = 1
Then R = 4 + 1 = 5 or R = 4 - 1 = 3
Assuming that the pink semi circle is bigger than the white one, then R = 5 (and the radius of the white semi circle is 3)
Finally the area of the pink semi circle is (Pi/2).(5^2 or
(25/2).Pi
Before watching my educated guess: 64pi ist the whole circle, thus 32pi is the area of the large semicircle. The diameter is 16. By guessing the the radii of the smaller circles could be 3 and 5 (adding nicely up to 16), the areas of the semicircles would be 9/2pi and 25/2pi. Both add up to 17pi (and 15+17=32). Check. The area of the pink semicircle is 12.5pi.
I feel a bit nerdy cz I did it in under a minute in my head
My way of solution ▶
the radius of the blue semicircle is equal to 8 cm
the radius of the white semicircle: a
the radius of the purple semicircle: b
⇒
Ablue= 15π cm²
⇒
15π = π*8²/2 - π*a²/2 - π*b²/2
15π= 32π - π*a²/2 - π*b²/2
15π= π/2(64 - a² - b²)
15*2= 64 - a² - b²
30= 64 - a² -b²
34= a²+b²
while the radius of the large circle is 8 length units, its diameter is 16 and this is equal to the large side of the rectangle:
2a+2b= 16
a+b= 8 cm
⇒
b= 8-a
if we put this in the equation above, we get:
a²+b²= 34
a²+(8-a)²= 34
a²+64-16a+a²= 34
2a²-16a+30=0
a²-8a+15=0
Δ= 64-4*1*15
Δ= 4
a₁= (8+2)/2
a₁= 5
b₁= 8-a₁
b₁= 3
a₂= (8-2)/2
a₂= 3
b₂= 8-a₂
b₂= 5
we can see that b > a
⇒
a= 3 cm
b= 5 cm
The area of the purple semicircle, Apurple:
Apurple= πb²/2
Apurple= π*5²/2
Apurple= 25π/2
Apurple≈ 39,27 cm²
Beginning @ 10:29 is perfect example of why DEI (Diversity Equity and Inclusion) Doesn't work for fabricated social disparities. Much like explaining to a child that a stork brought him/her in reference to childbirth. 🙂
The whole thing is painfully slow. When dividing both sides of an equation by 2, you really don't need to rewrite the whole thing out again with a 2 underneath each term (including 0/2) and then show how to divide each term by 2. Also, when you get to the stage at 9:37 where we see that R = 3 or R = 5, it's clear that these two results are the sizes of the 2 small circles.
The Dr.'s wise words should be a priority issue for every government and person in the world.
Thanks for the feedback ❤️
8*8*π*1/2=32π 32π-15π=17π
radius of purple semicircle : x radius of white semicircle : y
2x+2y=16 2(x+y)=16 x+y=8
x*x*π*1/2=x²π/2 y*y*π*1/2=y²π/2 x²π/2+y²π/2=17π
x²+(8-x)²=34 2x²-16x+30=0 x²-8x+15=0 (x-3)(x-5)=0 x>y , x=5
Purple semicircle area = 5*5*π*1/2 = 25π/2 = 12.5πcm²
If drawing is not to scale, how do you get R>r?
That was stated near the beginning as part of the initial information.
π8^2/2-πr^2/2-πR^2/2=15π (2r+2R=16...r+R=8...r=8-R)...32π-15π=(π/2)(R^2+(8-R)^2)=(π/2)(2R^2-16R+64)..34=2R^2-16R+64..R^2-8R+15=0...R=4+1=5...R=4-1=3..quindi R=5,r=3...Apurple=25π/2
STEP-BY-STEP RESOLUTION PROPOSAL :
01) Big (Blue) Semicircle Radius = 8 cm
02) Medium (Purple) Semicircle Radius = Y cm
03) Small (White) Semicircle Radius = X cm
04) 2X + 2Y = 16 cm ; X + Y = 8 cm
05) Rectangle Area = 8 * 16 = 128 sq cm
06) 64Pi - (X^2Pi + Y^2Pi) = 30Pi
07) X^2Pi + Y^2Pi = 64Pi - 30Pi
08) X^2Pi + Y^2Pi = 34Pi
09) Pi(X^2 + Y^2) = 34Pi
10) X^2 + Y^2 = 34 and X + Y = 8
11) X^2 + (8 - X)^2 = 34 ; X^2 + (64 - 16X + X^2) = 34 ; 2X^2 - 16X + 30 = 0 ; X^2 - 8X + 15 = 0
12) As : 5 + 3 = 8 and 5 * 3 = 15 we can factor the Quadratic Equation.
13) X^2 - 3X - 5X + 15 = 0 ; X(X - 3) - 5(X - 3) = 0 ; (X - 3) * (X - 5) = 0 ; X = 3 or X = 5
14) So : X = 3 and Y = 5 or X = 5 and Y = 3
15) As proposed by the Diagram : X < Y
16) White Semicircle Area = 9Pi / 2
17) Purple Semicircle Area = 25Pi / 2
18) Cheking Solutions : 64Pi - 30Pi = 9Pi + 25Pi ; 34Pi = 34Pi. OK!
Therefore,
OUR BEST ANSWER :
The Purple Semicircle Area is equal to 25Pi/2 Square Centimeters. Approximately 39,3 Square Centimeters.
P. S . - Best Regards from The Islamic Institute of Mathematical Sciences. Cordoba Caliphate.