Can you find area of the Yellow shaded region? | (Rectangle) |
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- เผยแพร่เมื่อ 3 ก.ค. 2024
- Learn how to find the area of the Yellow shaded region. Important Geometry and Algebra skills are also explained: area of the circle formula; area of the rectangle formula. Step-by-step tutorial by PreMath.com
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Good explanation
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It was quite easy 😊😊
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Thanks for always original and exciting puzzles,🎉. a+b+c=19, a=c-7, b=c-4, so 3c-11=19, c=10, a=3, b=6, therefore the area is 380-1/2(100+36+9)pi=380-145/2 pi.😊
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Very nice and enjoyable
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With my glades
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The math is easier if you define r from the *smallest* circle. That way, you do not have to be concerned with negative values.
2(r+(r+3)+(r+7) = 38
2(3r+10)=38
6r+20=38
6r=18
r=3
r1=3, r2=6, r3=10
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Nice to meet you again,Mr PreMath. Thanks for your lesson!
Very nice, many thanks, Sir!
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Radii of R, R-4, and R-7.
2R + 2(R-4) + 2(R-7) = 38
6R - 22 = 38, so R=10
R=10=height of rectangle.
Total rectangle area = 38*10=380
Semicircle areas are ((100pi)+(36pi)+9pi))/2 = 72.5pi.
Yellow shaded area = 380-72.5pi un^2.
380-227.77=152.23un^2.
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FO=OB=r EP=PF=r-4 AQ=QE=r-7
2(r-7)+2(r-4)+2r=38 2r-14+2r-8+2r=38 6r=60 r=10
Yellow shaded area = 38*10 - (3r*3*π*1/2 + 6*6*π*1/2 + 10*10*π*1/2) = 380 - 72.5π
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Let the radius of semicircle O be r. Let T be the point of tangency between the circumference of semicircle O and CD. As CD is tangent to semicircle O at T, ∠CTO = 90°, so as BC and OT are thus parallel, then BC = OT = r.
Carrying this over the other two semicircles, we can see that the radius of semicircle P is r-4 and the radius of semicircle Q is r-7. Looking at AB and CD, which as opposite sides of the rectangle will be equal in length, the length of AB is 2(r-7)+2(r-4)+2r and the length of CD is 38.
2(r-7) + 2(r-4) + 2r = 38
6r - 14 - 8 = 38
6r = 60
r = 60/6 = 10
The yellow shaded area eill be equal to the area of the rectangle ABCD minus the areas of the three semicircles.
Yellow shaded area:
Aₛ = hw - πr₁²/2 - πr₂²/2 - πr₃²/2
Aₛ = BC(CD) - (π/2)(r₁²+r₂²+r₃²)
Aₛ = 10(38) - (π/2)((r-7)²+(r-4)²+r²)
Aₛ = 380 - (π/2)(3²+6²+10²)
Aₛ = 380 - (π/2)(9+36+100)
Aₛ = 380 - 145π/2 ≈ 152.23 sq units
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Ok, let the circle radii, from left to right, be A, B, and C. So, we have
A + B + C = 19
-A + C = 7
- B + C = 4
I hope it's obvious how we got those. The solution is A = 3, B = 6, C = 10. So the full yellow rectangle would have area 38*10 = 380. We need to subtract pi/2 times 3^2 = 9 plus 6^2 = 36 plus 10^2 = 100, so 145. So the answer is 380 - 72.5*pi = 152.23.
2(a - 7) + 2(a - 4) + 2a = 38
6a = 60
a = 10
area sought = 38 x 10 - 9π/2 - 36π/2 - 100π/2 = 380 - 145π/2
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@@PreMath Thank you 🙂
S=5(152-29π)/2≈152,14
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380- (pi / 2) x 145
Which is approximately 152.13
İm Azerbaijani but im understand this so well thank you so much❤
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Label the radius of the semicircle with center Q (small) as r.
The gaps between the top of the semicircles and the ceiling are 7 (small) and 4 (medium).
So, the semicircle with center O (large) has a radius of r + 7.
Using the radius of the large semicircle, we can find the radius of the semicircle with center P (medium).
Medium semicircle radius = Large semicircle radius - 4
= (r + 7) - 4
= r + 3
The ceiling is the length of rectangle ABCD. So, by the Parallelogram Opposite Sides Theorem, AB = 38.
The bottom side of ABCD also is formed by the diameters of the semicircles. So:
2r + 2(r + 3) + 2(r + 7) = 38
2r + (2r + 6) + (2r + 14) = 38
6r + 20 = 38
6r = 18
r = 3
So, the radii of the semicircles are as follows:
Small: 3
Medium: 6
Large: 10
But the radius of the large semicircle is also the width of the rectangle. Find the area of the yellow region.
Yellow region area = Rectangle ABCD Area - Total Area of Three Semicircles
A = lw
= 38 * 10
= 380
A = (πr²)/2
= (π * 3²)/2
= (9π)/2
A = (π * 6²)/2
= (36π)/2
= 18π
A = (π * 10²)/2
= (100π)/2
= 50π
Total Area = (9π)/2 + 18π + 50π
= 4.5π + 18π + 50π
= 72.5π
= (145π)/2
Yellow region area = 380 - (145π)/2
So, the area of the yellow region is 380 - (145π)/2 square units, a. w. a. (760 - 145π)/2 square units (exact), or about 152.23 square units (approximation).
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wrong.
if r1=3, r2=6, r3=10,
total area of the 3 semicircles will be
= π/2 * (9+36+100)
or, 22/(7*2) * (9+36+100)
dividing this by 2, we have
= 11/7 * (9+36+100)
solving this will give you 1295/7
= 185
now, Area of the shaded region
= Area of Rect - total area of 3 semicircles
= 380 - 185
= 195 sq units
if you are getting your answer in decimals, you've got it wrong.
that's the idea behind this question.... I saw this question in an MCQ earlier somewhere... none of the options were in decimal.
interesante
380 - 72.5 pi or 152.2345
Let the width of the rectangle = n
then the diameter of the largest semi-circle = 2n
the diameter of the middle semi-circle = n-4 + n-4 = 2n-8
the diameter of the smallest semi-circle = n-7 + n-7 = 2n -14
Hence, 2n + 2n-8 + 2n-14 =38
6n - 22 =38
6n = 60
n=10
Hence, the radius of the largest semi-circle = 10. Hence , its area = 10^2pi/2 = 50pi
Hence, the radius of the middle semi-circle = 6 (10-4), Hence, its area = 36pi/2 = 18pi
Hence, the radius of the smallest semi-circle =3 (10-7). Hence, its area = 9 pi/2 = 4.5 pi
Hence, the TOTAL area of all the semi-circles is 72.5 pi (50 + 18 + 4.5)
Since n= 10, then the width of the rectangle = 10
Hence, its area = 380
The shaded region is the difference between the two
Hence, area of shaded region = 380 - 72.5 pi ot 152.2345
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R1=10. R2=6. R3=3 S=152.35
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My way of solution ▶
radius small semicircle: r₁
radius middle semicircle: r₂
radius large semicircle: r₃
⇒
2r₁+2r₂+2r₃= 38
r₁+r₂+r₃= 19
r₃= r₂+4
r₂+4= r₁+7
r₂= r₁+3
⇒
r₁+r₂+r₃= 19
r₁+ r₁+3 + r₂+4= 19
2r₁+ (r₁+3)= 19-7
3r₁= 9
r₁= 3
r₂= 6
r₃= 10
for the rectangular:
a= 38 length units
b= r₃
b= 10
Arectangular= a*b
= 38*10
= 380 square units
Ayellow= 380- πr₁²/2 - πr₂²/2 - πr₃²/2
= 380 - π/2(3²+6²+10²)
= 380- π/2(9+36+100)
Ayellow = 380- 145π/2
Ayellow = 380 - 72,5 π
Ayellow ≈ 152,23 square units
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Let large circle radius =R3
Middle R2,small. R1
R3=R2+4
R2=R1+3
2(R1+R2+R3)= 38
2(R1+R1+3+R1+3+4)=38
3R1+10=19
3R1=19-10=9
R1=9/3=3
Now area of Rectangle
=38x10=380 square unit
Deduction =
Pi /2(3^2+6^2+10^2)
Pi/2(9+36+100)=pi/2x 145=
227.765 square unit
Net yellow area=380-227.765
=152.235 square unit
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I would've been the Appointed Water Commissioner of the Roman Empire building Aqueducts. Now I accept the post Era Titular position. Knowing how to do this problem I think I've earned it. 🙂
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152.20 Sq. Units
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R3=R1+7, R3=R2+4, 2R3=R1+R2+11, 2R3=38-2(R1+R2), 3(R1+R2)=38-11, 3(R1+R2)=27, R1+R2=9, 2R3=R1+R2+11, 2R3=20, R3=10, R1=R3-7=3, R2=R3-4=6, Area of the Yellow shaded region = 380-(100+36+9)pi/2=380-145pi/2=152,234.
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I did this in grade six.
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@@PreMath Math Teachers rule.
152.35
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380-72.5π
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But we were not told the main figure was a rectangle.
r1+7=r2+4=R...2r1+2r2+2R=38..r1+r2+R=19..(2R-11)+R=19..R=10..Ay=38R-(π/2)(R^2+(R-7)^2+(R-4)^2)=38R-(π/2)(3R^2-14R+49-8R+16)..Ay=380-(π/2)145
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I'm proud that i did all the same in my mind just whilst looking at the picture in 5 minutes. 51y.o.
@@Jorick_73
Bravo!
R1=3×2=6
R2=6×2=12
R3=10×2=20
H=10×38=380 Area
152.24
After finding it myself before the video, I found out the decimal actually went 2345 which was funny
Sorry. Switched off. "SemI". Chalk scraping across a blackboard.
Let's find the area:
.
..
...
....
.....
From the known length of AB we can conclude:
AB = 38
AQ + QE + EP + PF + FO + OB = 38
R(left) + R(left) + R(middle) + R(middle) + R(right) + R(right) = 38
2*R(left) + 2*R(middle) + 2*R(right) = 38
R(left) + R(middle) + R(right) = 19
[R(right) − 7] + [R(right) − 4] + R(right) = 19
3*R(right) − 11 = 19
3*R(right) = 30
⇒ R(right) = 10
R(middle) = R(right) − 4 = 10 − 4 = 6
R(left) = R(right) − 7 = 10 − 7 = 3
Now we are able to calculate the size of the yellow area:
A(yellow)
= A(rectangle) − A(semicircle,left) − A(semicircle,middle) − A(semicircle,right)
= AB*BC − π*R²(left)/2 − π*R²(middle)/2 − π*R²(right)/2
= AB*R(right) − (π/2)*[R²(left) + R²(middle) + R²(right)]
= 38*10 − (π/2)*(3² + 6² + 10²)
= 380 − (π/2)*(9 + 36 + 100)
= 380 − (145/2)π
≈ 152.23
Best regards from Germany
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da = r1 + 7
da = r2 + 4
da = r3
3*da = r1 + r2 + r3 + 11
3*r3 = r1 + r2 + r3 + 11
2*r3 = r1 + r2 + 11 -> r1 + r2 = 2*r3 - 11
ab = 2*(r1 + r2 + r3)
ab = 2*(r1 + r2 + r3)
38 = 2*(r1 + r2 + r3)
r1 + r2 + r3 = 19
r1 + r2 + r3 = 19
2*r3 - 11 + r3 = 19
3*r3 = 30
r3 = 10
r3 = r2 + 4
r2 = 10 - 4
r2 = 6
r3 = r1 + 7
r1 = 10 - 7
r1 = 3
ret_area = da*38 = r3*38 = 10*38 = 380
yellow_area = 380 - (9*p1 + 36*pi + 100*pi)/2
yellow_area = 380 - (pi + pi*49 + pi*225)/2
yellow_area = 380 - (145*pi)/2
yellow_area = 152,3
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2a + 2b + 2c = 38 (÷2)
a + b + c = 19 ... ¹
a + 7 = b + 4
a - b = 4 - 7
a - b = -3 ... ²
a + 7 = c ... ³
a + b + c = 19
a - b = -3
===========
2a + c = 16 ... ⁴
2a + a + 7 = 16
3a = 9
*a = 3*
*b = 6*
*c = 10*
*A Rectangle = 380 Square Units*
A SCa = π 3²/2 = 9/2 π
A SCb = π 6²/2 = 18π
A SCc = π 10²/2 = 50π
Yellow Shaded Region = 380 - (9/2 π + 18π + 50π)
*YSR = 380 - 72,5π Square Units*
YSR = 152,2345326147 Square Units
*YSR = 152,2345 Square Units*
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It was quite juicy 😋 😊😊
STEP-BY-STEP RESOLUTION PROPOSAL :
01) DC = AB = 38
02) Let Radius of Big Semicircle = R(1) = CB = R
03) Radius of Medium Semicircle = R(2) = (R - 4)
04) Radius of Small Semicircle = R(3) = (R - 7)
05) AE = 2 * (R - 7) = 2R - 14
06) EF = 2 * (R - 4) = 2R - 8
07) FB = 2 * R = 2R
08) AB = 2R - 14 + 2R - 8 + 2R
09) 2R - 14 + 2R - 8 + 2R = 38 ; 6R - 22 = 38 ; 6R = 60 ; R = 60 / 6 ; R = 10
10) Rectangle [ABCD] Area = 38 * 10 = 380 sq un
11) Big Semicircle Area with Radius equal 10 = 50Pi sq un
12) Medium Semicircle Area with Radius equal 6 = 18Pi sq un
13) Small Semicircle Area with Radius equal 3 = 9Pi / 2 sq un
14) Sum of all 3 Semicircles Area = (50Pi + 18Pi + 9Pi/2) = (50 + 18 + 9/2)*Pi = 145Pi / 2 ~ 228 sq un (~ 227,77 sq un)
15) Yellow Are = 380 - 228 ~ 152 sq un
ANSWER : The Yellow Shaded Area approx. equal to 152 Square Units. (Exact Form = (380 - 145Pi/2) Square Units or more precisely 152,2345 Square Units).
Greetings from Cordoba Caliphate!!
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@@PreMath , I am here to learn Maths, wich is the Love of My Life. Thanks for sharing your knowledge and the Importance of Mathematical Thinking in our Lives.
@@LuisdeBritoCamacho
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NO!
152.35 Sq. Units
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152.35
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