Good explanation

@PreMath

22 күн бұрын

Thanks for liking❤️

Nice to meet you again,Mr PreMath. Thanks for your lesson!

wow very nice sharing sir✨

@PreMath

21 күн бұрын

Thanks for visiting🌹

It was quite easy 😊😊

@PreMath

22 күн бұрын

Thanks for the feedback ❤️

Excelente!!!

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.😊

@PreMath

22 күн бұрын

Great job! You are very welcome! Thanks for the feedback ❤️

Very nice and enjoyable Thanks PreMath Thanks prof. ❤❤❤❤❤ With my glades

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

@PreMath

21 күн бұрын

Excellent! Thanks for the feedback ❤️

Very nice, many thanks, Sir!

@PreMath

21 күн бұрын

Glad to hear that! You are very welcome! Thanks for the feedback ❤️🙏

Thank you!

@PreMath

21 күн бұрын

You are very welcome! Thanks, my dear friend James ❤️🙏

interesante

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).

@PreMath

21 күн бұрын

Excellent! Thanks for sharing ❤️

@harryloyi5644

3 күн бұрын

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.

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

@PreMath

22 күн бұрын

Excellent! Thanks for sharing ❤️

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.

@PreMath

21 күн бұрын

Excellent!👍 Thanks for sharing ❤️

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

@PreMath

21 күн бұрын

Excellent! Thanks for sharing ❤️

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.

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

@PreMath

21 күн бұрын

Super!👍 Thanks for sharing ❤️

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

@PreMath

22 күн бұрын

Excellent! Thanks for sharing ❤️

@Jorick_73

22 күн бұрын

I'm proud that i did all the same in my mind just whilst looking at the picture in 5 minutes. 51y.o.

@PreMath

22 күн бұрын

@@Jorick_73 Bravo!

@guerrino50

19 күн бұрын

R1=3×2=6 R2=6×2=12 R3=10×2=20 H=10×38=380 Area

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

@PreMath

22 күн бұрын

Excellent! Thanks for sharing ❤️

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

@PreMath

21 күн бұрын

Excellent Thanks for sharing ❤️

@cyruschang1904

21 күн бұрын

@@PreMath Thank you 🙂

İm Azerbaijani but im understand this so well thank you so much❤

@PreMath

21 күн бұрын

Glad to hear that! You are very welcome! Thanks dear❤️

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

@PreMath

21 күн бұрын

Excellent! Thanks for sharing ❤️

380- (pi / 2) x 145 Which is approximately 152.13

R1=10. R2=6. R3=3 S=152.35

@PreMath

21 күн бұрын

Excellent! Thanks for sharing ❤️

S=5(152-29π)/2≈152,14

@PreMath

22 күн бұрын

Excellent! Thanks for sharing ❤️

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

@PreMath

21 күн бұрын

Excellent! Thanks for sharing ❤️

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.

@PreMath

22 күн бұрын

Excellent! Thanks for sharing ❤️

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. 🙂

@PreMath

22 күн бұрын

👍😀 Thanks for the feedback ❤️

But we were not told the main figure was a rectangle.

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

@PreMath

21 күн бұрын

Excellent! You are the best! 😀 Thanks for sharing ❤️

152.24

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*

@PreMath

21 күн бұрын

Excellent! Thanks for sharing ❤️

152.20 Sq. Units

@PreMath

21 күн бұрын

Excellent! Thanks for sharing ❤️

After finding it myself before the video, I found out the decimal actually went 2345 which was funny

380-72.5π

@PreMath

22 күн бұрын

Excellent! Thanks for sharing ❤️

152.35

@PreMath

21 күн бұрын

Excellent! Thanks for sharing ❤️

Sorry. Switched off. "SemI". Chalk scraping across a blackboard.

I did this in grade six.

@PreMath

21 күн бұрын

Excellent!👍 Glad to hear that! Thanks for sharing ❤️

@jackreacher.

20 күн бұрын

@@PreMath Math Teachers rule.

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!!

@PreMath

22 күн бұрын

Super job! Thanks for sharing ❤️

@LuisdeBritoCamacho

22 күн бұрын

@@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.

@PreMath

21 күн бұрын

@@LuisdeBritoCamacho Thank you so much, my dear friend. Take care🌹

NO!

152.35 Sq. Units

@PreMath

22 күн бұрын

Excellent! Thanks for sharing ❤️

152.35

@PreMath

21 күн бұрын

Thanks for sharing ❤️