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)
Let's find the area: . .. ... .... ..... 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
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
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.
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
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.
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
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.❤❤❤
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.
