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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2037076

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Let the side length of the inscribed yellow square be 1 unit. Area yellow square is; 1² = 1 square unit. Side length of the small regular pentagon is 1 unit. a² = 2-2cos108 a = 1.61803398875 units. a is the side length of the big regular pentagon. b² = 1.61803398875²+1.61803398875²-2*1.61803398875*1.61803398875cos108 b = 2.61803398875 units. c = ½(2.61803398875-1) c = 0.80901699438 units. cos36 = 0.80901699438/e e = 1 unit sin36 = f/1 f = 0.58778525229 units. g = 1.61803398875-1 g = 0....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2035754

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Sir Mike Ambrose is the author of the question. a² = 2²+2² a = 2√(2) units. 2b² = 4 b = √(2) units. c = 3-√(2) units. Let the side of the inscribed regular pentagon be d. Calculating d. e = ⅕(180*3) e = 108° f = ½(180-108) f = 36° (g/sin108) = (d/sin36) g = 1.61803398875d units. (1.61803398875d)² = (3-√(2))²+(d/2)²  d = 1.03050649521 units. Where d is the side length of the regular pentagon.  g = 1.61803398875d g = 1.61803398875*1.03050649521  g = 1.66739453488 units. h = ½(2√(2)-1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2034047

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EG = 3AG Let AG = 1 unit. Therefore; EG = 3 units. The side length of the regular pentagon is; AG+EG = 1+3 = 4 units. AE = 4 units. It implies, Area Blue is; 0.5*2*4sin108 = 4sin108 = 3.80422606518 square units. Calculating Area Green. (DG)² = 3²+4²-2*3*4cos108 DG = 5.6935408899 units. It implies; 5.6935408899² = 2(GH)² (GH)² = 0.5(5.6935408899)² (GH)² = 16.20820393248 square units.  Area Green= ½(GH)² = 0.5*16.20820393248  = 8.10410196624 square units. Therefore; Area Blue ÷ Area G...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2033280

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Sir Mike Ambrose is the author of the question. Let the side of the blue square (length CD) be 1 unit. Area Blue is; 1² = 1 square unit. a = 180-20-40 a = 120° (b/sin40) = (1/sin20) b = 1.87938524157 units. (c/sin120) = (1/sin20) c = 2.53208888624 units. d² = 1²+1.87938524157²-2*1.87938524157cos60 d = 1.62871226577 units. Area Yellow is; (1.62871226577)² = 2.65270364467 square units. (a) Area Yellow ÷ Area Blue to 2 decimal places is; 2.65270364467 ÷ 1 = 2.65270364467 ≈ 2.65 (b) Calcul...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2032546

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Let the side length of the regular pentagon be 2 units. (CF)² = 5-4cos108 CF = 2.49721204096 units. CF is the side length of the blue regular triangle and radius of the sector CGF. Area Blue is; 0.5*2.49721204096*2.49721204096sin60 = 2.70029664413 square units. Area Green is; (⅙*2.49721204096*2.49721204096π)- Area Blue = 3.26519755757-2.70029664413 = 0.56490091344 square units. Blue Area ÷ Green Area is; 2.70029664413÷0.56490091344 = 4.78012440746

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2030918

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Let AO be 1 unit. AB will be 2 units. Therefore, radius r of the ascribed semi circle is; = 3 units. Area semi circle is; ½(3²)π = ½(9π) square units. tana = (1/3) a = atan(⅓)° b = (180-2atan(½))° c = b-90 c = (180-2atan(⅓)°))-90 c = (90-2atan(⅓))° sin(90-2atan(⅓)) = d/1 d = (4/5) units. cos(90-2atan(⅓)) = e/1 e = (3/5) units. f = r-e f = 3-(3/5) f = (12/5) units. g² = 3²-(3/5)² g² = 9-(9/25) g = ⅕(6√(6)) units. h = g-d h = ⅕(6√(6))-(4/5) h = ⅕(6√(6)-4) units. j² = d²+f² j² = (4/5)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2029469

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Sir Mike Ambrose is the author of the question. Let the side of the inscribed square be a. Calculating a. ¾ = a/b b = ⅓(4a) It implies; ½*a*((⅓(4a))+3a) = 78 ⅙(13a²) = 78 ⅙(a²) = 6 a = 6 cm. b = ⅓(4*6)+6+6 b = 20 cm. ¾ = c/20 c = 15 cm. Let the side of the regular hexagon be d. Calculating d. e² = 2d²-2d²cos120 e = √(3)d cm. (¾) = √(3)d/f f = ⅓(4√(3)d) cm. It implies; d+e+f = 15 d+√(3)d+⅓(4√(3)d) = 15 3d+3√(3)d+4√(3)d = 45 d = 45/(3+7√(3)) cm It implies; The side of the regular...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2028126

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Calculating x. Let the two congruent lengths of the plane shape be 1 unit. (a/sin11) = (1/sin22) a = 0.50935834748 units. (b/sin147) = (1/sin22) b = 1.45389601942 units. c = 1+a c = 1.50935834748 units. d = 1+b d = 2.45389601942 units. It implies; e² = 2.45389601942²+1.50935834748²-2*2.45389601942*1.50935834748cos11 e = 1.01402672537 units. Required angle x is; (1.01402672537/sin11) = (1.50935834748/sinx) x =16.5°

Class with Adesuwa

https://onlineedumath.com/update/class-with-adesuwa/2028034

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Please, browse through the link below to review our teaching and learning activities with super smart and clever Adesuwa. She is in year 6. Thank you. https://www.facebook.com/100085657676431/posts/pfbid0J4K6L27WpGtcJYLU672sE95EwoA4dLyCgMJ7TdfSyBEVZyKYErZNeDWbBVyVTivFl/?app=fbl

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2027792

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Sir Mike Ambrose is the author of the question. ⅓(4) = a/9 a = 12 cm. let alpha be b. Calculating b. (2tanb/(1-tan²b)) = (4/3) 2tan²b+3tanb-2= 0 b = atan(½)° ½ = c/9 c = 4.5 cm. d = ½(12-c) d = 3.75 cm. ½ = 3.75/e e = 7.5 cm f² = 8.25²+7.5² f = 11.14955156049 cm. g = ½(90-atan(½))° g = 31.71747441146° h = 90-31.71747441146-atan(8.25/7.5) h = 10.55621459463°  Let r be the radius of the inscribed circle. sin31.71747441146 = r/i i = 1.90211303259r  j²...
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