Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2646580

Sir Mike Ambrose is the author of the question. Let the single side length of the square be 2 units. Therefore; Area square is; 2*2 = 4 square units. Area red is; Area triangle with height (5/3) units and base (5/2) units + Area parallelogram with base (5/2) units and height (1/3) units. = ½((5/3)*(5/2)) + ((5/2)*(1/3)) = ½(25/6) + (5/6) = (25/12)+(5/6) = (35/12) square units. It implies; Area Red ÷ Area Square in its exact simplest form is; (35/12) ÷ 4 = 35/48...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2646329

The area of the flower is; 8(area quarter circle with radius 10 cm - area of triangle with height and base, 10 cm respectively) = 8(¼(100π)-½(100)) = 8(25π - 50) cm² = 200(π-2) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2645043

Let the single side length of the regular hexagon be 1 unit. Therefore; Area Regular Hexagon is; (6*1)/(4tan(180/6)) = ½(3√(3)) square units. Area Blue/Shaded is; 2(area triangle with two equal length ⅓(2√(3)) units and angle 120°, the angle they form). = 2(½(⅓(2√(3)))²sin120) = 2(½(12/9)sin120) = ⅓(2√(3)) square units. Therefore shaded/blue area fraction is; Area Blue ÷ Area Regular Hexagon. = ⅓(2√(3)) ÷ ½(3√(3)) = (⅔)÷½(3) = 4/9

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2645033

Area Yellow is; Area quarter circle with radius 6 cm - Area sector with radius 6 cm and angle 20° - Area triangle with two equal length 6 cm and angle 40°, the angle they form - Area sector with radius 6 cm and angle 40° + Area triangle with two equal length 6 cm and angle 40°, the angle they form. = ¼(36π) - (20*36π÷360) - ½(36sin40) - (40*36π÷360) + ½(36sin40) = 9π - 2π - 4π = 3π cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2645027

Calculating R, radius of the quarter circle. ¼πR²=64π² R = 16√(π) cm Calculating x, radius of the green inscribed semi circle. (2x)²+x²=(16√(π))² Therefore; 5x² = 256π x = (16√(5π))/5 cm It implies; Area green inscribed semi circle is; ½π((16√(5π))/5)² = (256π²)/10 cm² Calculating y, radius of the blue inscribed semi circle. (2y)²+(y+x)²=(16√(π))² Therefore; 4y²+y²+(32√(5π)y)/5+(256π)/5=256π 25y²+32√(5π)y-1024π= 0 y = √(25856π/625)- √(256*5π/625) y = 8.863752...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2645013

Let the single side length of the hexagon be 2 units. Therefore; AG =√(8-8cos150) units. Calculating AH. AH = √((8-8cos150)+4-4√(8-8cos150)cos105) AH = 4.78834034195 units Therefore; MH = ½AH MH = 2.39417017098 units. Calculating MG. MG = √(2.39417017098²+2²-4(2.39417017098)cos51.2060231127) MG = 1.93185165257 units. Calculating angle MGH Let it be x. x = asin((2.39417017098sin51.2060231127)/1.93185165257) x = 75° Therefore; Angle FGM (angle alpha)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2643746

a² = 2(15)² a = √(2(15)²) a = 15√(2) units. b²+21² = a² b² = 450-441 b² = 9 b = 3 units. tanc = 15/15 c = 45° tand = 3/21 d = atan(1/7)° e = 180-c-d e = 135-atan(1/7) e = (45+atan(7))° Therefore, Area Blue Inscribed Triangle is; 0.5*3*15sin(45+atan(7)) = ½*3*15*⅘ = 3*3*2 = 18 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2643538

cos30 = a/10 a = 5√(3) units. sin30 = b/10 b = 5 units. cos40 = c/10 c = 7.6604444312 units. sin40 = d/10 d = 6.4278760969 units. e = 180-30-40 e = 110° (10/sin110) = (f/sin30) f = 5.3208888624 units. g = 90-30 g = 60° h = 90-40 h = 50° j = 180-50-50 j = 80° k = 180-60-80 k = 40° (10/sin110) = (l/sin40) l = 6.8404028665 units. m = a-l m = 5√(3)-6.8404028665 m = 1.8198511713 units. n = c-f n = 7.6604444312-5.3208888624 n = 2.3395555688 units....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2643485

Sir Mike Ambrose is the author of the question. Let the single side length of the inscribed regular hexagon be 2 units. Therefore; Area rectangle ABCD is: 2√(3) * 4 = 8√(3) square units. Area Blue will be; Area rectangle with length 4 units and width 2√(3) units - Area trapezium with two parallel length sum (1+4) units and height √(3) units - Area rectangle with length √(3) units and width 1 unit - Area triangle with height √(3)sin(atan√(3)) units and base √(3)cos(atan√(3)) u...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2643477

Let the radius of the inscribed red semi circle be r. Therefore; (8-r)²+(9-r)²=r² 64-16r+r²+81-18r+r²=r² r²-34r+145=0 (r-17)²=-145+(17)² r = 17±√(144) r = 17±12 It implies ; r ≠ 29 cm. r = 5 cm. Therefore area red inscribed semi circle is; ½π(5)² = ½(25π) cm² Area rectangle (green) is; 9 * 8 = 72 cm² It implies; Area red inscribed semi circle ÷ Area green rectangle is; ½(25π)÷72 = 25π/144 = 275/504

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