Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3183735

Calculating Area Blue. Let a be the side length of the ascribed quarter circle. tan30 = 2/b b = 2√(3) units. c = (a-2) units. d = (2√(3)+½(a)) d = ½(4√(3)+a) units. Calculating a. (a-2)² = (½(4√(3)+a))²+2² a²-4a+4 = ¼(48+8√(3)a+a²)+4 4a²-16a = 48+8√(3)a+a² 3a²-(16+8√(3))a-48 = 0 3a²-29.8564064606a-48 = 0 Therefore; a = 11.3605 units. Again, a is the radius of the ascribed quarter circle. (11.3605)² = e²+f² f = √(129.06096025-e²) e is the side length of the inscribed blue square....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3182303

Calculating Area Yellow, Area of The Inscribed Half Circle. 4r+3r = 4*3 7r = 12 r = ⅐(12) units. r is the radius of the inscribed yellow half circle. Area Yellow Inscribed Half Circle is; ½(πr²) = ½*π(12/7)² = ½*π(144/49) = (72/49)π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3182297

Calculating Area Green. sin30 = a/12 a = 6 units. b²+6² = 12². b = √(144-36) b √(108) b = 6√(3) units. sin60 = c/12 c = 6√(3) units. d² = 12²-(6√(3))² d = √(144-108) d = √(36) d = 6 units. Therefore, green area is; (60π*12²/360)-(30π*12²/360) = 24π-12π = 12π square units. Another Approach is; Area green is; Area trapezoid with parallel lengths 6√(3) units and 6 units, and height (6√(3)+6) units+Area second with radius 12 units and angle 30°-Area triangle with height 12 units and b...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3181225

Calculating Shaded Green Area. sin30 = a/2 a = 1 unit. (a')²+1² = 2² a' = √(4-1) a' = √(3) units. a'' = 2a' a'' = 2√(3) units. a'' is the side length of the inscribed regular triangle. b² = 2²+2²-2*2*2cos30 b² = 8-8cos30 b = 1.03527618041 units. c²+a² = b² c² = 1.03527618041²-1² c = 0.26794919243 units. d = 2c d = 2*0.26794919243 d = 0.53589838486 units. Shaded Green Area is; Area quarter circle with radius 2 units+Area equilateral triangle with side length 2√(3) units-3(area triangle...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3181185

Calculating Length x. x ~ ½(2) (2+7) ~ x Cross Multiply. x² = 9 x = √(9) x = 3 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3181184

Calculating Area Blue Triangle. a² = 2(2)² a = 2√(2) units. a is the diagonal of a square. b = ½(180-30) b = 75° c = b-45 c = 75-45 c = 30° sin30 = d/a ½ = d/(2√(2)) d = √(2) units. d is the base of the blue triangle. cos30 = e/a ½√(3) = e/(2√(2)) 2e = 2√(6) e = √(6) units. e is the height of the blue triangle. Area Blue Triangle is; ½*de = ½*√(2)*√(6) = ½*2√(3) = √(3) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3179895

Calculating r/R. Let let the ascribed square side length be 2 units. Calculating R. a = ½(2) a = 1 unit. b = (2-R) units. Therefore; (2-R)² = R²+1 4-4R+R² = R²+1 4-1 = 4R R = ¾ units. Calculating r. c²+1² = 2² c = √(3) units. d = (2-r) units. e = (2+r) units. Therefore; (2+r)² = 1²+(2-r)² 4+4r+r² = 1+4-4r+r² 8r = 1 r = ⅛ units. Therefore; r/R is; ⅛ ÷ ¾ = ⅙

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3179893

Calculating Shaded Fraction. Let the ascribed square side length be 5 units. a = 1 unit. a is the side length of the big inscribed yellow square. b = ½(5-a) b = ½(5-1) b = 2 units. b is the side length of the bigger inscribed yellow square. c² = 2²+1² c = √(4+1) c = √(5) units. c is the side length of the biggest inscribed yellow square. Therefore, total shaded yellow area is; 1²+2²+√(5)² = 1+4+5 = 10 square units. It implies, shaded yellow area fraction is; Area shaded ÷ Area of the as...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3178822

Calculating the required angle. Let it be x. a = ⅑*180(9-2) a = 20*7 a = 140° a is the single interior angle of the regular nonagon. b = ½(140)-65 b = 5° c = ½(180(5-2)-3(140)) c = ½(540-420) c = ½(120) c = 60° d = 140-c d = 80° e = ½(140)-c e = 70-60 e = 10° Let the side length of the regular nonagon be 2 units. f² = 2(2)²-2(2)²cos140 f = 3.75877048314 units. g = 70-½(180-140)+5 g = 70-20+5 g = 55° f = 140-20-½(360-2(140)) f = 140-20-40 f = 80° g = 180-55-80 g = 45° (h/sin80) = (3...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3178623

Calculating Length x. Let a be the side length of the blue square. b² = 2a² b = √(2)a units. b is the diagonal of the blue square. It implies; x ~ 3 √(2)a ~ a Cross Multiply. ax = 3√(2)a It implies; x = 3√(2) units. x = 4.24264068712 units.

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