Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3174279

Calculating Red Area. A nice and careful examination of the irregular pentagon, makes a regular quadrilateral (square) with side length 1 cm. It implies; Red area is; 1² = 1*1 = 1 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3173559

Calculating Yellow Area. Let a be the side length of the square. b = ½(a) units. c² = a²+(½(a))² c² = ¼(5a²) c = ½√(5)a units. c is the radius of the half circle. It implies; a ~ ½√(5)a d ~ a Cross Multiply. ½√(5)d = a d = 2a/√(5) d = ⅕(2√(5)a) units. e = ½(d) e = ⅕√(5)a units. f = c-e f = ½√(5)a-⅕√(5)a f = ⅒(3√(5))a units. Calculating a. ½*a*½(a)+½*⅒(3√(5))a*⅕(2√(5)a) = 15 ¼(a²)+⅒(3a²) = 15 5a²+6a² = 300 11a² = 300 a = 10√(3)/√(11) a = 10√(33)/11 units. g = 2f g = 2*⅒(3√(5))a g = ⅕(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3173553

Calculating Yellow Area. a = ½(24) a = 12 units. b² = 12²+c² c = √(b²-144) units. Where; b is the ascribed half circle's radius. c is the inscribed half circle's radius. Area yellow is; ½(πb²)-½(πc²) = ½(πb²)-½(π√(b²-144)²) = ½(πb²)-½(πb²)+½(144π) = 72π square units.

Mathematics Question and Solution

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Calculating R, radius of the circle. Notice. 3 units is the side length of the inscribed yellow square. 2 units is the side length of the inscribed red square. 1 unit is the side length of the inscribed blue square. And; Length BT = Length BC. Therefore; a² = 2(3)² a = 3√(2) units. a is BC. b² = 2(1.5)² b = 1.5√(2) units. b is OB. It implies: R² = a²+b² R² = (3√(2))²+(1.5√(2))² R² =18+4.5 R = √(22.5) R = ½(3√(10)) units. R = 4.74341649025 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3172117

Calculating Shaded Area. Notice. a = ½(4) a = 2 cm. a is the radius of the inscribed circle. tanb = 4/2 b = atan(2)° c = 90-b c = atan(½)° d² = 2²+4² d = √(20) cm. d = 2√(5) cm. Therefore, shaded area is; ¼(2*2π)+2(½*√(20)*2sin(atan(½))) = π+4 = (π+4) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3172109

Calculating x. a = 23+x+3 a = (26+x) units. b = 23-3 b = 20 units. Length x is; 21²+20² = (26+x)² 441+400 = 676+52x+x² x²+52x-165 = 0 (x+26)² = 165+26² (x+26)² = 676+165 (x+26)² = 841 x = -26±√(841) x = -26±29 It implies; x ≠ -26-29 x = -26+29 x = 3 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3170573

Area Pink is; ¼(π*3²)-½(3²)+3((¼π)-½) = ¼(9π)-½(9)+¼(3π)-½(3) = ¼(12π)-½(12) = 3π-6 = 3(π-2) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3170568

Calculating x. Notice. AC = BC Let a be alpha. It implies; sina = 0.5x/(1+4) sina = ⅒(x) --- (1). Again. sina = 1/x --- (2). Equating (1) and (2). ⅒(x) = 1/x x² = 10 x = √(10) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3170567

Calculating the area of the ascribed half circle. Let a be the radius of the circle. b² = a²+(3√(3))² b = √(a²+27) units. c = ½(3+b) c = ½(3+√(a²+27)) units. d = c-3 d = ½(3+√(a²+27))-3 d = ½(√(a²+27)-3) units. e² = (3√(3))²-(½(√(a²+27)-3))² e² = 27-¼(a²+27-6√(a²+27)+9) e² = ¼(108-a²-27+6√(a²+27)-9) e² = ¼(72-a²+6√(a²+27)). It implies; Calculating a. e²+c² = a² ¼(72-a²+6√(a²+27))+(½(3+√(a²+27)))² = a² ¼(72-a²+6√(a²+27))+¼(9+6√(a²+27)+a²+27) = a² 72-a²+6√(a²+27)+36+6√(a²+27)+a² = 4a²...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3168936

Calculating red Area. Notice. 3x = 4-x 4x = 4 x = 1 unit. Where x is the radius of the red quarter circle and the inscribed white half circle. y = 4-x And x = 1 units. y = 4-1 y = 3 units. y is the radius of the bigger quarter circle. Area red is; ¼(πy²)-¼(πx²) = ¼*π(3²-1²) = ¼*π(9-1) = ¼*π*8 = 2π square units.

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