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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3175040

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Calculating the area of the ascribed regular octagon. Let the equal longest length of the both inscribed yellow triangles be x. a = ⅛*180(8-2)) a = ⅛(180*6) a = ½(45*6) a = 135° a is the single interior angle of the ascribed regular octagon. sin(0.5*135) = (0.5x)/b b = 0.54119610015x units. b is the side length of the regular octagon. c = 135-(0.5*45)-½(360-2(135)) c = 135-22.5-45 c = 135-67.5 c = 67.5° c is the second bigger interior angle of inscribed triangle 21 square units. 2e² = (0....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3174823

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Calculating Length AB. Let AB = (a+7+b) units. Calculating a and b. c = (7+b) units. Therefore; a(7+b) = 60 a = 60/(7+b) --- (1). Again. d = (a+7) units. Therefore; b(a+7) = 5*12 ab+7b = 60  ab = 60-7b a = (60-7b)/b --- (2). Equating (1) and (2) to get b. 60/(7+b) = (60-7b)/b Cross Multiply. 60b = 420-49b+60b-7b² 7b²+49b-420 = 0 b²+7b-60 = 0 b²+12b-5b-60 = 0 b(b+12)-5(b+12) = 0 (b+12)(b-5) = 0 Therefore; b ≠ -12  b = 5 units. Calculating a, us...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3174819

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Shaded Area is; Area circle with radius 4 cm - 4(area quarter circle with radius 4 cm - area isosceles right-angled triangle with equal adjacent lengths 4 cm). = π*4²-4(¼(4²π)-½(4²)) = 16π-4(4π-8) = 16π-16π+32 = 32 square units.

Mathematics Question and Solution

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

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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

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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

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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

https://onlineedumath.com/update/mathematics-question-and-solution/3173552

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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

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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

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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

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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.
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