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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3198573

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Calculating Alpha, let it be x. a = 180-24-54 a = 180-78 a = 102° Let the side length opposite/facing angle 24° (the shortest length) of the inscribed triangle with angles 24°, 54° and 102° be 1 unit. It implies; (1/sin24) = (b/sin102) b = 2.40486717237 units. (1/sin24) = (c/sin54) c = 1.98904379074 units. Therefore; d² = 1.98904379074²+2.40486717237²-2*1.98904379074*2.40486717237cos(24+24) d = 1.82709091528 units. (1.82709091528/sin(24+24)) = (2.40486717237/sine)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3198062

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Calculating Shaded Area Blue. Notice. 8 cm is the side length of the ascribed square. Therefore; ½(8) = 4 cm is the radius of the both inscribed half circles. tana = 4/8 a = atan(½)° b = 2a b = 2atan(½)° tanc = 8/4 c = atan(2)° d = 2c d = 2atan(2)° Area Blue is; Area sector with radius 8 cm and angle 2atan(½)°-Area triangle with height 8 cm and base 8sin(2atan(½)) cm+Area sector with radius 4 cm and angle 2atan(2)°-Area triangle with height 4 cm and base 4sin...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3198060

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Calculating Area Red Circle. Let a be the radius of the inscribed red circle. b = (4+a) m. c = (21+a) m. Therefore; Calculating a. a(4+a)+a(21+a)+a(4+21) = (4+a)(21+a) 4a+a²+21a+a²+25a = 84+25a+a² 50a+2a² = 84+25a+a² a²+25a-84 = 0 a²+28a-3a-84 = 0 a(a+28)-3(a+28) = 0 (a-3)(a+28) = 0 It implies; a ≠ -28 a = 3 m. Therefore, area red inscribed circle is; πa² = π(3)² = 9π m²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3197241

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Calculating Area Ascribed Rectangle. Let x be the width of the ascribed rectangle. Let y be the length of the ascribed rectangle. a = 1+2+1 a = 4 units. b = (x-2) units. c = (y-2) units. 4² = (x-2)²+(y-2)² 16 = x²-4x+4+y²-4y+4 8 = x²+y²-4x-4y  x²+y² = 8+4(x+y) --- (1). d²+1² = 2² d = √(3) units. e = (x-1) units. f = 2d+2e f = (2x+2√(3)-2) units. f = (2x+1.46410161514) units. f is the diagonal of the ascribed rectangle. It implies; f² = x²+y² (2x+1.464...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3196588

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Calculating The Required Angle. Let it be x. a = ½(12)+b a = (6+b) cm. b is the radius of the inscribed sector. Calculating b. c² = 6²+12² c² = 36+144 c = √(180) c = 6√(5) cm. It implies; a = c 6+b = 6√(5) b = (6√(5)-6) cm. b = 7.416407865 cm. Again, b is the radius of the inscribed sector. tand = 12/6 d = 63.4349488229° e = 180-d e = 116.565051177° f = ½(d) f = 31.7174744115° cos31.7174744115 = g/12 g = 10.2078097002 cm. 7.416407865² = 12²+12²-2*...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3196265

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Calculating Shaded Fraction. Let the radius of the inscribed circle be 1 unit. Therefore, the radius of the inscribed half circle is 2 units. a² = 2²+(2+2)² a² = 20 a = 2√(5) units. a is the radius of the ascribed quarter circle. Shaded Area is; ½*π(2)²+π(1)² = 2π+π = 3π square units. Area ascribed quarter circle is; ¼*π(2√(5))² = ¼(20π) = 5π square units. Therefore, shaded fraction area is; Shaded Area ÷ Area ascribed quarter circle = 3π÷5π = ⅗

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3196253

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Calculating Inscribed Blue Rectangle Area. a = 15+5 a = 20 units. Let b be the length of the blue inscribed rectangle. b² = 2(20)²-2(20)²cos(90-x) b² = 800-800(cos90cosx+sin90sinx) b² = 800-800sinx --- (1). b² = 2(15)²-2(15)²cos(90+x) b² = 450-450(cos90cosx-sin90sinx) b² = 450-450(-sinx)  b² = 450+450sinx --- (2). Equating (1) and (2). 800-800sinx = 450+450sinx 350 = 1250sinx sinx = 7/25 x = 16.2602047083° It implies; c² = 15²+20²-2*15*20cos16.2602047083...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3195459

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Calculating x. Let a be the side length of the square. b² = 2a² b = √(2)a units. c = ½(b) c = ½√(2)a units. a ~ 1 0.5 ~ d Cross Multiply. d = 1/(2a) units. Calculating a. c+d = a ½√(2)a+(1/(2a)) = a √(2)a²+1 = 2a² (2-√(2))a² = 1 a = √(1/(2-√(2))) a = 1.30656296488 units. Again, a is the side length of the square. Recall. d = 1/(2a) units. And a = 1.30656296488 units. d = 1/2(1.30656296488) d = 0.38268343236 units. e² = d²+d² e² = 2(0.38268343236)² e...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3195313

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Calculating x. 10² = 18²+x²-2*18xcosy 36xcosy = 224+x² cosy = (224+x²)/(36x) --- (1). 10² = 6²+x²-2*6xcosy 12xcosy = x²-64 cosy = (x²-64)/(12x) --- (2). Equating (1) and (2). (224+x²)/(36x) = (x²-64)/(12x) 3(x²-64) = 224+x² 3x²-192 = 224+x² 2x² = 416 x² = 208 x = √(208) x = 4√(13) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3195312

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Calculating Ratio Purple Area and Orange Area. Let 2 be the radius of the inscribed orange area. It implies; Area orange inscribed circle is; π(2)² = 4π square units. tan30 = 2/a a = 2√(3) units. b = 2a b = 4√(3) units. b is the side length of the inscribed regular triangle. sin30 = 2/c c = 4 units. c is the radius of the ascribed circle. Calculating area purple. ⅓((½*(4√(3))²sin60)-π(2)²)+(120π*4²/360)-(½*4²sin120) = ⅓(12√(3)-4π)+⅓(16π)-(4√(3)) = 4√(3)-⅓(4π)+⅓(16π)-4√(3) = ⅓(16π)-...
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