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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3120228

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a = ½(2+7) a = 4.5 units. a is the radius of the ascribed half circle. Let b be the radius of the inscribed circle. c = (4.5-2b) units. d = (4.5-b) units. e = 4.5-2-b e = (2.5-b) units. Calculating b. 4.5 ~ (4.5-b) (4.5-2b) ~ (2.5-b) Cross Multiply. 4.5(2.5-b) = (4.5-2b)(4.5-b) 11.25-4.5b = 20.25-4.5b-9b+2b² 2b²-9b+9 = 0 (b-¼(9))² = -(9/2)+(-9/4)² (b-¼(9))² = (81/16)-(9/2) (b-¼(9))² = (9/16) b = (9/4)±√(9/16) b = (9/4)±(3/4) Therefore; b ≠ 3 units....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3120171

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Let the side length of the regular heptagon be 1 unit. Calculating angle x. a = ⅐*180(7-2) a = ⅐(900)° a is the single interior angle of the regular heptagon. b = ½(360-2*⅐(900)) b = 180-⅐(900) b = ⅐(360)° c = ⅐(900)-90 c = ⅐(270)° sin(⅐(270)) = d/1 d = 0.62348980186 units. e = 1+2d e = 2.24697960372 units. f = ⅐(360)-⅐(270) f = ⅐(90)° Therefore, the required angle, x is; x = 180-g Calculating g. (2.24697960372/sing) = (1/sin(⅐(90))) g = 30° It imp...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3120166

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Calculating x. a = ⅙*180(6-2) a = 120° a is the single interior angle of the green regular hexagon. b = ½(x) units. c = ⅙(x) units. It implies; d ~ b b ~ c Therefore; d ~ ½(x) ½(x) ~ ⅙(x) Cross Multiply (Calculating d). ⅙(dx) = ¼(x²) 2d = 3x d = ½(3x) units. Calculating x, the side length of the green regular hexagon. 13² = d²+b²-2dbcos120 169 = ¼(9x²)+¼(x²)-2*½(3x)*½(x)cos120 169 = ¼(9x²)+¼(x²)+¼(3x²) 676 = 9x²+x²+3x² 13x² = 676 x² = 13*4 x = 2√(1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3118678

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Let a be the side length of the two equal inscribed lengths. r = 2+1 r = 3 units. r is the radius of the ascribed half circle. b²+(a/2)² = 3² b²= 9-¼(a²) b = √(9-¼(a²)) b = ½√(36-a²) units. c = a-b c = a-½√(36-a²) c = ½(2a-√(36-a²)) units. Calculating a. 2² = c²+(a/2)² 4 = (½(2a-√(36-a²)))²+¼(a²) 4 = ¼(4a²-4a√(36-a²)+(36-a²))+¼(a²) 16 = 4a²-4a√(36-a²)+36-a²+a² 16 = 4a²-4a√(36-a²)+36 4 = a²-a√(36-a²)+9 a√(36-a²) = a²+5 a²(36-a²) = (a²+5)² 36a²-a⁴ =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3117610

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Let the side length of the ascribed square be 1 unit. Therefore, the side length of the inscribed green regular triangle 1 unit. a = 90-60° a = 30° b² = 2-2cos30 b = 0.51763809021 units. c = ½(180-30) c = 75° d = c+60 d = 135° (1/sin135) = (0.51763809021/sine) e = 21.4707014327° f = 180-135-e f = 45-21.4707014327 f = 23.5292985673° (1/sin135) = (g/sin23.5292985673) g = 0.56457945531 units. h = 1-g h = 1-0.56457945531 h = 0.43542054469 units. j² = 0.4354...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3117503

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sina = 4/(4+1) a = asin(4/5)° b² = 1²+4²-2*1*4cos(asin(4/5)) b = ⅕√(305) units. b = 3.49284983931 units. (3.49284983931/sin(asin(4/5))) = (4/sinc) c = asin(0.9161573349) c = 113.629377731° d = c-90 d = 23.629377731° e = 180-2d e = 132.741244538° tan(0.5asin(4/5)) = f/1 f = ½ units. f is the radius of the inscribed circle. Calculating length x. x² = 2(0.5)²-2(0.5)²cos132.741244538 x = 0.9161573349 units. x = (16√(305)/305) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3117353

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a² = (3+5)²+x² a² = a²-64 a = √(64+x²) cm. b = (8-x) cm. It implies; 5 ~ 8 (8-x) ~ √(64+x²) Cross Multiply. 8(8-x) = 5√(64+x²) (64-8x)² = 25(64+x²) 4096-1024x+64x² = 1600+25x² 2496-1024+39x² = 0 Resolving the above quadratic equation. x = 2.71909 cm. Therefore; (8-2.71909)² = 5²+c² c = 1.69941473105 cm. Recall. a = √(64+x²) And x = 2.71909 cm. a = √(64+(2.71909)²) a = 8.44946450541 cm. d = a+c d = 8.44946450541+1.69941473105 d = 10.1488792365 cm. e...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3117326

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Let single side of the square be x. Notice; PB=¼(AB)=¼(x). Calculating single side length of square ABCD. It will be; Area triangle ADP of height x and base (x-¼(x)) + area triangle DOP of base (√(x²+(x-¼(x))²)) and height 4 unit + 2(area triangle DOC of height 4 unit and base x) + area triangle OPB of height (x-4) and base ½(x) = area square ABCD with length and width x respectively. Therefore; ½(x*(3x/4))+½(4√(x²+(3x/4)²)+½(8x)+½(¼(x)(x-4))=x² (3x²)/8+2√(x²+(9x²/16))...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3117316

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Sir Mike Ambrose is the author of the question. Let the bigger inscribed square be 2 units. BC = 1+2+x = (3+x) units. Calculating x. (3+x)²=1²+8-4√(2)cos135 3+x=√(13) x = (√(13)-3) units. Therefore; BC (side of the ascribed square ABCD) is; BC = 3+√(13)-3 BC = √(13) units. Therefore; Area ABCD is; √(13)² = 13 square units. Area Yellow is; Area trapezium with parallel sides √(13) units and (√(13)-3) unit respectively, and height (√(13)-2) unit - Ar...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3117079

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Notice. The ascribed right-angled triangle is not drawn to scale. Let a be the side length of the inscribed square. Calculating Area Inscribed Square. tanb = 3√(3)/(9√(3)) b = atan(⅓)° tanb = a/c tan(atan(⅓)) = a/c c = 3a units. d = 9√(3)-c-a d = (9√(3)-4a) units. e = (3√(3)-a) units. Calculating a. (3√(3))² = d²+e² 27 = (9√(3)-4a)²+(3√(3)-a)² 27 = 243-72√(3)a+16a²+27-6√(3)a+a² 17a²-78√(3)a+243 = 0 a²-(78√(3)/(2*17))a = -243/17 (a-(39√(3)/17))² = (-...
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