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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2426738

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Let a be the radius of the inscribed semi circle. b = 2+4+a b = (6+a) cm. b is the diameter of the ascribed circle. c = ½(b) c = ½(6+a) cm. c is the radius of the ascribed circle. e = c-2 e = ½(6+a)-2 e = ½(6+a-4) e = ½(2+a) cm. Therefore, calculating a, radius of the inscribed semi circle. c² = a²+e² (½(6+a))² = a²+(½(2+a))² ¼(36+12a+a²) = a²+¼(4+4a+a²) 36+12a+a² = 4a²+4+4a+a² 0 = 4a²-8a-32 4a²-8a-32 = 0 a²-2a-8 = 0 Resolving the above quadratic equation via factorization approach to ge...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2426678

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Calculating R, radius of the circle. It implies; 10²+11²-2*10*11cosa = 7²+8²+2*7*8cosa 221-220cosa = 113+112cosa 221-113 = 112cosa+220cosa 108 = 332cosa cosa = 108/332 a = acos(108/332) a = 71.016175615° b² = 10²+11²-2*10*11cos71.016175615 b = 12.2243091805 units. Therefore, R, radius of the circle is; sin71.016175615 = (0.5*12.2243091805)/R R = (0.5*12.2243091805)/sin71.016175615 R = 6.4637130226 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2426635

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a = ⅙(180(6-2)) a = ⅙(180*4) a = 120° a is the single interior angle of the regular hexagon. b² = 2²+4²-2*2*4cos120 b = √(20+7) b = 2√(7) units. (2√(7)/sin120) = (2/sinc) c = 19.1066053509° d = 120-2c d = 120-2*19.1066053509 d = 81.7867892983° sin30 = e/4 e = 2 units. f = 2+e f = 2+2 f = 4 units. g² = 4²+4²-2*4*4cos120 g = √(32+16) g = √(48) g = 4√(3) units. Let r be the radius of the inscribed circle. Calculating r. sin60 = r/(4-r) r = 4sin60-rsin60...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2425871

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Calculating x. a = 9+3 a = 12 units. Therefore the ascribed plane shape (quadrilateral or irregular quadrilateral) is an equilateral triangle of side length 12 units with a subtraction of a smaller equilateral triangle of side length 3 units. It implies; a² = x²+6² 12² = x²+6² x² = 144-36 x = √(108) x = 6√(3) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2425859

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Notice. The ascribed triangle is equilateral. Let r be the radius of the inscribed half circle. Calculating r. sin60 = r/6 √(3)/2 = r/6 r = 3√(3) cm. It implies, area orange (inscribed semi circle) is; ½(r²)π = ½(3√(3))²π = ½(27)π = 13.5π cm² = 42.4115008235 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2423961

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Let the side length of the regular octagon be 1 unit. a = ⅛(180(8-2)) a= ⅛(180*6) a = 135° a is the single interior angle of the regular octagon. 2b² = 1² b = √(1/2) b = ½√(2) units. c = 2b+1 c = 2(½√(2))+1 c = (√(2)+1) units. c is the side length of the regular triangle (equilateral triangle). d² = 2-2cos135 d = Let the side length of the regular octagon be 1 unit. a = ⅛(180(8-2)) a= ⅛(180*6) a = 135° a is the single interior angle of the regular octagon. 2b² = 1² b = √(1/2) b = ½√(2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2423757

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Calculating red length. Let r be the radius of the circle. sin30 = a/2 a = 1 unit. cos30 = b/2 b = √(3) units. Or b = ½√(2²+2²-2*2*2cos120) b = ½√(8+4) b = ½√(12) b = ½*2√(3) b = √(3) units. c = 2b c = 2√(3) units. d = 2+1+1 d = 4 units. r = ½(d) r = ½(4) r = 2 units. Again, r is the radius of the circle. e = 3r-a e = 3*2-1 e = 5 units. It implies, red length is; Let it be f. f² = 5²+b² f² = 25+√(3)² f² = 25+3 f = √(28) f = 2√(7) units. Again, f is the red length.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2423144

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8 = 0.5b²sin113.7 16 = b²sin113.7 b = √(16/sin113.7) b = 4.1801537416 units. Therefore; x² = 2*4.1801537416²-2*4.1801537416²cos113.7 x = 6.9995991687 units. x = 7 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2422932

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Notice. Since x+y = 1 cm. It explains that the pentagon or irregular pentagon can be presented in form of a regular quadrilateral (square) with side length 1 unit. Therefore; Area red is; 1² = 1*1 = 1 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2422470

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a = 180-72-36 a = 180-108 a = 72° Notice. Triangle BCD is isosceles. Let AD = BC = 1 unit. Therefore, calculate b. ((b+1)/sin72) = (1/sin36) bsin36+sin36 = sin72 bsin36 = sin72-sin36 b = (sin72-sin36)/sin36 b = 0.6180339887 units. b is AB. c² = 0.6180339887²+1²-2*0.6180339887*1cos72 c = 1 unit. c is AC. d = 180-72 d = 108° Therefore; (a/sin108) = (1/sin36) a = 1.6180339887 units. a is CD. It implies; a/b is; 1.6180339887/0.6180339887 = 2.618033989  ≈ 2.62 to 2 decimal places.
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