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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2439064

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Calculating angle x. Let AB be 1 unit. a = 9-20  a = 70° a is angle ABH. b = 180-10-70 b = 100° (c/sin70) = (1/sin100) c = 0.9541888941 units. (d/sin10) = (1/sin100) d = 0.1763269807 units. sin20 = e/1 e = 0.3420201433 units e is BH. cos20 =f/1 f = 0.9396926208 units. f is AH. g = e-d g = 0.3420201433-0.1763269807 g = 0.1656931626 units. h = 180-20-30 h = 130° (1/sin130) = (j/sin30) j = 0.6527036447 units. k = f-j k = 0.9396926208-0.6527036447 k = 0.2869889761 units. Therefore, the...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2438789

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Let BC be 1 unit. a = ½(180-45) a = ½(135) a = 67.5° a is angle ABC = angle ACB. (1/sin45) = (b/sin67.5) b = 1.3065629649 units. b is AB. c = 180-45-67.5 c = 67.5° c is angle BEC. It implies; BE = 1 unit. d = 67.5-30 d = 37.5° d is angle BCD. e = 180-67.5-37.5 e = 75° e is angle BDC. (1/sin75) = (f/sin37.5) f = 0.630236207 units. f is BD. g = 67.5-45 g = 22.5° g is angle DBE. h² = 1²+0.630236207²-2*1*0.630236207cos22.5 h = 0.4823619098 units. h is DE...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2437851

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Let the side length of the regular pentagon be 1 unit. a = ⅕(180(5-2)) a = 108° a is the single interior angle of the regular pentagon. b² = 2-2cos108 b = 1.6180339887 units. c = ½(180-108) c = 36° d = 108-36 d = 72° e = 180-72-30 e= 78° (1.6180339887/sin78) = (f/sin30) f = 0.8270909153 units. g = 180-78 g = 102° Therefore, the required angle alpha is; Let it be h. (1/sin102) = (0.8270909153/sinh) h = 54°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2437741

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Notice! Radius of the ascribed semi circle is 6 cm. It implies; Radius, a of the inscribed bigger circle is; a = ½(6) a = 3 cm. Calculating b, radius of the smaller inscribed circle. c = (6-b) cm. c² = b²+e² (6-b)² = b²+e² e² = 36-12b e = √(36-12b) cm. f = (3-b) cm. g = (3+b) cm. Therefore; g² = f²+e² (3+b)² = (3-b)²+36-12b 9+6b+b² = 9-6b+b²+36-12b 12b = 36-12b 24b = 36 b = (36/24) b = (3/2) cm. Again, b is the radius of the smaller inscribed circle. Painted Area is; Area semi circle...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2437696

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Calculating the area of the inscribed rectangle. Let the length and width of the inscribed rectangle be x and y respectively. Therefore, observing similar plane shapes (right-angled) side length ratios. 6 - y x - 5 Cross Multiply. xy = 30 square units. Where xy is the area of the inscribed rectangle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2437685

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Calculating area of the inscribed blue square. Let the inscribed blue square side be x. a² = 2x² a = √(2)x units. a is the diagonal of the inscribed blue square. x² = 2b² b² = x²/2 b = ½√(2)x units. It implies; 1² = a²+b² 1 = 2x²+(x²/2) 1 = ½(5x²) 2 = 5x² x² = (2/5) square units. x² = 0.4 square units. Where x² is the area of the inscribed blue square.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2436610

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Notice! AQ = 6 cm. a² = 6²+3²-2*6*3cos108 108° is the single interior angle of the regular pentagon. a = 7.4916361229 cm. a is AP. cosb = 6/7.4916361229 b = acos(6/7.4916361229)° b is angle PAQ. Therefore, area triangle PAQ (area blue) is; 0.5*6*7.4916361229sin(acos(6/7.4916361229)) = 13.4581390311 cm² ≈ 13.46 cm² to 2 decimal places.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2436205

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(4+5+3)² = 2a² a² = 72 a = 6√(2) cm. a is the congruent adjacent sides of the right-angled triangle. b² = √(72)²+3²-2*√(72)*3cos45 b² = 81-36 b = √(45) b = 3√(5) cm. (3√(5)/sin45) = (3/sinc) c = 18.4349488229° d² = √(72)²+4²-2*√(72)*4cos45 d² = 88-48 d = √(40) d = 2√(10) cm. (2√(10)/sin45) = (4/sine) e = 26.5650511771° f = 90-c-e f = 90-18.4349488229-26.5650511771 f = 45° 2g² = 5² g² = 25/2 g = √(25/2) g = ½(5√(2)) cm. g = 3.5355339059 cm. It implies; Shaded Area is; Area quarter circl...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2435756

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Let the base of the ascribed triangle be 1 unit. a = 180-20-20-10-10 a = 120° (1/sin120) = (b/sin20) b = 0.3949308436 units. c = 180-120-20 c = 40° (0.3949308436/sin40) = (d/sin120) d = 0.5320888862 units. e = 180-40-60 e = 80° (0.3949308436/sin80) = (f/sin60) f = 0.3472963553 units. g² = 0.3472963553²+0.5320888862²-2*0.3472963553*0.5320888862cos20 g = 0.2375646984 units. (0.2375646984/sin20) = (0.5320888862/sinh) h = 50° j = 180-h j = 180-50 j = 130° Therefore the required angle x,...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2435595

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a = 180-20-50-40 a = 70° It implies the triangle is isosceles. Let the base of the isosceles triangle be 1 unit. 1² = 2b²-2b²cos40 1 = 0.4679111138b² b = √(1/0.4679111138) b = 1.4619022 units. b is the two equal lengths of the isosceles triangle. c = 180-40-20 c = 120° (1.4619022/sin120) = (d/sin20) d = 0.5773502692 units. (1.4619022/sin120) = (e/sin40) e = 1.0850635751 units. f = e-d f = 1.0850635751-0.5773502692 f = 0.5077133059 units. g = b-d g = 1.461902...
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