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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2571637

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It implies; Calculating x. (½*x*3x)+(½(2x+3x)*3x) = 12 3x²+15x² = 24 18x² = 24 x² = 4/3 x = ⅓(2√(3)) units. Therefore, 3x, the side length of the ascribed square  is; 3*⅓(2√(3)) = 2√(3) units  tana = 2√(3)/⅓(2√(3)) a = atan(3)° b = 180-45-a b = (135-atan(3))° It implies; (⅓(2√(3))/sin(135-atan(3))) = (c/sin(atan(3))) c = 1.2247448714 units. Yellow Area is; ½(2√(3))²-0.5*1.2247448714*⅓(2√(3))sin45 = 6-0.5 = 5½ square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2571617

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Calculating x. 7² = x²+5²+2*5*xcosa 49 = x²+25+10xcosa 24 = x²+10xcosa 10xcosa = 24-x² cosa = (24-x²)/(10x) --- (1). b = (x+5) units. 7² = x²+(x+5)²-2x(x+5)cosa 49 = x²+x²+10x+25-(2x²+10x)cosa  24 = 2x²+10x-(2x²+10x)cosa --- (2). Substituting (1) in (2). 24 = 2x²+10x-(2x²+10x)cosa And cosa = (24-x²)/(10x) 24 = 2x²+10x-(2x²+10x)(24-x²)/(10x) 240= 20x²+100x-(2x+10)(24-x²) 240 = 20x²+100x-(48x-2x³+240-10x²) 240 = 20x²+100x-48x+2x³-240+10x² 480 = 2x²+30x²+52x x³+15x²+26x-240= 0 It implies;...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2571557

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The required angle, x is; a = ½(90) a = 45° it implies; x = ½(180-a) x = ½(180-45) x = ½(135) x = 67.5°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2571545

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Notice! The ascribed polygon is a regular nonagon. The inscribed polygon is a regular hexagon. Calculating the required angle, x. a = ⅙*180(6-2) a = 120° a is the single interior angle of the regular hexagon. b = ⅑*180(9-2) b = 140° b is the single interior angle of the regular nonagon. c = ½(360-140+140) c = ½(360-280) c = ½(80) c = 40° d = ⅕*180(5-2) d =108° d is the single interior angle of the regular pentagon. e = 5d e = 540° e is the sum of interior angles of a regular pentagon....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2571534

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Sir Mike Ambrose is the author of the question. Let the side of the regular nonagon be 1 unit. Therefore; Area nonagon is; ½(9)*(1/(2tan(180/9))) = ¼(9)*(1/tan(20)) = 6.18182419377 square units. Calculating r, radius of the inscribed circle. 2(½(r) + ½(r)√(2-2cos140)) = 2(½(√(2-2cos140))cos120) r + √(2-2cos140)r = √(2-2cos140)cos120 r = √(2-2cos140)cos120 ÷ (1+√(2-2cos140)) r = 0.56525793742 unit. Area of the inscribed circle is; π(0.56525793742)² = 1.00379080162 square units. It im...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2571525

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Sir Mike Ambrose is the author of the question. Let the side of the regular nonagon be 1 unit. Therefore; Length Green is; ((1+2sin50)sin20/sin80) + √(1.08506357513²+1²) = 0.87938524157 + 1.47558902208  = 2.35497426365 units. Length Blue is; √(1.1371580426²+1²-2(1.1371580426)cos100) = 1.63953019414 units. It implies; Length Blue ÷ Length Green to 2 decimal places is;  1.63953019414 ÷ 2.35497426365 = 0.69619877357  ≈ 0.70

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2571520

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a = 8+9 a = 17 units. a is the radius of the half circle. b²+8² = 17² b = √(225) b = 15 units. c²+d² = 17² c = √(289-d²) units. e = (17-d) units. f = ½(c) f = ½√(289-d²) unit. g = 17+17-9 g = 34-9 g = 25 units. Observing similar plane shape (right-angled) side length ratios. (17-d) - 25 ½√(289-d²) - 15 Cross Multiply. 15(17-d) = 25*½√(289-d²) 30(17-d) = 25√(289-d²) 6(17-d) = 5√(289-d²) 102-6d = 5√(289-d²) (102-6d)² = 25(289-d²) 10404-1224d+36d² = 7225-25d² 11d²-1224d+3179 = 0 It impl...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2571502

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Let the equal side lengths be a. 14² = a²+16²-2*16acosb 196 = a²+256-32acosb --- (1). 21² = 2a²+2a²cosb cosb = (441-2a²)/2a² --- (2). Substituting (2) in (1). 196 = a²+256-32a(441-2a²)/2a² 32(441-2a²)/2a = a²+256-196 32(441-2a²)/2a = a²+60 14112-64a² = 2a³+120a 2a³+64a²+120a-14112 = 0 It implies; a = 12 units. a is the required length.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2570794

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Notice! r, radius of the inscribed quarter circle is also the side length of the ascribed square. Let alpha be a. Notice again. 2a = 60° a = 30° Alpha = 30° It implies; sin30 = b/1 b = ½ units. c = (r-½) units. d = ½(r) units. Calculating r, radius of the inscribed quarter circle and the side length of the ascribed square. r² = (r-½)²+(½(r))² r² = r²-r+¼+¼(r²) ¼(r²)-r+¼ = 0 r²-4r+1 = 0 Calculating the above quadratic equation via completing the square ap...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2570036

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Let the square side length be a. Observing similar plane shape (right-angled) side length ratios. 8 - ½(a) b - a Cross Multiply. ½(ab) = 8a b = 16 units. c² = 8²+16² c = √(64+256) c = 8√(5) units. d = ½(c) d = 4√(5) units  d = 8.94427191 units. d is the radius of the inscribed quarter circle and also the side length of the ascribed square. d = a = 4√(5) units. a = 8.94427191 units. a is the height/length of the blue inscribed rectangle, the square side length, and also the radius of the i...
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