Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2510950

Let the side of the regular pentagon be 1 unit. a = 360-2(108)-90 a = 54° tan72 = b/0.5 b = 1.53884176859 units. tan54 = 1.53884176859/c c = 1.11803398875 units. d = c-0.5 d = 0.61803398875 unit. (e/tan36)+(e/tan27) = d Where e is the radius of the inscribed circle. e = 0.61803398875/((1/tan36)+(1/tan27)) e = 0.18509595408 unit. f = 180-72-54 f = 54° Therefore; g = d = 0.61803398875 unit. Shaded Area Red is; Area triangle with height 0.61803398875 units a...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2510927

Let the side length of the square be 2 units. Area square is; 2² = 4 square units. a² = 2(1²) a = √(2) units. b = ½(a) b = 0.5√(2) units. c² = 1²+(0.5√(2))²-2*0.5√(2)cos135 c = 1.5811388301 units. c = ½√(10) units. (1.5811388301/sin135) = (0.5√(2)/sind) d = 18.4349488229° e = 90-d e = 71.5650511771° f² = 1²+1.5811388301²-2*1.5811388301cos71.5650511771 f = 1.5811388301 units. it implies; cos18.4349488229 = 1/g g = 1.0540925534 units. Area inscribed gree...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2510881

Let the side length of the regular pentagon be 1 unit. x² = 2-2cos30 x = 0.5176380902 units. y = 108-90 108° is the single interior angle of the regular pentagon. y = 18° c = y+½(180-30) c = 18+75 c = 93° d² = 1²+0.5176380902²-2*0.5176380902*1cos93 d = 1.1498397111 units. d is a plus b (a+b). (1.1498397111/sin93) = (1/sine) e = 60.2841662413° f = 180-75-e f = 44.7158337587° Calculating a. (a/sin75) = (0.5176380902/sin44.7158337587) a = 0.7106400231 units....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2510229

Let FC = x Let CE = y Therefore; 5+x = y+4 x = y-1 --- (1). Notice; Triangle AEF is similar to triangle EFC. It implies; 5+x - y 4 - x Cross Multiply. x²+5x = 4y --- (2). Substituting (1) in (2). (y-1)²+5(y-1) = 4y y²-2y+1+5y-5 = 4y y²-y-4 = 0 Resolving the above quadratic equation via completing the square approach to get y. (y-½)² = 4+(¼) y = ½±½√(17) Therefore; y ≠ ½-½√(17) y = ½+½√(17) y = ½(√(17)+1) units. It implies; x will be; At (1). x = y-1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2509323

a² = 2(3+2√(3))²-2(3+2√(3))²cos120 a = 11.1961524227 units. b = a-(3+2√(3)) b = 4.7320508076 units. tan30 = c/(0.5(3+2√(3))) c = 1.8660254038 units. cos30 = (0.5(3+2√(3)))/d d = 3.7320508076 units. e = (3+2√(3))-d e = 2.7320508076 units. tan60 = f/2.7320508076 f = 4.7320508076 units. g² = 4.7320508076²+4.7320508076² g = 6.6921304299 units. h² = 6.6921304299²-(2+2√(3))² h = 1.7320508076 units. h = √(3) units. j = (3+2√(3))-2√(3) j = 3 units. It implies, are...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2508834

a² = 2b² a = √(2)b cm. (8+b)² = 2(4+a)² (8+b)² = 2(4+√(2)b)² 64+16b+b² = 2(16+8√(2)b+2b²) 64+16b+b² = 32+16√(2)b+4b² 3b²+16√(2)b-16b-32 = 0 3b²+(16√(2)-16)b-32 = 0 It implies; b = (8-4√(2)) cm. c = 8+b c = 8+(8-4√(2)) c = (16-4√(2)) cm. Recall. a = √(2)b And b = (8-4√(2)) cm. a = √(2)(8-4√(2)) a = (8√(2)-8) cm. d = 4+a d = 4+(8√(2)-8) d = (8√(2)-4) cm. Therefore, area of the figure (quadrilateral) is; Area triangle with height (8√(2)-4) cm and base (16-4√(2))s...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2508440

Considering the quadrilateral cyclic. a² = 2*10²-2*10²cosx a² = 200-200cosx --- (1). a² = 9²+7²-2*9*7cosx a² = 130+126cosx --- (2). Equating (1) and (2). a² = a² 200-200cosx = 130+126cosx 200-130 = 126cosx+200cosx 326cosx = 70 x = acos(70/326) x = 77.6006688181° y = 180-x y = 102.3993311819° Therefore, area cyclic quadrilateral is; Area triangle with height 10 units and base 10sin77.6006688181 units + Area triangle with height 9 units and base 7sin102.3993311819...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2506798

Let the base of the ascribed triangle be 1 unit. a = 180-12-42-12-36 a = 180-102 a = 78° (1/sin(12+36)) = (b/sin78) b = 1.3162274259 units. c = 180-12-42 c = 180-54 c = 126° (1.3162274259/sin126) = (d/sin12) d = 0.3382612127 units. f² = 0.3382612127²+1-2*0.3382612127cos12 f = 0.6728163648 units. It implies, the required angle x, is; (0.6728163648/sin12) = (0.3382612127/sinx) x = 6.0442188698°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2506755

Let AC be 2 units. a = 180-15-30 a = 135° b = 180-135 b = 45° (1/sin15) = (c/sin30) c = 1.9318516526 units. d² = 1.9318516526²+1²-2*1.9318516526*1cos45 d = √(2) units. d = 1.4142135624 units. Therefore, the required angle x is; (1.4142135624/sin45) = (1/sinx) x = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2506708

a² = 2²+1² a = √(5) units. b² = 2(a)² b² = 2√(5)² b² = 10 b = √(10) units. b is the diagonal of the blue square. tanc = (2/1) c = atan(2)° d = c-45 d = (atan(2)-45)° Calculating x. It implies; x² = 1+√(10)²-2√(10)*1cos(atan(2)-45) x = 2.2360679775 units.

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