Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3341152

Area Orangle ÷ Area Large Square to 3 d. p. Let the side length of the large square be x. Calculating x. 25=x²+64-(16xcos20) Therefore; x = 11.70244 cm It implies; Area large square is; 11.70244² = 136.947101954 cm² Area orange is; Area triangle with height and base 8 cm each - Area triangle with two side 2.91176187413 cm and 2.27177411013 cm, and angle 70° = ½(8)² - ½(2.91176187413*2.27177411013)sin70 = 32 - 3.107970027 = 28.892029973 cm² Therefore; Area Orangle ÷ Area Large S...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3319726

Calculating R, radius of the circle. (4√(2))² = 1+5²-2*5cosa 32 = 26-10cosa 16 = 13-5cosa cosa = (-3/5) a = 126.869897646° b = 360-2a b = 360-2*126.869897646 b = 106.260204708° c = ½(b) c = 53.130102354° It implies, R is; sin53.130102354 = (0.5*4√(2))/R R = 2√(2)/sin53.130102354 R = ½(5√(2)) units. R = 2.5√(2) units. R = 3.53553390594 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3318889

Calculating r, radius of the circle. Let theta be x. Notice. BC = CD, let it be a. a² = 2²+13²-2*2*13cosx a² = 173-52cosx --- (1). a² = 11²+13²-2*11*13cosx a² = 290-286cosx --- (2). Calculating x. Equation (1) and (2). 173-52cosx = 290-286cosx 234cosx = 117 cosx = ½ x = 60° Calculating a. Substituting x in (1). a² = 173-52cos60 a² = 173-26 a = √(147) units. Therefore, r is; sin60 = ½√(147)/r √(3)r = √(147) r = 7 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3317494

Calculating area of the ascribed circle. Let r be the radius of the ascribed circle. a = 3+2 a = 5 units. a is the side length of the inscribed regular triangle (equilateral triangle). sin60 = b/2 ½√(3) = b/2 b = √(3) units. cos60 = c/2 ½ = c/2 c = 1 units. d = a-c d = 4 units. Therefore, r is; r² = √(3)²+4² r² = 19 r = √(19) units. Again, r is the radius of the ascribed circle. Area ascribed circle is; πr² = π√(19)² = 19π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3294835

Calculating Area Orange. tana = 3/6 a = atan(½)° sin(atan(½)) = b/3 b = 1.3416407865 units. cos(atan(½)) = c/3 c = 2.683281573 units. d = √(6²+3²) d = √(45) d = 3√(5) units. e² = 3²+1.3416407865²-2*3*1.3416407865cos(90+atan(½)) e = 3.7947331922 units. (3.7947331922/sin(90+atan(0.5))) = (1.3416407865/sinf) f = 18.4349488231° g = 180-(90+atan(0.5))-18.4349488231 g = 45° Therefore, Area Orange is; (0.5(2.683281573+3√(5))*2.683281573)-(0.5*2.683281573²)-(0.5*2.6...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3294682

Notice. The composite plan shape is not drawn to scale. Calculating the inscribed orange triangle area. Notice Again. 3 units is the diameter of the inscribed circle. a = ½(3) a = 1.5 units. a is the radius of the inscribed circle. tanb = 3/4 b = atan(3/4)° tan(atan(¾)) = c/1.5 ¾ = ⅓(2c) c = 9/8 units. d = (3/2)-(9/8) d = 3/8 units. e = (180-atan(4/3))° e = 126.869897646° (1.5/sin(180-atan(4/3))) = ((3/8)/sinf) f = 11.5369590328° g = 180-2f g = 156.9260819...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3294509

Calculating R, radius of the inscribed circle. cosa = BA/BC a = acos(4/8) a = 60° a is angle ABC. b = ½(a) b = 30° b is angle ABD = angle CBD. sin30 = c/8 c = 4 cm. c is CD. cos30 = d/8 d = 4√(3) cm. d is BD. Calculating R. 4R+8R+4√(3)R = 4*4√(3) (12+4√(3))R = 16√(3) (3+√(3))R = 4√(3) R = (4√(3))/(3+√(3)) R = 2√(3)-2 R = 2(√(3)-1) units. R = 1.46410161514 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3294072

Calculating length BP. Let CQ = a. b = (16+a) units. b is CD = AC, the radius of the ascribed half circle. c = b-5 c = (11+a) units. c is BC. e² = a²+(11+a)²-2a(11+a)cos120 e² = 2a²+22a+121+11a+a² e² = 3a²+33a+121 e = √(3a²+33a+121) units. e is BP, the required length. Notice. AB = CQ = a = 5 units. Therefore; BP (e) is; √(3(5)²+33(5)+121) BP = √(361) BP = 19 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3292949

Calculating angle alpha, let it be x. Let the equal squares side length be 2 units. 2² = 1²+a² a = √(3) units. b = a-1 b = (√(3)-1) units. b = 0.73205080757 units. sinc = 1/2 c = 30° tan30 = d/0.73205080757 d = 0.42264973081 units. cos30 = 0.73205080757/e e = 0.84529946162 units. f = 2-e f = 1.15470053838 units. g² = 1.15470053838²+2² g = 2.30940107676 units. h = g+d h = 2.73205080757 units. j = h-2 j = 0.73205080757 units. tank = 1.15470053838/2 k = 30° tan30 = l/0.73205080757 l =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3292074

Calculating r, radius of a circle. a = (r-2) cm. b = (r-1) cm. It implies; r² = (r-2)²+(r-1)² r² = r²-4r+4+r²-2r+1 r²-6r+5 = 0 r²-r-5r+5 = 0 r(r-1)-5(r-1) = 0 (r-1)(r-5) = 0 It implies; r ≠ 1 cm. r = 5 cm. Again, r is the radius of the circle. The square side length is; 2r-2 = 2*5-2 = 10-2 = 8 cm.

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