Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2778301

Shaded inscribed area is; Area regular hexagon with side 1 unit - 6(Area sector with radius 1 unit and angle 60°) + 6(Area equilateral triangle with side 1 unit). = 6/(4tan(180÷6)) - 6(60π÷360) + 6*½(sin60) = (3√(3)/2) - 6(π/6) + 6*¼(√(3)) = ½(3√(3)) - π + ½(3√(3)) = (3√(3)-π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2778284

Sir Mike Ambrose is the author of the question. Let the single side length of the regular octagon be 2 units. Therefore; Area green is; Area triangle with height and base √(2) units + Area triangle with height 2√(2) units and base 1 unit - Area triangle with height √(2) units and base ½ units. = ½(√(2))² + ½(2√(2)) - ½(½*√(2)) = 1 + √(2) - ¼√(2) = ¼(4+4√(2)-√(2)) = ¼(4+3√(2)) space units. Area blue is; Area square with side 2 units + Area rectangle with length 2 units an...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2777592

Sir Mike Ambrose is the author of the question. 18² = 14²+16²-2*14*16cosa 448cosa = 14²+16²-18² a = acos((14²+16²-18²)/448) a = 73.39845040098° (18/sin73.39845040098) = (14/sinb) b = 48.18968510422° c = 180-48.18968510422-73.39845040098 c = 58.4118644948° cotd = cota+cotb+cotc cotd = 1.80748828181 d = acot(1.80748828181) d = 28.95373454979° e = 58.4118644948-d e = 29.45812994501° f = 180-28.95373454979-73.39845040098 f = 77.64781504923° (14/sin77.64781504923) = (...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2776792

Green Area is; 2(area square with side 2 cm - area quarter circle with radius 2 cm) + 3(area sector with radius 1 unit and angle 300°) + 4(area equilateral triangle with side 1 unit) = 2(4 - ¼(4π)) + 3(300π÷360) + 4(½sin60) = 8 - 2π + ½(5π) + √(3) = ½(16-4π+5π+2√(3)) = ½(16+π+2√(3)) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2776777

Sir Mike Ambrose is the author of the question. a = 6-2 a = 4 units. b² = 6²-4² b = 2√(5) units. c = 6-b c = (6-2√(5)) units. d = asin(⅔)° tan(asin⅔) = e/(6-2√(5)) e = ⅕(12√(5)-20) units. f = 4+e f = ⅕(12√(5)) units. g² = 6²+(⅕(12√(5)))² g = ⅕(18√(5)) units. 6² = (⅕(18√(5)))²+(⅕(12√(5)))²-2*⅕*⅕*18*12*5cosh 36 = ⅕(468)-⅕(432)cosh ⅕(432)cosh = ⅕(468)-36 ⅕(432)cosh = ⅕(288) h = acos(288/432) h = 48.18968510422° i = 90-h i = 41.81031489578° Therefore; A...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2776225

½(xy)=24 xy = 48 ------ (1) x²+y²=10² ------- (2) And; x²+y²=(x+y)²-2xy Therefore; (x+y)²-2xy=100 And; xy = 48 (x+y)²-2(48)=100 (x+y)²=196 x+y=14 ----- (3) x = 14-y ------ (4) Substitute (4) in (1) y(14-y)=48 y²-14y+48=0 (y-7)²=-48+49 y = 7±1 Therefore; y = 8 units or 6 units. It implies; For y = 8 units, x = 14-y x = 14-8 x = 6 units. Calculating r, radius of the inscribed circle. ½(6r)+½(8r)+½(10r)=½(8*6) 3r+4r+5r=24 12r = 24...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2775410

Let the radius, r of the circle be 7 units. Therefore perimeter of the circle is; 2πr = 44 units. Calculating the length of the square. 14² = (14-2x)²+(14-4x)² 196 = 196-56x+4x²+196-112x+16x² 20x²-168x+196=0 5x²-42x+49=0 (x-(21/5))²= (-49/5)+(-21/5)² (x-(21/5))²= (196/25) x = (21/5)±(14/5) x = (21±14)/5 x ≠ (35/5) ≠ 7 x = (7/5) units. It implies; The square length is; 14-2x, and x = (7/5) units. = 14-2(7/5) = (70-14)/5 = 56/5 units. Therefore perimeter of...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2775407

Let r be the radius of the semi circle. Therefore; A right-angled triangle with hypotenuse 6 units, and adjacent height and base, (r-7) units and √(r²-49) units respectively is derived. It implies; 6²=(r-7)²+√(r²-49)² 36=r²-14r+49+r²-49 2r²-14r-36=0 r²-7r-18=0 (r-(7/2))²=18+(7/2)² (r-(7/2))²=(121/4) r = (7/2)±(11/2) r = ½(7±11) Therefore; r ≠ ½(-4) ≠ -2 r = ½(18) = 9 unit.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2774057

Let the side length of the regular pentagon be 1 unit. tan36 = a/0.5 a = 0.363271264 units. b² = 0.363271264²+0.5² b = 0.61803398875 units. tan72 = c/0.5 c = 1.53884176859 units. Area Green is; 0.5sin108+0.5*1.53884176859*0.5-0.5*0.5*0.363271264-0.5*0.61803398875*0.61803398875sin36 = 0.65716389015 square units. Calculating Area Blue. Let d be the radius of the inscribed blue circle. Calculating d. e = 0.5(90+36) e =63° (d/tan63) + (d/tan45) = a d = 0.3632...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2774044

Let r be the radius of the circle. (6√(2))²=r²+(6+r)² 72=r²+36+12r+r² 2r²+12r-36=0 r²+6r-18=0 (r+3)²=18+9 r = -3±3√(3) Therefore; r ≠ - 3 - 3√(3) r = (3√(3)-3) units. Therefore; Area triangle AOB is; Area triangle with two side 6√(2) unit each, and angle (45-atan((3√(3)-3)/(3√(3)+3)))° = ½(6√(2))²sin(45-atan((3√(3)-3)/(3√(3)+3))) = ½*72*½ = ¼(72) = 18 square units.

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