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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2364184

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Let the side of the inscribed square be a. Calculating a. tan60 = a/b √(3) = a/b b = a/√(3) units. It implies; b+b+a = 10 2(a/√(3))+a = 10 ⅓(2√(3)a)+a = 10 2√(3)a+3a = 30 (2√(3)+3)a = 30 a = 30/(2√(3)+3) units. a = 4.6410161514 units. c = ½(a) c = ½(30/(2√(3)+3)) c = (15/(2√(3)+3)) units. c = 2.3205080757 units. It implies; b = a/√(3) b = 4.6410161514/√(3) b = 2.6794919243 units. d = a+b d = 4.6410161514+2.6794919243 d = 7.3205080757 units. Therefore, the required angle, x is; tany =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2364009

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Let the length of the ascribed regular hexagon be 2 units. Therefore it area will be; (2*2*6)/(4tan(180/6)) = 6/tan30 = 6√(3) square units. The green areas is; Area rectangle with length 2√(3) units and width 1 unit + Area regular hexagon with side ⅔√(3) units + Area regular hexagon with side ⅓√(3) unit. = (2√(3)*1) + (6*(⅔√(3))²/(4tan(180/6)) + (6*(⅓√(3))²/(4tan(180/6)) = 2√(3) + 2√(3) + ½(√(3)) = 4√(3) + ½(√(3)) = ½(9√(3)) square units. It implies; Shaded fraction is; Area Green ÷ Are...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2363725

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Let the side of the ascribed regular hexagon be 2 units. Area regular hexagon is; (2*2*6)/(4tan(180/6)) = 6/tan30 = 6√(3) square units. Area green is; Area kite with perpendicular lengths 4 units and ⅓(4√(3)) units respectively + Area kite with perpendicular lengths 2 units and ⅓(2√(3)) units respectively. = ½(4*⅓(4√(3))) + ½(2*⅓(2√(3))) = ⅓(8√(3)) + ⅓(2√(3)) = ⅓(10√(3)) square units. It implies; Shaded fraction is; Area Green ÷ Area Regular Hexagon. = ⅓(10√(3)) ÷ 6√(3) = (10√(3))÷(18√...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2363133

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a = asin(1/6) b = 2a b = 2asin(1/6) Where b is angle BAC cos(2asin(1/6)) = c/6 c = ⅓(17) units. Where c is AD. d = 6-c d = 6-⅓(17) d = ⅓ units Where d is CD. sin(2asin(1/6)) = e/6 e = 1.97202659437 units. Where e is BD. f = asin((1/3)/2) f = 9.59406822686° Where f is angle EBC = angle CAE. tan(9.59406822686) = g/(17/3) g = 0.95784148869 units. Or  g = 17√(35)/105 units. Where g is DE.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2361981

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a = acos(3/5) Where a is angle GCA. tan(acos(3/5)) = b/3 b = 4 units. Where b is AG. c = atan(⅔) Where c is angle GBE. d = 180-acos(3/5)-atan(⅔) d = 93.17983011986° Where d is angle BEC. (6/sin93.17983011986 ) = (e/sin(acos(⅗))) e = 4.80740170062 units. Where e is BE. Area Parallelogram BCDE is; 2(area triangle with height 4.80740170062 units and base 6sin(atan(⅔)) units. = 2(0.5*4.80740170062*6sin(atan(⅔))) = 16 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2361212

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Let the inscribed square side be a. tan60 = a/b a = √(3)b cm. Where b is BS = CR It implies; b+√(3)b+b = 10 (2+√(3))b = 10 b = 10(2-√(3)) cm. Therefore; a = √(3)*10(2-√(3)) a = 10(2√(3)-3) cm. a = 4.64101615138 cm. sin60 = 10(2√(3)-3)/c c = 20(2-√(3)) cm. Where c is CQ. d² = 10²+(10(2-√(3)))²-20*10(2-√(3))cos60 d = 8.96575472168 cm. Where d is AS. (8.96575472168/sin60) = (10(2-√(3))/sine) e = 15° Where e is angle PAS. f = 180-60-15 f = 105° Where f is angle BSA. g = 105-90 g = 15° Whe...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2360210

mathematics-question-and-solution_1707324361DrfE2d.jpeg
Let the inscribed square side be a. tan60 = a/b a = √(3)b cm. Where b is BS = CR It implies; b+√(3)b+b = 10 (2+√(3))b = 10 b = 10(2-√(3)) cm. Therefore; a = √(3)*10(2-√(3)) a = 10(2√(3)-3) cm. sin60 = 10(2√(3)-3)/c c = 20(2-√(3)) cm. Where c is CQ. d² = 10²+(10(2-√(3)))²-20*10(2-√(3))cos60 d = 8.96575472168 cm. Where d is AS. (8.96575472168/sin60) = (10(2-√(3))/sine) e = 15° Where e is angle PAS. f = 180-60-15 f = 105° Where f is angle BSA. g = 105-90 g = 15° Where g is angle PSA. ta...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2359383

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Let a be the side of the inscribed square. Calculating a. tan60 = a/b b = a/(tan60) cm. Where b is BS = CR. Therefore; 2b+a = 10 2a/(tan60)+a = 10 a((2/tan60)+1) = 10 a = 10(2√(3)-3) cm. a = 4.64101615138 cm. c = ½(10-a) c = 0.5(10-4.64101615138) c = 2.67949192431 cm. Where c = b. d = c+a d = 2.67949192431+4.64101615138 d = 7.32050807569 cm. Where d is BR. e = atan(4.64101615138/7.32050807569) e = 32.37365912705° It implies; Calculating TI, let it be f. tan32.37365912705 = f/g g = f/...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2359159

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a² = 6²-1² a = √(35) units. b = atan(√(35)) cos(atan(√(35))) = c/2 c = ⅓ unit. Where c is CE. d = atan(√(35))-60 d = 20.40593177314° Where d is angle FCE. tan20.40593177314 = e/(1/3) e = 0.12400483955 units. Where e is EF.  f = 6-(1/3) f = ⅓(17) units. Where f is AE. Therefore; Length AF is; √((⅓(17))²+(0.12400483955)²) = 5.66802331605 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2358759

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Sir Mike Ambrose is the author of the question. Let the side of the regular pentagon be 1 unit. It implies; Area red is; ½*sin54*cos54 = 0.23776412908 square units. Area orange is; ½*((sin54-cos54sin72)/sin72)((sin54sin72sin108+sin²54)/sin72sin108) = 0.20143700267 square units. Area green is; ½*((sin54)⁴/(sin108(sin72)²)) = 0.24898982848 square units. Therefore; Area Red ÷ (Area Green - Area Orange) is; 0.23776412908 ÷ (0.24898982848 - 0.20143700267) = 0.23776412908 ÷ 0.04755282581 = 5....
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