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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2602498

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Sir Mike Ambrose is the author of the question. Please, move the question left or right to review the solution. Thank you!

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2602470

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Sir Mike Ambrose is the author of the question. Yellow Area exactly in square units is; ½*√(3)*(4√(3)-4) = ½(12-4√(3)) = 2(3-√(3)) square units. = 2.5358983849 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2601207

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Let the inscribed square side length be a. b²+1² = a² b = √(a²-1) units. c = 2b c = 2√(a²-1) units  d = 1+7 d = 8 units. Observing similar plane shape (right-angled) side length ratios. 8 - √(a²-1) 2√(a²-1) - 1 Cross Multiply. 2(a²-1) = 8 a²-1 = 4 a = √(5) units. Again, a is the side length of the inscribed square. Area inscribed square is; a² = √(5)² = 5 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2598670

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Mazana Job Shared The Questions. Number 1. a = 180-30-45 a = 105° It implies; (4/sin105) = (b/sin30) b = 2.0705523608 units. c = 45+90 c = 135° Therefore, the required length, ? is; ?² = 4²+2.0705523608²-2*4*2.0705523608cos135 ? = 5.6568542495 units. Number 2. Calculating yellow area. Let a be the radius of the inscribed quarter circle, and also, the side length of the ascribed square. tanb = a/(0.5a) b = atan(2)° sin(atan(2)) = c/4 c = 3.577708764...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2598600

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Let the side length of the regular octagon be a. b = ⅛*180(8-2) b = 135° b is the single interior angle of the regular octagon. 2c² = a² c = √(a²/2) c = ½(√(2)a) units. c = 0.7071067812a units. c is AQ = BQ  d = a+c d = 1.7071067812a units. d is GQ. tane = 1.7071067812a/a e = 59.6388065954° e is angle GFQ. f = e-45 f = 14.6388065954° f is angle GFP. g = 180-(90-f) g = 104.6388065954° g is angle FPQ. h = 90-e h = 30.3611934046° h is angle FQG. cos14.6388...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2598199

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r2 will be; Let it be x. Therefore; (4+x)²=(8-x)²+(4-x)² 16+8x+x²=64-16x+x²+16-8x+x²  x²-32x+64=0 (x-16)²=-64+256 x = 16±8√(3) x ≠ 16+8√(3) x = 16-8√(3) Therefore; x = r2 = 8(2-√(3)) unit.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2597388

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The radius, r if the small circle will be; (5+r)²=2(5-r)² 25+10r+r²=50-20r+2r² Therefore r²-30r+25=0 (r-15)²=-25+225 r = 15±√(200) r = 15±10√(2) It implies; r ≠ 15+10√(2) r = 15-10√(2) unit. Therefore area blue will be; ¼(100-25π) - (15-10√(2))²π = 3.05304958693 square unit.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2597040

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Sir Mike Ambrose is the author of the question. Notice! AB = BC Calculating AB. AB = (20sin60)/sin104 AB = BC = 17.85075169007 cm. Calculating BD BD = (20sin16)/sin104 BD = 5.68151231326 cm. Calculating CE. CE = (17.85075169007sin44)/sin60 CE = 14.31848768674 cm. Therefore; Half DE is; ½(CE=DE) = ½(14.31848768674) = 7.15924384337 cm. It implies; Area Blue (Big) is; Area triangle with height 7.41361402363 cm and base 7.15924384337 cm. = ½(7.15...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2595419

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Let the side length of the inscribed green square be a. b = ½(10-a) units. c = b+a c = ½(10-a+2a) c = ½(10+a) units. Therefore, a² (area green inscribed triangle) is; 10² = a²+(½(10+a))² 100 = a²+¼(100+20a+a²) 400 = 4a²+100+20a+a² 5a²+20a-300 = 0 a²+4a-60 = 0 Resolving the above quadratic equation via factorization approach to get a, side length of the inscribed green square. a²+10a-6a-60 = 0 a(a+10)-6(a+10) = 0 (a+10)(a-6) = 0 It implies; a ≠ -10 a = 6 units....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2595378

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Let OC be a. Let the radius of the inscribed blue half circle be b. Calculating area inscribed blue half circle. 20² = 2c² c = √(200) c = 10√(2) units. d = ½(c) d = 5√(2) units. d is OD. Observing similar plane shape (right-angled) side length ratios to get a (OC). a - 5√(2) 5√(2) - 10√(2) Cross Multiply. 10√(2)a = 50 a = ½(5√(2)) units. e = b-a e = ½(2b-5√(2)) units. It implies; b² = (½(2b-5√(2)))²+(5√(2))² b² = ¼(4b²-20√(2)b+50)+50 b² = b²-5√(2)b+(2...
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