Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3223465

Calculating x. Let r be the radius of the inscribed green circle. Let R be the radius of the ascribed quarter circle. (R-r)² = r²+5² --- (1). (R-r)²+5² = a² --- (2). a² = x²+r² --- (3). Substituting (3) in (2). (R-r)²+5² = x²+r² (R-r)² = x²+r²-5² --- (4). Substituting (4) in (1). x²+r²-5² = r²+5² x²-5² = 5² x² = 5²+5² x² = 25+25 x² = 50 x = √(50) x = 5√(2) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3223450

Calculating radius of the semi circle. Radius, r of the semi circle is; Cos(2atan(1/(4+√(7))))=(5+√(7))/d Notice; d = 2r Therefore, d = (5+√(7))/Cos(2atan(1/(4+√(7)))) d = 8 units. And d = 2r 2r = 8 r (radius semi circle) is; 4 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3223441

Calculating Area Blue Fraction. Let the radius of inscribed blue semi circle be 2 units. Therefore the radius of the inscribed blue quarter circle will be 2 units. The length of the ascribed rectangle will be 4 units. The width of the ascribed rectangle will be; Let the width be x. x²+2²=4² x²=16-4 x = √(12) x = 2√(3) units. Area ascribed rectangle is 4*2√(2) = 8√(3) square units. Area blue is; Area quarter circle with radius 2 units + Area semi circ...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3222910

Calculating Area Quadrilateral ABCD. Notice. A right-angled triangle is derived considering quadrilateral ABCD with interior angles 40°, 50° and 90°. AD = 1 unit. BC = 3 units. Let AB = CD = x. sin40 = a/3 a = 3sin40 units. b = (3sin40-x) units. sin50 = c/3 c = 3sin50 units. d = (3sin50-x) units. Calculating x. (3sin50-x)²+(3sin40-x)² = 1² 9sin²50-6xsin50+x²+9sin²40-6xsin40+x² = 1 2x²+9-8.45299231683x = 1 2x²-8.45299231683x+8 = 0 x²-4.22649615842x+...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3222173

Sir Mike Ambrose is the author of the question. Calculating Area R ÷ Area large square. Let the single side length of the large square be 2 units. Therefore; Area large square is; 2 x 2 = 4 square units. Calculating Area R. It will be; Area trapezoid of two parallel lengths (√(2)-1+2) units and height (3-√(2)) units + Area rectangle of length 2 units and width (√(2)-1) units - 2( area triangle of height (√(2)-1) units and base 1 unit) - Area sector of radius 1 unit an...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3222166

Calculating BD/AC. Let 1 unit be AB = AD. a = 180-30-30 a = 120° a is angle BAD. b² = 2-2cos120 b = √(2-2cos120) b = √(3) units. b is length BD. c = 180-15-45 c = 120° c is angle BCD. (√(3)/sin120) = (d/sin45) d = √(2) units. d is length BC. e² = 1²+√(2)²-2*√(2)*1cos(30+15) e² = 3-2 e = √(1) e = 1 unit. e is length AC. It implies; Length BD÷Length AC is; √(3)÷1 = √(3)

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3221444

Calculating shaded yellow area. Let x be the side length of the ascribed regular decagon. Let y be the side length of the two equal inscribed regular pentagons. a = ⅒*180(10-2) a = 144° a is the single interior angle of the ascribed regular decagon. b = ⅕*180(5-2) b = 108° b is the single interior angle of the both equal inscribed regular pentagons. c = (1+y) units. It implies; (1+y)² = 2x²-2x²cos144 y = 1.90211303259x-1 --- (1). d² = 2y²-2y²cos108 d = 1.61803398...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3221344

Calculating x, the required angle. Let the side length of the regular pentagon be 1 unit. a = ⅕*180(5-2) a = 108° a is the single interior angle of the regular pentagon. b² = 1²+0.5²-2*1*0.5cos108 b = 1.24860602048 units. (1.24860602048/sin108) = (1/sinc) c = 49.6138224406° d = 180-108-c d = 72-49.6138224406 d = 22.3861775594° e = 180-c-d e = 180-49.6138224406-22.3861775594 e = 108° e = a. (0.5/sin108) = (f/sin49.6138224406) f = 0.40044657145 units. g = b-f g...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3220781

Calculating the required area. Let it be y. a = x cm² b = (50-x) cm² c = 60-b c = 60-(50-x) c = (10+x) cm² d = 40-c d = 40-(10+x) d = (30-x) cm² It implies; y = d+x y = (30-x)+x y = 30 cm² Again, y is the required area. Or 40+50 = 60+y y = 90-60 y = 30 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3220773

Calculating y, the red required angle. It implies; ½*πr² = 32π r = √(64) r = 8 cm. r is the radius of the half circle. 2r = 16 cm, the diameter of the half circle. Let x be the radius of the inscribed circle. a = (8-x) cm. Calculating x. 2x² = (8-x)² 2x² = 64-16x+x² x²+16x-64 = 0 (x+8)² = 64+(8)² (x+8)² = 128 x = -8±√(128) x = -8±8√(2) It implies; x ≠ (-8-8√(2)) cm. x = (8√(2)-8) cm. x = 3.31370849898 cm. Again, x is the radius of the inscribed circle....

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Mathematics Question and Solution

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Mathematics Question and Solution

Mathematics Question and Solution

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