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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2586274

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6² = a²+(2+3)² a² = 36-25 a = √(11) units. b² = 3²+c² b = √(9+c²) units. b is the diameter of the circle. d = c-a d = (c-√(11)) units. e² = 2²+(c-√(11))² e = √(4+c²-2c√(11)+11) e = √(c²-2c√(11)+15) units. Therefore; √(9+c²)² = 6²+√(c²-2c√(11)+15)² 9+c² = 36+c²-2c√(11)+15 2√(11)c = 51-9 2√(11)c = 42 √(11)c = 21 c = (21√(11))/11 units. c = 6.3317382361 units. Notice! PQ = e = √(c²-2c√(11)+15) PQ = √((6.3317382361)²-2(6.3317382361)√(11)+15) PQ = 3.6181361349...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585818

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a = 0.5*5*5sin40 a = 8.0348451211 square units. (b/sin40) = (5/sin70) b = 3.4202014333 units.   c = 0.5*3.4202014333*3.4202014333sin40 c = 3.7595933295 square units. Shaded Area is: a-c = 8.0348451211-3.7595933295 4.2752517916 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585632

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sin60 = 0.5d/a a = ⅓√(3)d units. a is the radius of the half circle. Therefore, total blue area in terms of d is; 2(60π*⅓√(3)d*⅓√(3)d/360)+(0.5*⅓√(3)d*⅓√(3)dsin120)-(0.5*⅓√(3)d*⅓√(3)dsin60) = 2(πd²/18) = ⅑πd² square units.l

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585623

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a = ½(d) units. a is the radius of the green half circle. b = d+a b = ½(3d) units. c² = (½(3d))²-(½(d))² c² = ¼(8d²) c = √(2(d)²) c = √(2)d units. c is the diameter of the small blue semi circle. d = ½(c) d = ½√(2)d units. d = 0.7071067812d units. d is the radius of the small blue semi circle. tane = ½(d)/√(2)d e = atan((½)/√(2)) e = atan(¼√(2)) f² = (3d)²+(√(2)d)²-2*3d*√(2)dcos(0.25√(2)) f = 1.5858373728d units. f is the diameter of the big semi circle. g = ½(f...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585604

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Notice! Triangle ABM is equilateral. Calculating a:b. Let AB be 1 unit. c = 90-60 c = 30° cos30 = (0.5*1)/d d = ½÷½√(3) d = 1/√(3) d = ⅓√(3) units. d is the radius of the semi circle. e = 2d e = ⅔√(3) units. e is d, diameter of the semi circle. Therefore; a : b is; 1:⅔√(3) = 3:2√(3) = 3/(2√(3)) = ½√(3)

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585239

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a = ⅓(90) a = 30° sin30 = b/4 b = 2 cm. sin60 = c/4 c = 2√(3) cm. d = √(4²-2²)-√(4²-(2√(3))²) d = √(12)-√(4) d = (2√(3)-2) cm. d = 1.4641016151 units. Area Blue is; Area trapezoid with parallel sides 2 cm and 2√(3) cm respectively and height (2√(3)-2) cm + Area sector with radius 4 cm and angle 30° - Area triangle with height 4 cm and base 4sin30 cm. = (½*(2+2√(3))*(2√(3)-2))+(30π*4*4/360)+(½*4*4sin30) = (½*8)+⅓(4π)-4 = 4+⅓(4π)-4 = ⅓(4π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585177

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Let a be 1 unit. b = 1+(2*1)+1 b = 4 units. c = ½(b) c = 2 units. 2d² = 4² d² = 8 d = 2√(2) units. e² = 1²+(2√(2))²-2*2√(2)cos45 e = √(9-4) e = √(5) units. (√(5)/sin45) = (1/sinf) f = 18.4349488229° g = 90-2f g = 53.1301023542° h = ½(g) h = 26.5650511771° J² = 2k² j = √(2)k units. k is the radius of the circle. l = (2-√(2)k) units. k² = 1²+(2-√(2)k)² k² = 1+4-4√(2)k+2k² k²-4√(2)k+5 = 0 Calculating k via completing the square approach. (k-2√(2)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585127

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Let the side length of the inscribed square be a. Calculating the area of the inscribed red square. 2b² = a² b = ½√(2)a unit. c² = 2a² c = √(2)a unit. c is the diagonal of the inscribed red square. It implies; 10² = (½√(2)a)²+(√(2)a)² 100 = (a²/2)+2a² 100 = ½(5a²) 5a² = 200 a² = 40 a = 2√(10) units. Therefore, area red inscribed square is; a² = 40 square units. Again, a = 2√(10) units is the side length of the inscribed red square.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585091

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Calculating area of the inscribed blue half circle. a = ½(6+4) a = 5 units. a is the radius of the ascribed semi circle. b = a-4 b = 1 unit. c²+1² = d² d = √(c²+1) units. c is the radius of the inscribed blue half circle. 5² = c²+√(c²+1)² 25 = c²+c²+1 2c² = 24 c² = 12 c = 2√(3) units. Again, c is the radius of the inscribed blue half circle. Therefore; Area inscribed blue half circle is; ½(c²)π = ½*2√(3)*2√(3)π = 6π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2584218

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Notice! Triangle ABC is equilateral. Let AB be 1 unit. a = 180-42-12 a = 126° a is angle BDC. (1/sin126) = (b/sin42) b = 0.8270909153 units. b is CD. c² = 1+0.8270909153²-2*0.8270909153cos48 c = 0.759747123 units. c is AD. Therefore, the required angle, angle CAD is; Let it be d. (0.759747123/sin48) = (0.8270909153/sind) d = 54° Again, d is the required angle, angle CAD.
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