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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2548386

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Let the square side length be 2 units. Therefore, r, radius of the circle is; r = 1 unit. Calculating the required angle, let it be a. cosb = 1/2 b = acos(1/2) b = 60° c² = 2²+2² c = √(8) c = 2√(2) units. c is a diagonal of the square. d = ½(c) d = √(2) units. e = 45+60 e = 105° It implies; f² = √(2)²+1²-2√(2)cos105 f = 1.9318516526 units. (1.9318516526/sin105) = (1/sinh) h = 30° Therefore, the required angle, angle a is; a = h+45 And h = 30° It im...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2548180

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Notice! The side length of the inscribed biggest square is 5 units. Calculating inscribed red area. tana = 2.5/5 a = atan(½)° b = 2atan(½)° tanc = 5/2.5 c = atan(2)° d = 2atan(2)° Therefore, area red is; Area sector with radius 2.5 units and angle 2atan(2) - Area triangle with height 2.5 units and base 2.5sin(2atan(2)) units + Area sector with radius 5 units and angle 2atan(½) - Area triangle with height 5 units and base 5sin(2atan(½)) units - 2(area quarter circle wit...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2548054

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Let the side length of the three equal regular triangles be 2a units each. b² = 2a²-2a²cos120 b = √(3)a unit. (6√(7))² = (2a)²+(√(3)a)²-2*2a*√(3)acos90 252 = 4a²+3a² 252 = 7a² a² = 36 a = 6 units. It implies; 2a = 2*6 = 12 units. Where 12 units is the side length of the three equal regular triangles each. b = √(3)a And a = 6 units. b = 6√(3) units. Therefore, paint area is; 3(area triangle with height 6√(3) units and base 12 units + area triangle with height 6 u...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2547722

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Sir Mike Ambrose is the author of the question. Area R ÷ Area Square exactly in its simplest decimal form is; Area R is; ½(5x7.5) - ½(√(81.25)x√(65.25))sin(tan–¹(5/7.5) - tan–¹(3/7.5)) - ½((4.5)²xtan(tan–(3/7.5)) - ½(9sin(tan–¹(5/7.5))xsin(tan–(7.5/3))÷sin(180-(tan–¹(5.7.5))-(tan–¹(7.5/3)))) = 18.75 - 7.5 - 4.05 - 2.36842105263 = 4.83157894737 square unit. Area square is; 3² = 9 square unit. Therefore; Area R ÷ Area Square exactly in its simplest decimal form is; 4.83157...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2547705

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Sir Mike Ambrose is the author of the question. Let the square side be 1 unit. Therefore; Area B is; Area triangle with height 1 unit and base sin(53.13010235416) unit - Area triangle with height 1 unit and base (0.71428571429sin(53.13010235416)) unit. = ½*1(sin(53.13010235416)) - ½*1*(0.71428571429sin(53.13010235416)) = 0.4 - 0.28571428572  = 0.11428571428 square unit. Area A is; Area triangle with height 1 unit and base 0.5 unit + Area triangle with height 0.5√(5) unit...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2547271

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a*9 = 6*3 a = 2 units. b = ½(9+2) b = 5.5 units. c = ½(6+3) c = 4.5 units. d = c-3 d = 1.5 units. Calculating e, radius of the circle. e² = b²+d² e² = 5.5²+1.5² e = 5.7008771255 units. Again, e is the radius of the circle. Calculating green area. tanf = 5.5/1.5 f = atan(1.1/0.3)° g = 2f g = 2atan(1.1/0.3)° g = 149.4897625939° h = 5.5-2 h = 3.5 units. tanj = 4.5/3.5 j = atan(0.9/0.7)° k = 2j k = 2atan(0.9/0.7)° k = 104.2500326978° l² = 3²+2²...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2546667

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a²+8² = 10² a = 6 units. b = ½(10) b = 5 units. b is the radius of the ascribed semi circle. 6² = 5²+5²-2*5*5cosc 50cosc = 50-36 cosc = 14/50 c = acos(7/25)° c = 73.7397952917° d = 2c d = 147.4795905834° e² = 5²+5²-2*5*5cos147.4795905834 e = 9.6 units. f = ½(e) f = 4.8 units. cosg = 4.8/6 g = acos((24/5)/6) g = acos(⅘)° g = 36.8698976458° h = ½(g) h = ½(acos(⅘))° h = 18.4349488229° tan(½(acos(⅘))) = j/(4.8) j = ⅓*(24/5) j = ⅕(8) units. Where j is the...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2546616

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a² = 24²+7² a = 25 units. a is AC. b = 7+a b = 32 units. c² = b²+24² c = √(32²+24²) c = 40 units. c is the diameter of twice the quarter circle (semi circle (2AB)). d = ½(c) d = 20 units. d is AB, radius of the quarter circle. It implies, length BC is; (BC)²+20² = 25² BC = √(625-400) BC = √(225) BC = 15 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2546585

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(½(r))²+(3/2)² = r² (9/4) = r²-¼(r²) (9/4) = ¾(r²) r² = 3 r = √(3) units. Where r is the radius of the circle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2546114

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a² = 2²+2² a = √(8) cm. a = 2√(2) cm. b = 4+2 b = 6 cm. c = 2+6-4 c = 4 cm. d² = 6²+4² d = √(52) cm. d = 2√(13) cm. Therefore, shaded portion perimeter is; 4+4+2√(2)+6+2√(13) = 14+2√(2)+2√(13) = 2(7+√(2)+√(13)) cm. = 24.0395296757 cm.
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