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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2646670

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Let the three equal lengths of the quadrilateral be 1 unit each. Therefore; a² = 1²+1²-2cos100 a = 1.5320888862 units. a is BD. b = 60-½(180-100) b = 60-40 b = 20° b is angle ABD. c² = 1²+1.5320888862²-2*1.5320888862cos20 c = 0.6840402866 units. c is AD. Therefore, the required angle, angle BAD (?) is; Let it be e. (0.6840402866/sin20) = (1.5320888862/sind) d = 50° It implies; e = 180-d e = 130° Again, e is the required angle, angle BAD.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2646650

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Calculating Area of the Circle. Let a be the radius of the circle. Calculating a. a² = 8²+6²-2*8*6(-cosb) a² = 100+96cosb --- (1). a² = 8²+16²-2*8*16cosb a² = 320-256cosb --- (2). Equating (1) and (2) to get b. 100+96cosb = 320-256cosb 352cosb = 220 b = acos(220/352) b = 51.3178125465° Recall. a² = 320-256cosb at (2). And b = 51.3178125465° It implies; a² = 320-256cos51.3178125465 a = 12.6491106407 units. Or Recall again. a² = 100+96cosb at (1) And...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2646614

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Shaded Area is; Area sector with radius 1 unit and angle 2atan(½)° - Area triangle with two lengths 1 unit each and angle 2atan(½)°, the angle they form + Area sector with radius ½ units and angle 2atan(2)° - Area triangle with two lengths ½ units each and angle 2atan(2)°, the angle they form. = (2atan(½)π/360)-(½sin(2atan(½)))+(2atan(2)*0.25π/360)-(½*0.25sin(2atan(½))) = 0.063647609+0.17678717945 = 0.24043478845 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2646597

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Area triangle ABC is; ½(3*(36/5)) = (54/5) square units. Calculating Area Yellow. The point coordinates are; ((54/23), (36/23)) (3, 2) ((3/2), (18/5)) Therefore Area Yellow is; = (39/46) square units. Therefore; Area Yellow ÷ Area Triangle ABC is; (39/46)÷(54/5) = 65/828

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2646585

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Area Triangle Blue (S2) is; ½(12) = 6 square units. Area Triangle Yellow (S1) is; ½(1.4)sin(atan(4/3)) = (14/25) square units. Therefore; S1/S2 = (14/25)/6 = 7/75

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2646580

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Sir Mike Ambrose is the author of the question. Let the single side length of the square be 2 units. Therefore;  Area square is; 2*2 = 4 square units. Area red is; Area triangle with height (5/3) units and base (5/2) units + Area parallelogram with base (5/2) units and height (1/3) units. = ½((5/3)*(5/2)) + ((5/2)*(1/3)) = ½(25/6) + (5/6) = (25/12)+(5/6) = (35/12) square units. It implies; Area Red ÷ Area Square in its exact simplest form is; (35/12) ÷ 4 = 35/48...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2646329

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The area of the flower is; 8(area quarter circle with radius 10 cm - area of triangle with height and base, 10 cm respectively) = 8(¼(100π)-½(100)) = 8(25π - 50) cm² = 200(π-2) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2645043

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Let the single side length of the regular hexagon be 1 unit. Therefore; Area Regular Hexagon is; (6*1)/(4tan(180/6)) = ½(3√(3)) square units. Area Blue/Shaded is; 2(area triangle with two equal length ⅓(2√(3)) units and angle 120°, the angle they form). = 2(½(⅓(2√(3)))²sin120) = 2(½(12/9)sin120) = ⅓(2√(3)) square units. Therefore shaded/blue area fraction is; Area Blue ÷ Area Regular Hexagon. = ⅓(2√(3)) ÷ ½(3√(3)) = (⅔)÷½(3) = 4/9

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2645033

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Area Yellow is; Area quarter circle with radius 6 cm - Area sector with radius 6 cm and angle 20° - Area triangle with two equal length 6 cm and angle 40°, the angle they form - Area sector with radius 6 cm and angle 40° + Area triangle with two equal length 6 cm and angle 40°, the angle they form. = ¼(36π) - (20*36π÷360) - ½(36sin40) - (40*36π÷360) + ½(36sin40)  = 9π - 2π - 4π = 3π cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2645027

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Calculating R, radius of the quarter circle. ¼πR²=64π² R = 16√(π) cm Calculating x, radius of the green inscribed semi circle. (2x)²+x²=(16√(π))² Therefore; 5x² = 256π x = (16√(5π))/5 cm It implies; Area green inscribed semi circle is; ½π((16√(5π))/5)² = (256π²)/10 cm² Calculating y, radius of the blue inscribed semi circle. (2y)²+(y+x)²=(16√(π))² Therefore; 4y²+y²+(32√(5π)y)/5+(256π)/5=256π 25y²+32√(5π)y-1024π= 0 y = √(25856π/625)- √(256*5π/625) y = 8.863752...
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