Mathematics Question and Solution

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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...

Mathematics Question and Solution

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Shaded red area is; Area quarter of radius 10 units - Area sector of radius 10 units and angle 30° - Area equilateral triangle of single side length 5 units - Area sector of radius 5 units and angle 120°. = ¼(100π)-(100π)/12-¼(25√(3))-⅓(25π) = 25π - ⅓(50π) - ¼(25√(3)) = (300π-200π-75√(3))/12 = (100π-75√(3))/12 square units. = 15.3546212326 square units.

Mathematics Question and Solution

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Let the radius of the quarter circle be 2 units. Radius, r of the small inscribed semi circle is; (2+r)²=1+(2-r)² r = ⅔ units. Calculating angle x; ½(atan(3/4)) = 18.43494882292°

Mathematics Question and Solution

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a = ½(360-90) a = ½(270) a = 135° It implies; b² = 1²+√(2)²-2√(2)cos135 b = √(5) units. Calculating c, radius of the quarter circle. 2c² = b² 2c² = √(5)² c = √(5/2) units. Again, c is the radius of the quarter circle. It implies, area yellow (quarter circle) is; ¼(πc²) = ¼*√(5/2)²*π = ⅛(5π) square units. = 0.625π square units.

Mathematics Question and Solution

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Green Area is; Area square with length 4 units - Area quarter circle with radius 2 units - Area square with length 2 units - 3(Area square with length 1 unit) - Area square with length 3 units + Area quarter circle with radius 3 units - Area square with length 2 units + Area quarter circle with radius 2 units - Area quarter circle with radius 1 unit. = 16 - ¼(4π) - 4 - 3 - 9 + ¼(9π) - 4 + ¼(4π) - ¼(π) = 16 - 20 + ¼(9π - π) = ¼(8π) - 4 = 2π - 4 = 2(π-2) square units.

Mathematics Question and Solution

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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 circle with radius 2 units. = ¼(4π)+½(4π) =...

Mathematics Question and Solution

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Radius of the ascribed quarter circle is; 1+1 = 2 cm. a² = 1²+2² a = √(5) cm. Observing similar plane shape (right-angled) side length ratios. √(5) - 4 1 - b Cross Multiply. b = ⅕(4√(5)) cm. c²+b² = a² c² = √(5)²-(⅕(4√(5)))² c² = 5-(80/25) c = ⅕√(45) cm. Therefore, area inscribed green quadrilateral is; Area triangle with height 1 cm and base 2 cm + Area triangle with height ⅕(4√(5)) cm and base ⅕√(45) cm. (½*1*2)+(½*⅕√(45)*⅕(4√(5))) = 1+ ⅕(6) = 2⅕ cm² =...

Mathematics Question and Solution

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

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...

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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2593252

Calculating x (length OD). a = (x+4) units. a is the radius of the quarter circle (length OA = length OB). It implies; (x+4)² = 8²+x² x²+8x+16 = 64+x² 8x = 48 x = 6 units. Again, x is required length OD.

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