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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3096792

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Radius R of the big circle is 50 unit. (R-10)²=R(R-18) R²-20R+100=R²-18R 2R=100 R=50 units. Radius r of the small inscribed unshaded circle is 41 unit. 2r=2R-18, and R=50 2r=100-18 2r=82 r=41 units. Therefore shaded area will be; 50²π - 41²π = 819π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3096525

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Let the radius of the ascribed quarter circle be 2x. Therefore the radius of the inscribed yellow half circle is x. Calculating x. ¼(π(2x)²)-½(πx²) = 17 πx²-½(πx²) = 17 2πx²-πx² = 34 πx² = 34 x² = 34/π x = √(34/π) cm. Therefore, shaded yellow area is; ½(πx²) And x = √(34/π) cm. = ½(π√(34/π)²) = 17 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3096012

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a = 180-30-45 a = 105° (1/sin105) = (b/sin30) b = 0.51763809021 units. c = √(2)-b c = 0.89657547217 units. (0.89657547217/sin105) = (d/sin30) d = 0.46410161514 units. It implies, area green is; 0.5*0.46410161514*0.89657547217sin45 = 0.14711431703 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3096008

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a = (r+R) b = (R-r) It implies; 8²+b² = a² R²+2Rr+r² = 64+R²-2Rr+r² 4Rr = 64 Rr = 16 R = 16/r --- (1). It implies; a = r+(16/r) a = (16+r²)/r b = (16/r)-r b = (16-r²)/r Therefore; Calculating r. (2r)²+((16+r²)/r)² = 10²+((16-r²)/r)² 4r⁴+(16+r²)² = 100r²+(16-r²)² 4r⁴+r⁴+32r²+256 = 100r²+256-32r²+r⁴ 4r⁴+64r² = 100r² 4r² = 36 r² = 9 r = 3 units. Calculating R. At (1), R = 16/r. And r = 3 units. It implies; R = 16/3 units. Calculating x. x = 2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3095856

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Each of the triangles is equilateral with single side length; (√(24)÷3^(1/4)) units. Total yellow area will be; (√(24)÷3^(1/4))²x(3/4)x0.5sin60 + (√(24)÷3^(1/4))²x(2/3)x(0.5)²sin60 + (√(24)÷3^(1/4))²x(1/4)(1/3)x0.5sin60 = 4.5 + 2 + 0.5 = 7 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3095852

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Diameter of circle Z will be; ½(Diameter of the ascribed circle - length of the inscribed square) = ½(√(16²+16²) - 16) = ½(16√(2)-16) = (8√(2)-8) cm The radius of circle Z will be; ½ the diameter of circle Z. = ½(8√(2)-8) = (4√(2)-4) cm. Radius of circle X will be; 2(area triangle with height r cm and base 16 cm) + Area triangle with height r cm and base 16√(2) cm = Area triangle with height and base 16 cm respectively. 2*½*16r + ½*16√(2)r = ½*16² 16r + 8√(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3095850

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The required angle is; 180° - (11.25°+33.75°) = 180° - 45° = 135°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3095849

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Let the single side length of the regular pentagon be 4 unit. Therefore angle x degree will be; Sinx = (2/4) Sinx = ½ Therefore; x = sin–¹(½) x = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3095137

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Let R be radius of the quarter circle. Let r be radius of the inscribed circle. R will be; √(3²+4²) R=√(25) R= 5 units. Therefore r will be; (5-r)²=(3+r)²+r² 25-10r+r²=9+6r+r²+r² r²+16r-16=0 Using quadratic equation app to resolve the quadratic equation above. Therefore; r = 0.94427 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3093925

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