Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2571545

Notice! The ascribed polygon is a regular nonagon. The inscribed polygon is a regular hexagon. Calculating the required angle, x. a = ⅙*180(6-2) a = 120° a is the single interior angle of the regular hexagon. b = ⅑*180(9-2) b = 140° b is the single interior angle of the regular nonagon. c = ½(360-140+140) c = ½(360-280) c = ½(80) c = 40° d = ⅕*180(5-2) d =108° d is the single interior angle of the regular pentagon. e = 5d e = 540° e is the sum of interior angles of a regular pentagon....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2571525

Sir Mike Ambrose is the author of the question. Let the side of the regular nonagon be 1 unit. Therefore; Length Green is; ((1+2sin50)sin20/sin80) + √(1.08506357513²+1²) = 0.87938524157 + 1.47558902208 = 2.35497426365 units. Length Blue is; √(1.1371580426²+1²-2(1.1371580426)cos100) = 1.63953019414 units. It implies; Length Blue ÷ Length Green to 2 decimal places is; 1.63953019414 ÷ 2.35497426365 = 0.69619877357 ≈ 0.70

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2569608

a = 6 units. a is the side length of the square. Therefore; Area square is; 6*6 = 36 square units. Area inscribed shaded triangle is; ½(6*6) = 18 square units. It implies; The percentage of the square that is shaded is; (18/36)*100 = 50%

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2569002

Let the single side length of the square be (x+10) unit. Calculating x. 10²=(5+x)²+5² 100-25-25=10x+x² Therefore; x²+10x-50=0 Resolving the above quadratic equation via completing the square approach to get x. (x+5)²=50+25 x =-5±√(75) Therefore; x ≠ -5-5√(3) x = (5√(3)-5) cm. Square length is; (x+10) And x = (5√(3)-5) cm. = 10+5√(3)-5 cm. = (5+5√(3)) units. Area of the square is; (5+5√(3))² = 100+50√(3) = 50(2+√(3)) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2563885

a = 6+4 a = 10 cm. b²+6² = 10² b = √(100-36) b = √(64) b = 8 cm. c = b+4 c = 12 cm. d = c-r d = (12-r) cm. r is the ascribed half circle radius. e = 6+d e = 6+(12-r) e = (18-r) cm. Calculating r, radius of the ascribed half circle. r² = (18-r)²+6² r² = 324-36r+r²+36 36r = 360 r = 10 cm. Area Painted is Area half circle with radius 10 cm - Area half circle with radius 6 cm - Area half circle with radius 4 cm. = ½*π*10²-½*π*6²-½*π*4² = ½(100π-36π-16π) =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2562085

Equation of the curve is; y = ½(x^(3/2)) It implies; PR = √(4²+6²) PR = 2√(13) units QR = (2√(13))/(cos(90-atan(⅔))) = 13 units. Therefore; Shaded Area is; (Area under the curve at x = 4 and x = 10) - Area trapezium with parallel sides 4 units and 13 units, and height 6 units. = 20√(10) - (32/5) - ½*6(4+13) = 20√(10) - (32/5) - 51 = (100√(10)-287)/5 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2558608

Let the inscribed square side be a. b² = 2a² b = √(2)a cm. b is the diagonal of the inscribed square. c = 90+45 c = 135° It implies, calculating a, the inscribed square side length. 10² = 5²+(√(2)a)²-2*5*√(2)acos135 100 = 25+2a²+10a 2a²+10a-75 = 0 Resolving the above quadratic equation via completing the square approach to get a, the side length of the inscribed square. (a+(5/2))² = (75/2)+(5/2)² (a+(5/2))² = ¼(150+25) (a+(5/2))² = ¼(175) a = -(5/2)±√(¼(175)) a = -(5...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2556640

sin30 = 10/a a = 20 units. a is the side length of the square. sin30 = b/10 b = 5 units. Therefore; Area Yellow is; Area triangle with height 20 units and base 5 units. = ½*20*5 = ½(100) = 50 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2554627

Notice! BD is the diameter of the circle. a = ½(BD) a = ½(14) a = 7 cm. a is OB = OD. a is the radius of the circle. b = 2(45) b = 90° b is angle AOB. Calculating shaded area. It is; Area quarter circle with radius 7 cm - Area triangle with height and base 7 cm respectively. = ¼(7*7)π-½(7*7) = ¼(49π)-½(49) = ¼(49π-98) = ¼(49(π-2)) square units. = 13.9845100065 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2553292

Let a be the radius of the half circle. a²+8² = b² b = √(a²+64) units. b is OC. Observing similar plane shape (right-angled) side length ratios. 8 - c √(a²+64) - a Cross Multiply. c = 8a/√(a²+64) units. Again; √(a²+64) - a a - d Cross Multiply. d = a²/√(a²+64) units. It implies; e = (√(a²+64)+a) units. e is AC. f = ((8a/√(a²+64))+8) units. f is CH. g = (a+(a²/√(a²+64))) g is AH. It implies; Observing Pythagoras Rule on Triangle AHC. (√(a²+64)+a)² =...

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Mathematics Question and Solution

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Mathematics Question and Solution

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Mathematics Question and Solution

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