Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2829677

Sir Mike Ambrose is the author of the question. Let the side of the regular pentagon be 1 unit. Therefore; Area purple is; Area trapezium with two parallel side 0.07294901687 units and 0.38196601125 units, and height cos18 units. = 0.5*cos18*(0.07294901687+0.38196601125) = 0.21632495093 space units. Area yellow is; Area trapezium with two parallel side 0.27345747199 units and 1 unit, and height 1 unit - Area triangle with height 0.27345747199 units and base 0.198678483 uni...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2829667

Notice; Green regular hexagon's side is 8 cm. sin60 = a/8 a = 4√(3) cm Where a is the the height of the white triangle enclosed by the two green regular hexagon and the regular pentagon. b = 2a b = 8√(3) cm. c = 5+b c = (5+8√(3)) cm. d = 8+4 d = 12 cm. (a) Calculating Radius² exactly. Let r be radius. It implies; r² = 12²+(5+8√(3))² r² = (361+80√(3)) cm² Therefore; Radius, r of the ascribed circle is; √(361+80√(3)) r = 22.3509298376 cm. (b) Ca...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2828072

Radius of the circle, r is; √(2-2cos120) r = √(3) units. Therefore! Area yellow is; Area regular hexagon with side 2 units - 2(area triangle with height ½ units and base √(3) units) - 2(area sector with radius √(3) units and angle 60°) = (24/(4tan30)) - 2(½*½√(3)) - 2(60π(√(3))²/360) = 6√(3) - ½√(3) - π = ½(12√(3) - √(3) - 2π) = ½(11√(3)-2π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2828064

Let the base of the triangle ascribing the green and yellow area b 1 unit. tan55 = a/1 a = 1.42814800674 units. tan55 = b/(1-b) b = 0.58816349035 units. Where b is the side of the white and yellow square. c = 1-b c = 0.41183650965 units. sin55 = d/0.58816349035 d = 0.4817953255 units. tan35 = 0.4817953255/e Where e is the side of the blue area. e = 0.68807503377 units. Area Blue is; e² = 0.68807503377² e = 0.47344725209 square units. sin35 = 1/f f = 1.743446...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2826551

Sir Mike Ambrose is the author of the question. Notice; Square side is 4 units. Small regular hexagon side is 2 units. Big regular hexagon side is 4 units. Therefore; Area shaded, exactly in square units decimal is; Area square with side 4 units - Half area small regular hexagon + Area triangle with height 2.53589838486 units and base 4√(3) units + Area triangle with height 4 units and base 4sin60 units. = 4² - ½(6√(3)) + ½(2.53589838486*4√(3)) + ½(4*4sin60) = 16 - 3√(3)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2826532

Let the regular octagon side be 2 units. Area Octagon is; ½(8*2)*(2/(2tan(180/8))) = 8/(tan(180/8)) = 19.31370849898 square units. Calculating Area Red. a = ⅛(180*6) a = 135° Let the square side be b. Calculating b. (b/tan60)+(b/tan45)+b = 2 b = 2/(1+(1/tan60)+(1/tan45)) b = 0.77599076226 unit. Area Red is; b² = 0.77599076226² = 0.60216166311 square units. Therefore; Area Octagon ÷ Area Red to 2 decimal places is; 19.31370849898 ÷ 0.60216166311 =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2825213

a² = R²+6² a = √(R²+36) units. a is OB. b = a-4 b = (√(R²+36)-4) units. c = 6+4 c = 10 units. It implies; 10² = R²+(√(R²+36)-4)² 100 = R²+R²+36-8√(R²+36)+16 48 = 2R²-8√(R²+36) 8√(R²+36) = 2R²-48 4√(R²+36) = R²-24 16(R²+36) = R⁴-48R²+576 16R²+576 = R⁴-48R²+576 16R² = R⁴-48R² R² = 64 R = 8 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2825199

Let the single side length of the two congruent inscribed regular pentagon be 1 unit. Area Green is; 0.5*5(1/(2tan(180/5))) = 1.72047740059 square units. Calculating Area Shaded. a = ⅕(180*3) = 108° Where a is the single interior angle of the congruent inscribed regular pentagon. b² = 2-2cos108 b = 1.61803398875 units. c = 1+b c = 2.61803398875 units. d² = 2.61803398875²+1²-2*1*2.61803398875cos72 d = 2.49721204096 units. Where d is the radius of the inscribed a...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2824411

Let a be the big inscribed circle radius. Let b be the bigger inscribed circle radius. Let c be the biggest inscribed circle radius. It implies; a+b = 12 --- (1). a+c = 22 --- (2). b+c = 26 --- (3). Therefore, resolving (1), (2) and (3) simultaneously to get a, b and c. Subtracting (2) from (3). (b+c = 26)-(a+c = 22) It implies; b-a = 26-22 b-a = 4 a = b-4 --- (4). Substituting (4) in (1) to get b. a+b = 12 And at (4), a = b-4 b-4+b = 12 2b = 12+4 b...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2823634

(x⁶-1)/(x⁴-x) = 2 ((x³+1)(x³-1))/(x(x³-1)) = 2 (x³+1)/x = 2 x³-2x+1 = 0 For x = 1 x³-2x+1 = 0 Therefore; (x-1) is a factor of x³-2x+1 Calculating the other values of x. (x³-2x+1)/(x-1) = (x²+x-1) Remainder 0 Solving x²+x-1 = 0, using completing the square method. (x+(½))² = 1+(¼) x = -½±½(√(5)) It implies; x = -½(1+√(5)) Or x = ½(√(5)-1) The values of x are; -½(1+√(5)), ½(√(5)-1)

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