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)

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2823616

Calculating the area of the red rectangle. Let a be the radius of the half circle. b = ½(a) units. Calculating a. a² = (½(a))²+1² a² = ¼(a²)+1 (3/4)a² = 1 a² = 4/3 a = √(4/3) a = ⅓(2√(3)) units. a = 1.1547005384 units. Again, a is the radius of the half circle and also the width of the red rectangle. Notice! b = ½(a) b = ½(1.1547005384) b = 0.5773502692 units. c = a+b c = 1.7320508076 units. c = √(3) units. c is the length of the red rectangle. Therefore, a...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2822228

Calculating red area. a = (1+x) units. a is the diameter of the circle. b = ½(a) b = ½(1+x) units. b is the radius of the circle. c = ½(1-x) units. Calculating x. (½(1+x))² = (½(1-x))²+(½(1-x))² ¼(1+2x+x²) = ¼(1-2x+x²)+¼(1-2x+x²) 1+2x+x² = 1-2x+x²+1-2x+x² 1+2x+x² = 2-4x+2x² x²-6x+1 = 0 (x-3)² = -1+(-3)² (x-3)² = 8 x = 3±√(8) It implies; x ≠ 3+√(8) units. x = 3-√(8) units. x = 0.1715728753 units. Recall. b = ½(1+x) units. And x = 0.1715728753 units. b = ½(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2822218

Calculating x, the side length of the ascribed square. tana = 14/7 a = atan(2)° b = 90+a b = (90+atan(2))° c² = 14²+7² c² = 196+49 c = √(245) units. c = 7√(5) units. d² = √(245)²+9²-2*9*7√(5)cos(90+atan(2)) d = 24.0416305603 units. d is the diagonal of the square. Therefore, x is; d² = 2x² 24.0416305603² = 2x² x² = 289 x = √(289) x = 17 units. Again, x is the side length of the ascribed square.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2820591

a = 8+2 a = 10 units. a is OA, the radius of the half circle. b = 2a b = 20 units. b is AB, the diameter of the half circle. c = 2+a c = 12 units. c is BC. Notice. BC = BD = 12 units. d²+12² = 20² d² = 400-144 d = √(256) d = 16 units. d is AD. tane = 16/12 e = atan(4/3)° e is angle ABD. f = ½(e) f = ½(atan(4/3))° f is angle ABF Therefore, the required length, BF is; Let it be g. cos(½(atan(4/3))) = g/20 g = 20cos(½(atan(4/3))) g = 17.88854382 uni...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2820581

Calculating x. Let a be the radius of the ascribed quarter circle. b = 4+4 b = 8 units. Notice. BD = 4 units. OB = x+BD OB = (x+4) units. OB is the radius of the ascribed quarter circle. It implies; (x+4)² = x²+8² x²+8x+16 = x²+64 8x = 48 x = 6 units.

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