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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3165459

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Calculating Length MA. Let a be the AB, side length of the regular hexagon. b = (a-3) units. b is AM. c = ½(a) units. c is AK = FK. Let d be CM = KM. It implies; d² = a²+3²-2*3acos120 d² = a²+9+3a --- (1). Again. d² = (0.5a)²+(a-3)²-2*½(a)(a-3)cos120 d² = ¼(a²)+a²-6a+9+½(a²-3a) d² = ¼(a²+4a²-24a+36+2a²-6a) d² = ¼(7a²-30a+36) --- (2). Equating (1) and (2). a²+9+3a = ¼(7a²-30a+36) 4a²+12a+36 = 7a²-30a+36 42a = 3a² a = ⅓(42) a = 14 units. Again a is AB, the side length of the regular...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3164095

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Calculating Area Triangle ABC. Let a be the height of the triangle. b = √(a²+4) units. b is AC. c = √(a²+9) units. c is BC. Calculating a. ½(2+3)a = ½√(a²+4)*√(a²+9)sin45 10a = √(2(a²+4)(a²+9)) 100a² = 2a⁴+26a²+72 2a⁴-74a²+72 = 0 a⁴-37a²+36 = 0 a⁴-a²-36a²+36 = 0 a²(a²-1)-36(a²-1) = 0 (a²-1)(a²-36) = 0 It implies; a² = 1 or a² = 36 Therefore; a ≠ √(1) units. a = √(36) a = 6 units. Area Triangle ABC is; ½*a*(AB) = ½*6*(2+3) = 3(5) = 1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3164092

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Let AD be x. a² = x²+4²-2*4*xcos30 a = √(x²+16-4√(3)x) units. a is BD. b² = x²+3²-2*3*xcos30 b = √(x²+9-3√(3)x) units. b is CD. c² = 4²+3²-2*3*4cos60 c² = 25-12 c = √(13) units. c is BC. Calculating length AD (x). It implies; a+b = c √(x²+16-4√(3)x)+√(x²+9-3√(3)x) = √(13) x²+9-3√(3)x = (√(13)-√(x²+16-4√(3)x))² x²+9-3√(3)x = 13-2√(13)√(x²+16-4√(3)x)+x²+16-4√(3)x 20-√(3)x = 2√(13)√(x²+16-4√(3)x) (20-√(3)x)² = 52(x²+16-4√(3)x) 400-40√(3)x+3x² = 52x²-208√(3)x+832 49x²-168√(3)x+432 = 0...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3162934

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Calculating R, radius of the circle. a²+4² = 6² a = √(36-16) a = √(20) a = 2√(5) units. b²+4² = 10² b = √(100-16) b = √(84) b = 2√(21) units. Therefore; 2√(5)*2√(21) = 4c c = √(105) units. d = 4+c d = (4+√(105)) units. d = 14.246950766 units. e = ½(d) e = 7.123475383 units. f = ½(a+b) f = ½(2√(5)+2√(21)) f = (√(5)+√(21)) units. g = f-a g = (√(5)+√(21))-2√(5) g = (√(21)-√(5)) units. g = 2.34650771746 units. Therefore R, radius of the circle is; R² =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3162932

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Let the two equal lengths be 1 unit each. (1/sin20) = (a/sin40) a = 1.87938524157 units. b = a-1 b = 0.87938524157 units. (1/sin20) = (c/sin(180-40-20)) c = 2.53208888624 units. d² = 2(2.53208888624)²-2(2.53208888624)cos40 d = √(3) units. d = 1.73205080757 units. e² = 1.73205080757²+0.87938524157²-2*1.73205080757*0.87938524157cos50 e = 1.34729635533 units. (1.34729635533/sin50) = (√(3)/sinf) f = 80° Where f is the required angle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3161756

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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

https://onlineedumath.com/update/mathematics-question-and-solution/3161751

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Calculating Length x. 7² = x²+(x+5)²-2*(5+x)xcosa 49 = 2x²+10x+25-(10x+2x²)cosa (10x+2x²)cosa = 2x²+10x-24 cosa = (2x²+10x-24)/(10x+2x²) --- (1). 7² = x²+5²+2*x*5cosa 49-25-x² = 10xcosa 24-x² = 10xcosa cosa = (24-x²)/(10x) --- (2). Therefore, length x is; Equating (1) and (2). (2x²+10x-24)/(10x+2x²) = (24-x²)/(10x) 20x³+100x²-240x = 240x+48x²-10x³-2x⁴ 2x⁴+30x³+52x²-480x = 0 x³+15x²+26x-240 = 0 Therefore; x = 3 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3161747

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Calculating length c. cosd = 9/12 d = acos(3/4)° e = 2d e = 2acos(¾)° Therefore, length c is; cose = c/a cos(2acos(¾)) = c/12 c = 12cos(2acos(¾)) c = 12*⅛ c = ½(3) units. c = 1.5 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3159873

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Let a be the diameter of the circle. x² = 6²+6²+2*6*6cosb x² = 72+72cosb --- (1). x² = 14²+a²-2*14acosb x² = 196+a²-28acosb --- (2). Notice. x is length AC b is angle ADC. cosb = 14/a --- (3). Calculating a. Equating (1) and (2). 72+72cosb = 196+a²-28acosb --- (4). Substituting (3) in (4). 72+72(14/a) = 196+a²-28a(14/a) 72+72(14/a) = 196+a²-28(14) 72a+1008 = 196a+a³-392a a³-268a-1008 = 0 Therefore; a = 18 units. Again, a is the diameter of the circ...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3159871

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Let a = AB = AD = CD. b² = 2a² b = √(2)a units. b is BD. Notice. Triangle BCD is a right-angled triangle, where BC is the hypotenuse. c² =(√(2)a)²+a² c = 3a² c = √(3)a units. c is BC. cosd = (0.5*10)/a cosd = 5/a --- (1). d is angle CAD. From (1). sind = √(a²-25)/a --- (2). (√(3)a)² = a²+10²-2a*10cos(90-d) 3a² = a²+100-20a(cos90cosd+sin90sind) 2a² = 100-20a(sind) a² = 50-10asind --- (3). Calculating a. Substituting (2) in (3). a² = 50-10a√(a²-25)/a a² = 50-10√(a²-25) 10²(a²-25) =...
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