Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2722645

Sir Beg is the author of the question. Notice. The composite plane shape is not drawn to scale. 2a² = 2² a² = 2 a = √(2) cm. a is OC = AC. 2b² = √(2)² b² = 1 b = 1 cm. b is OF = CF = DE, the length of the inscribed rectangle CDEF. Calculating EF = CD, the width of the inscribed rectangle CDEF. Let it be d. c = (1+d) cm. It implies; 2² = 1+(1+d)² 4 = 1+1+2d+d² d²+2d-2 = 0 (d+1)² = 2+1² d = -1±√(3) Therefore; d = (√(3)-1) cm. d = 0.7320508076 cm. Again, d...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2722594

Let a be the radius of the circle and also the width of the rectangle. b = (a-x) units. c = (a-2x) cm. d = a+c d = (2a-2x) cm. d is the length of the rectangle. 12² = (2a-2x)²+(a-x)² 144 = 4a²-8ax+4x²+a²-2ax+x² 144 = 5a²-10ax+5x² --- (1). 2x - (a-x) (a-x) - (2a-2x) Cross Multiply. (a-x)² = 4x(a-x) a-x = 4x a = 5x --- (2). Substituting (2) in (1). 144 = 5(5x)²-10(5x)x+5x² 144 = 125x²-50x²+5x² 144 = 80x² 36 = 20x² 18 = 10x² x² = 9/5 x = √(9/5) x = ⅕(3√(5...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2722585

a = r+r a = 2r units. a is the diameter of the circle. b = 3+1 b = 4 units. c²+(2r)² = 4² c = √(16-4r²) units. It implies; 2√(3) - 4 r - √(16-4r²) Cross Multiply. 4r = 2√(3(16-4r²)) 2r = √(3(16-4r²)) 4r² = 3(16-4r²) 4r² = 48-12r² 16r² = 48 r = √(3) units. Therefore, area circle is; πr² = π√(3)² = 3π square units. Calculating the required angle, theta. Let it be x. c = √(16-4r²) And r = √(3) units. c = √(16-4√(3)²) c = √(16-12) c = √(4) c = 2 u...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2721819

Sir Mike Ambrose is the author of the question. Area yellow exactly in square cm is; Area triangle with height 10√(13)/9 cm and base 5√(13)/6 cm = ½(10√(13)/9*5√(13)/6) = ½((13*50)/(54)) = (13*25)/54 = 325/54 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2721532

Calculating shaded red area. Let a be the diameter of the inscribed big half circle. Therefore, 2a is the diameter of the bigger half circle or a is the radius of the bigger half circle. a² = 6²+b² b = √(a²-36) cm. 6 - c a - √(a²-36) Cross Multiply. c = (6√(a²-36))/a cm. Again. a - √(a²-36) √(a²-36) - d Cross Multiply. d = (a²-36)/a cm. e = a+d e = (2a²-36)/a cm. Calculating a. 10² = ((2a²-36)/a)²+((6√(a²-36))/a)² 100 = ((4a⁴-144a²+1296)/a²)+((36a²-1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2721256

a² = 4²+4²-2*4*4cos120 a² = 32+16 a² = 48 a = 4√(3) cm. b²+(0.5a²) = 4² b² = 16-12 b = √(4) b = 2 cm. Or cos60 = b/4 ½ = b/4 b = 2 cm. Therefore, total surface area of the hexagonal based prism is; 2(area rectangle with length 4√(3) cm and width 4 cm)+4(area triangle with base 4√(3) cm and height 2 cm)+6(area rectangle with length 12 cm and width 4 cm) = 2(4√(3)*4)+4(½*4√(3)*2)+6(12*4) = 32√(3)+16√(3)+288 = 288+48√(3) = 48(6+√(3)) cm² = 371.1384387633 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2721242

Let a be the radius of the inscribed green circle. tan30 = a/b b = √(3)a cm. c = ½(3√(3)-π) cm² d = 60π*a²/360 d = ⅙(πa²) cm² Therefore; 0.5*a*√(3)a = ½(3√(3)-π)+⅙(πa²) ½√(3)a² = ½(3√(3))-½(π)+⅙(πa²) 3√(3)a² = 9√(3)-3π+πa² 3√(3)a²-πa² = 9√(3)-3π a²(3√(3)-π) = (9√(3)-3π) a² = (9√(3)-3π)/(3√(3)-π) It implies; a = 1.7320508076 cm. a = √(3) cm. Again, a is the radius of the inscribed green circle. Area inscribed green circle is; πa² = π√(3)² = 3π cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2721240

Angle ABC = ½(angle BAC) since angle ACD and angle BCD are equal and length BC = length AD + length AC It implies; a +½(100)+100 = 180 a = 180-150 a = 30° a is angle ACB. b = ½(a) b = 15° b is angle BCD = angle ACD Therefore, the required angle, alpha is; Let it be c. It implies; 100+15+c = 180 c = 180-115 c = 65° Again, c is the required angle alpha.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2721224

Let AB = 1 unit. a = 180-15-30 a = 135° a is angle ADB. (1/sin135) = (b/sin15) b = 0.3660254038 units. b is BD = CD. c = 2b c = 0.7320508076 units. c is BC. d² = 0.7320508076²+1²-2*1*0.7320508076cos30 d = 0.5176380902 units. d is AC. e = 180-135 e = 45° e is angle ADC. Therefore, the required angle, x is; (0.3660254038/sinx) = (0.5176380902/sin45) x = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2720582

(9^(p)*81)-9^(p) = 240 Let 9^(p) = a. 81a-a = 240 80a = 240 a = 3 It implies; 9^(p) = 3 p = ½ Therefore; (8p)^(p) is; √(8*½) = √(4) = 2

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