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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3680794

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Calculating x, length DP.sina = 8/10a = asin(4/5)°a is angle BCP.b = 360-90-90-ab = (180-asin(4/5))°b = 126.869897646°b is angle ABC.c = ½(b)c = ½(126.869897646)c = 63.4349488229°c is theta, angle ABP equal angle CBP.d = 90-ad = 90-asin(4/5)d = acos(4/5)°d = 36.8698976458°e = c-de = 63.4349488229-36.8698976458e = 26.5650511771°tane = f/8f = 8tan26.5650511771f = 4 units.Ortan63.4349488229 = 8/ff = 8/tan63.4349488229f = 4 units.Therefore, the required length x is;x+f = 5x = 5-fx = 5-4x = 1 unit.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3679919

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Calculating Area Yellow.a = 2+20a = 22 units.b²+x² = 20²b = √(400-x²) units.cosy = b/20cosy = √(400-x²)/20 --- (1).10² = 22²+√(400-x²)²-2*22√(400-x²) cosy100 = 484+400-x²-44√(400-x²)cosy44√(400-x²)cosy = 784-x²cosy = (784-x²)/(44√(400-x²)) --- (2).Equating (1) and (2).√(400-x²)/20 = (784-x²)/(44√(400-x²))20(784-x²) = 44(400-x²)5(784-x²) = 11(400-x²)3920-5x² = 4400-11x²11x²-5x² = 4400-39206x² = 480x² = 80x = √(80)x = 4√(5) units.x is the shortest length (base) of the inscribed yellow area.Recall...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3679915

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Calculating ratio AM/BC.Let alpha be x.Let BC be 2 units.It implies;BM = CM = 1 unit.tan(2x) = a/1a = tan(2x)a = (2tanx)/(1-tan²x) --- (1).a is AB.tanx = 2/a --- (2).Therefore, equating (2) in (1).a = (2(2/a)/(1-(2/a)²)a = (4/a)/(1-(4/a²))a = (4/a)/((a²-4)/(a²))Cross Multiply.4/a = (a²-4)/a4 = a²-4a² = 8a = √(8)a = 2√(2) units.Again, a is AB.b² = 1²+ab² = 1²+(2√(2))²b² = 1+8b= √(9)b = 3 units.b is AM.It implies, ratio AM/BC is;3 : 2= 3/2= 1.5

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3678215

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Calculating length x.½*5*a = S5a = 2Sa = ⅕(2S) units.a is CE.x ~ 5b ~ aTherefore;x ~ 5b ~ ⅕(2S)Cross Multiply.⅕(2S)x = 5b2Sx = 25bb = (2Sx)/25 units.b is BC.Calculating x, length AB.½*x*b = 3S+S½*x*((2Sx)/25) = 4Sx²S = 100Sx² = 100x = √(100)x = 10 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3678212

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Calculating length x.a = 2+3 = 1+4a = 5 units.a is the radius of the ascribed quarter circle.b = a+3b = 5+3b = 8 units.It implies;c*c = 2*bc² = 2*8c = √(16)c = 4 units.d = c-1d = 4-1d = 3 units.Therefore, length x is;x² = 3²+d²x² = 3²+3²x² = 18x = √(18)x = 3√(2) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3677686

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Calculating x/y.tana = 4/3a = atan(4/3)°b = 180-ab = (180-atan(4/3))°b = 126.869897646°c² = 3²+1²-2*3*1cos126.869897646c = 3.68781778292 units.(1/sind) = (3.68781778292/sin(126.869897646))d = 12.5288077092°e = 90-de = 90-12.5288077092e = 77.4711922908°Calculating x.cose = x/4x = (4cos77.4711922908) units.Calculating y.sine = y/4y = (4sin77.4711922908) units.It implies, x/y is;(4cos77.4711922908)/(4sin77.4711922908)= (cos77.4711922908)/(sin77.4711922908)= 0.22222222222= 2/9

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3677287

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Calculating the required angle.Let it be x.Let the square side length be 2 units.a = ½(2)a = 1 unit.tanb = 2/1b = atan(2)°c = 180-bc = (180-atan(2))°(2/sin(180-atan(2))) = (1/sind)d = 26.5650511771°e = 180-(180-atan(2))-de = atan(2)-26.5650511771e = 36.8698976458°f² = 2²+1²-2*2*1cos36.8698976458f = 1.3416407865 units.g² = 1.3416407865²+1²-2*1.3416407865*1cos(atan(2))g = 1.26491106407°Therefore, the required angle x is;(1/sinx) = (1.26491106407/sin(atan(2)))x = asin(0.70710678119)x = 45°Again, x...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3677274

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Calculating Area ADPB.Notice.BD = 3 units.CD = 1 unit.a = ½(3)a = 1.5 units.a is AC.b = a+1b = 1.5+1b = 2.5 units.b is OB, the radius of the circle.cosc = a/bc = acos(1.5/2.5)c = 53.1301023542°c is angle OBD.d = 90-cd = 90-53.1301023542d = 36.8698976458°d is angle ABP.cosd = 4/ee = 4/cos36.8698976458e = 5 units.e is AP = BP.It implies, area green (area quadrilateral ADPB) is;Area triangle with height 4 units and base 5 units - Area triangle with height 1 unit and base 5 units.½(4*5)-½(1*5)= 10-...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3676028

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Calculating Area Yellow÷Area Red.Notice.Inscribed blue square area is 8 m².a = √(8)a = 2√(2) m.a = 2.82842712475 m.a is the side length of the inscribed blue square.Let r be the radius of the inscribed yellow circle.b = r-ab = (r-2√(2)) m.Calculating r.b²+b² = r²2(r-2√(2))² = r²2r²-8√(2)r+16 = r²r²-8√(2)r+16 = 0(r-4√(2))² = -16+(-4√(2))²(r-4√(2))² = -16+32r-4√(2) = √(16)r = 4√(2)±4It implies r ≠ (4√(2)-4) m.r = (4√(2)+4) m.r = 4(√(2)+1) m.r = 9.65685424949 units.Again, r is the radius of the in...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3676023

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Calculating area ADBC.Let let x be the side length of the small square.a = (2-x) cm.b² = x²+x²b = √(2x²)b = √(2)x cm.b is BD, the diagonal of the small square.c² = x²+a²c² = x²+(2-x)²c² = x²+4-4x+x²c = √(2x²-4x+4) cm.c is AD, the diagonal of the big square.d = 2bd = 2√(2)x cm.d is the side length of the big square.√(2x²-4x+4)² = (2√(2)x)²+(2√(2)x)²2x²-4x+4 = 16x²7x²+2x-2 = 0(x+(1/7))² = (2/7)+(1/7)²(x+(1/7))² = (15/49)x = -(1/7)±√(15/49)x = -(1/7)±(√(15)/7)x = -(1/7)+(√(15)/7)x = ⅐(√(15)-1)x =...
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