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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3578240

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Calculating Area Red.Let x be the radius of the circle.a = (x-7) cm.Calculating x.x²+(x-7) = 13²x²+x²-14x+49 = 1692x²-14x-120 = 0x²-7x-60 = 0x²-12x+5x-60 = 0x(x-12)+5(x-12) = 0(x-12)(x+5) = 0It implies;x ≠ -5x = 12 cm.Again, x is the radius of the circle.Recall.a = x-7And x = 12 cm.a = 12-7a = 5 cm.Therefore, area red is ;Area quarter circle with radius 12 cm - Area right-angled triangle with height 12 cm and base 5 cm.(¼*12²π) - (½*12*5)= 36π-30= 6(6π-5) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3577145

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Sir Mike Ambrose is the author of the question.Calculating Area Blue ÷ Area Green to 2 decimal places.Let the side length of the regular pentagon be 1 unit.tan36 = a/0.5a = 0.363271264 unit.b² = 0.363271264²+0.5²b = 0.61803398875 unit.tan72 = c/0.5c = 1.53884176859 units.Area Green is;0.5sin108+0.5*1.53884176859*0.5-0.5*0.5*0.363271264-0.5*0.61803398875*0.61803398875sin36= 0.65716389015 square units.Calculating Area Blue.Let d be the radius of the inscribed blue circle.Calculating d.e = 0.5(90+...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3576039

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Area shaded in pink (area A) is;Notice:The single side length of the square is;3√(2) units.Shaded yellow area height is;Height = 6/(3√(2))= √(2) units.Therefore area A is;Height = 3√(2)-√(2) = 2√(2) units.Base = ⅓(3√(2)) = √(2) units.Area A = ½*√(2)*2√(2)= ½(4)= 2 square units.Area shaded in pink (area A) is;2 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3574810

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Calculating x.Notice.The ascribed irregular quadrilateral is cut out a regular triangle (Equilateral Triangle).Therefore, resolving with the above fact/analysis.a = (x+6) units.a is the side length of the factual equilateral triangle.b = a-8b = (x+6)-8b = (x-2) unit.b is the shortest length of the cut out scalene triangle.It implies;A right-angled triangle is derived where b is the adjacent, 6 units is the hypotenuse and 60° is the angle that indicates the unknown opposite of the derived right-...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3573920

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Calculating the required red length.Let a be the radius of the half circle.b = (a-1) units.c²+1² = b²c² = a²-2a+1-1c = √(a²-2a) units.d = a+c d = (a+√(a²-2a)) units.d is the required length.e = (a+1) units.Calculating a.(a+√(a²-2a))²+(a-1)² = (a+1)²a²+2a√(a²-2a)+a²-2a+a²-2a+1 = a²+2a+12a√(a²-2a)+2a² = 6aa√(a²-2a)+a² = 3a√(a²-2a)+a = 3a²-2a = (3-a)²a²-2a = 9-6a+a²4a = 9a = ¼(9) units.a = 2.25 units.Recall.d = (a+√(a²-2a))And a = ¼(9) units.d = ((9/4)+√((9/4)²-2(9/4)))d = ¼(9)+√((81/16)-(9/2))d...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3572696

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Calculating Length x.Notice.4 units is the radius of the inscribed circle.8 units is the radius of the inscribed quarter circle.a² = 2(4)²a = 4√(2) units.a is half the diagonal of the ascribed square.Therefore;8² = (4√(2))²+4²-2*4*4√(2)cosyCalculating y.32√(2)cosy = 48-64cosy = -16/(32√(2))y = acos(-16/(32√(2)))°y = 110.704811055°b = 360-2yb = 360-2(110.704811055)b = 138.59037789°It implies, length x is;x² = 2(4²)-2(4²)cos138.59037789x = √(56) units.x = 2√(14) units.x = 7.48331477355 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3571581

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Calculating Length PQ.x+a = y+2a4+a = 3+2aa = 1 unit.b = y+ab = 4+aAnd a = 1 unit.b = 5 units.Orb = y+2ab = 3+2aAnd a = 1 unit.b = 3+2(1)b = 5 units.Orb² = 4²+3²b = √(25)b = 5 units.c = x-yc = 4-3c = 1 unit.Let PQ = d.Calculating db² = c²+d²5² = 1²+d²d² = 25-1 d = √(24)d = 2√(6) units.Again, d is the required length, PQ.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3570703

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Calculating x²/y².Let the inscribed blue square side length be 2 units.Therefore;y = 2 units.y² = 4 square units, the area of the blue inscribed square.Calculating x², the area of the yellow inscribed square.tana = 2/1a = atan(2)°b = 90-ab = atan(½)°c² = 2²+1²c = √(5) units.d = c-xd = (√(5-x) units.It implies;1 ~ (√(5-x)2 ~ x Cross Multiply.x = 2√(5)-2x3x = 2√(5)9x² = 20x² = 20/9 square units.Again, x² is the area of the yellow inscribed square.Therefore;x²/y² is;(20/9)/4= 20/36= 5/9

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3569490

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Calculating x/y.Let 10 units be the side length of the ascribed square.x+y = 10 y = (10-x) --- (1).Let a be each of the 2 equal angles.tana = x/(10-x) --- (2).tan(2a) = 10/(10-x)2tana/(1-tan²a) = 10/(10-x)10-10tan²a = 20tana-2xtana --- (3).Substituting (2) in (3).10-10(x/(10-x))² = 20(x/(10-x))-2x(x/(10-x))10(10-x)²-10x² = 20x(10-x)-2x²(10-x)10(100-20x+x²)-10x² = 200x-20x²-20x²+2x³1000-200x = 200x-40x²+2x³2x³-40x²+400x-1000 = 0x³-20x²+200x-500 = 0It implies;x = 3.104And x+y = 10y = 10-xy = 10-...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3568896

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Calculating Length x.AB = √(256+144)AB = √(400)AB = 20 units, the side length of square ABCD.tana = 16/12a = atan(4/3)°b = 90-ab = atan(3/4)°tan(b) = c/20¾ = c/20c = 15 units.c is length DF.d² = 20²+15²d = √(400+225)d = √(625)d = 25 units.d is length AF.It implies, length EF, x is;x = 25-12x = 13 units.
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