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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3606806

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Calculating R, radius of the circle.a = (7-R) units.b²+R² = a²b²+R² = (7-R)²b² = 49-14R+R²-R²b = √(49-14R) units.c = (R-1) units.Therefore R, radius of the circle is;b²+c² = R²√(49-14R)²+(R-1)² = R²49-14R+R²-2R+1 = R²49-14R-2R+1 = 050 = 16R25 = 8RR = ⅛(25) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3605706

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Calculating r, radius of the inscribed circle.Notice.The ascribed plane shape is a trapezium.a = 2r units.a is the diameter of the inscribed circle.b²+(2r)² = 10²b = √(100-4r²) units.c²+(2r)² = 17²c = √(289-4r²) units.d = b+3+cd = (√(100-4r²)+3+√(289-4r²)) units.d is AD.Calculating r.½(3r)+½(10r)½(17r)+½((√(100-4r²)+3+√(289-4r²))r) = ½(3+(√(100-4r²)+3+√(289-4r²)))*2r½(33r+√(100-4r²)r+√(289-4r²)r) = ½(6r+2r√(100-4r²)+6r+2r√(289-4r²))33r+√(100-4r²)r+√(289-4r²)r = ½(6r+2r√(100-4r²)+6r+2r√(289-4r²)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3604873

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Calculating the required inscribed orange triangle area.a = ½(3)a = 1.5 units.a is the radius of the inscribed circle.tanb = 3/4b = atan(3/4)°tan(atan(3/4)) = c/1.53/4 = c/1.54c = 4.5c = 1.125 units.d = (180-atan(4/3))°e = a-ce = 1.5-1.125e = 0.375 units.(1.5/sin(180-atan(4/3))) = (0.375/sinf)f = 11.5369590328°g = 180-2fg = 180-2(11.5369590328)g = 156.926081934°h² = 2(1.5)²-2(1.5)²cos156.926081934h = 2.93938769134 units.h is the longest length of the inscribed orange triangle.j = (180-atan(4/3)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3604589

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Calculating length d.Notice.3 units is the radius of the inscribed green circle.6 units is the radius of the inscribed purple circle.9 units is the radius of the ascribed blue circle.a = 6-3a = 3 units.b =9+9-3-6b = 9 units.c²+3² = 9²c² = 81-9c = √(72)c² = 6√(2) units.x ~ 6√(2)6 ~ 9Cross Multiply.9x = 36√(2)x = 4√(2) units.4√(2) ~ 6√(2)e ~ 32 ~ 3e ~ 3Cross Multiply.3e = 6e = 2 units.f = 3+ef = 5 units.g²+5² = 9²g² = 81-25g = √(56)g = 2√(14) units.Therefore, length d is;d = 2gd = 2*2√(14)d = 4√(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3603745

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Calculating Area Yellow (Triangle CDE).Let BE be x.Calculating x.½(9+6)*x = 8715x = 174x = (174/15) cm.a = x+ya = ((174/15)+y)a is BC, y is CE.Calculating y.9 ~ 6((174/15)+y) ~ yCross Multiply.9y = 6((174/15)+y)9y = ⅕(348)+6y3y = ⅕(348)y = ⅕(116) cm.Again, y is CE.It implies, area yellow is;½*DE*CE= ½*6*⅕(116)= ⅗(116) cm²= ⅕(348) cm²= 69.6 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3603742

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Calculating x, the required red length.a = (16-y) cm.y is the radius of the inscribed blue circle.b = ½(16)b = 8 cm.Calculating y.(16-y)² = y²+8²256-32y+y² = y²+64256-32y = 64192 = 32yy = 6 cm.Again, y is the radius of the blue inscribed circle.tanc = 8/6c = atan(4/3)°d = 2cd = 2atan(4/3)°Therefore, x, the required red length is;x² = 6²+6²-2*6*6cos(2atan(4/3))x² = 92.16x = √(92.16)x = 9.6 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3602466

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Calculating Angle x.Let 1 unit be the radius of the quarter circle.Let y be the radius of the inscribed circle.Notice.OB = BD.OC = 8 cm, radius of the quarter circle.a = ½(OC)a = ½(8)a = 4 units.a is half the radius of the ascribed quarter circle.sinb = 4/8b = asin(½)b = 30°b is angle BOD = angle BDO c = ½(180-b)c = ½(180-30)c = ½(150)c = 75°c is angle CDO = angle DCO.Therefore, the required angle, x isx = c-bx = 75-30x = 45°Again, x is the required angle.Calculating r, radius of the inscribed...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3602054

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Calculating Area Green Inscribed Half Circle.Notice.Length AM = Length CM.AN = 2 cm.Let x be the radius of the inscribed half circle.a = 2+x+xa = (2+2x) cm.a is the radius of the the ascribed sector.b = ½(a)b = (1+x) cm.b is AM = CM.c = (2+x) cm.Calculating x.x²+(1+x)² = (2+x)²x²+1+2x+x² = 4+4x+x²x²+1+2x = 4+4xx²-2x-3 = 0x²-3x+x-3 = 0x(x-3)+1(x-3) = 0(x-3)(x+1) = 0It implies;x ≠ -1x = 3 cm.Therefore, area inscribed half circle is;½(πx²)= ½*π*3*3= ½(9π) cm²= 4.5π cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3600309

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Calculating Area BCDA, Area of the Ascribed Square.Notice.Length EF = 1 cm.Let x be the FD, the radius of the congruent circles.a² = 2x²a = √(2)x cm.b = ½(a)b = ½√(2)x cm.c = x+bc = (x+½√(2)x) cm.Calculating x.1² = c²+b²1 = (x+½√(2)x)²+(½√(2)x)²1 = x²+√(2)x²+½(x²)+½(x²)1 = x²+√(2)x²+x²1 = 2x²+√(2)x²1 = (2+√(2))x²x² = 1/(2+√(2))x = √(1/(2+√(2)))x = 0.54119610015 cm.Again, x is FD, the radius of the congruent circles.AB = 2(FD)FD = FC = xAB = 2xAB = 2(0.54119610015)AB = 1.0823922003 cm.AB = CD, t...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3600296

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Calculating Area Red.a = 6-2a = 4 units.b = 10-3b = 7 units.It implies;Red Area is;Area rectangle with length 10 units and width 6 units - Area triangle with height 4 units and base 3 units - Area triangle with height 6 units and base 7 units.(10*6)-½(4*3)-½(7*6)= 60-6-21= 60-27= 33 square units.
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