Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3620111

Calculating R, radius and circumference of the ascribed circle.a = √(3) units.a is the side length of the blue inscribed square.b = √(5) units.b is the side length of the green inscribed square.c = √(7) units.c is the side length of the red inscribed square.√(5)² = √(3)²+√(7)²-2√(3)*√(7)cosd2√(21)cosd = 10-5cosd = 5/(2√(21))d = 56.9381038653°d is the angle opposite √(5) units.(√(5)/sin56.9381038653) = (√(7)/sine)e = 82.5824442091°e is the angle opposite √(7) units.f = 180-df = 180-56.9381038653...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3620109

Calculating Angle alpha and the circumference of the inscribed circle.Let alpha be x.Notice.32π cm² is the area of the ascribed half circle.Calculating R, radius of the ascribed half circle.½*πR² = 32πR² = 64R = 8 cm.Calculating r, radius of the inscribed circle.a = (8-r) cm.It implies;2r² = a²2r² = (8-r)²2r² = 64-16r+r²r²+16r-64 = 0(r+8)² = 64+(-8)²r+8 = ±√(128)r = -8±8√(2)It implies;r ≠ -8-8√(2)r = 8√(2)-8r = 8(√(2)-1) cm.r is the radius of the inscribed circle.b = 8+rb = 8+8√(2)-8b = 8√(2) c...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3618677

Calculating Area Blue.Notice.The inscribed triangle is equilateral.Area Blue is;Area sector with radius 10 cm and angle 60° - Area sector with radius 5 cm and angle 120° - Area equilateral triangle with height 5 cm and base 2sin60 cm.(60π*10²/360)-(120π*5²/360)-(½*5*5sin60)= ⅓(50π)-⅓(25π)-¼(25√(3))= (200π-100π-75√(3))/12= (100π-75√(3))/12 cm²= (25(4π-3√(3)))/12 cm²= 15.3546212326 cm²

Mathematics Question and Solution

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Calculating Blue Area.Let x be the side length of the big inscribed blue square.a = ½(10)a = 5 units.a is the radius of the half circle.b = ½(x) units.Calculating x.5² = x²+(½*x)²25 = x²+¼(x²)25 = ¼(5x²)x² = 20x = √(20)x = 2√(5) units.Again, x is the side length of the big inscribed blue square.Recall.b = ½(x)b = ½(2√(5))b = √(5) units.Let y be the side length of the small inscribed blue square.c = b+yc = (y+√(5)) units.Calculating y.5² = y²+(y+√(5))²25 = y²+y²+2√(5)y+520 = 2y²+2√(5)yy²+√(5)y-1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3617678

Calculating Green Area.Let x be the side length of the square.a = ½(x) units.a is the radius of the inscribed half circle.tanb = ½(x)/xb = atan(½)°c = 2bc = 2atan(½)°d = 90-cd = (90-2atan(½))°d = 36.8698976458°cos36.8698976458 = e/xe = xcos36.8698976458 units.Calculating x.½*x*xcos36.8698976458sin36.8698976458 = 6x²cos36.8698976458sin36.8698976458 = 120.48x² = 12x = √(12/0.48)x = 5 units.Recall.e = xcos36.8698976458And x = 5 units.e = 5cos36.8698976458e = 4 units.sin36.8698976458 = f/5f = 3 uni...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3616483

Calculating x, the radius of the big quarter circle.a = ½(90)a = 45°a is half angle DBE.b = ½(180-45)b = 67.5°b is angle BDF.c = (x+1) units.c is AB.d = (x-1) units.Therefore, x is;tan67.5 = (x+1)/(x-1)x+1 = (tan67.5)x-tan67.51+tan67.5 = (tan67.5)x-x((tan67.5)-1)x = 1+tan67.5x = (1+tan67.5)/((tan67.5)-1)x = (1+√(2)) units.x = 2.41421356237 units.Again, x is the radius of the big quarter circle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3615400

Calculating Area Pink Regular Hexagon.Let a be the side length of the pink regular hexagon.Let b be the side length of the red regular hexagon with area 6 square units.A nice analysis of the composite plane shape implies that;c = a+bWhere c is the side length of the yellow regular hexagon.Notice.120° is each of the interior angle of the regular hexagons.Calculating b.c² = 2b²-2b²cos120c² = 3b²c = √(3b²)c = √(3)b units.c is the height of the red regular hexagon.It implies;(2*½*b*bsin120)+(√(3)b*...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3614564

Calculating Area Triangle APC, Area Yellow.Let x be AP = CPa = ½(10)a = 5 cm.a is the mid-point of AC.Notice.Since angle BAP = angle CAP is equal and length AP = length CP.It implies;AB = 5 cm.cosb = (AB)/(AC)cosb = 5/10cosb = ½b = acos(½)b = 60°b is angle BAC.c = ½(b)c = ½(60)c = 30°c is angle BAP.Therefore;Calculating x, AP = CP.cosc = (AB)/xcos30 = 5/x½√(3) = 5/x√(3)x = 10x = 10/√(3)x = ⅓(10√(3)) cm.It implies, yellow area triangle APC is;½*(AB)*(CP)= ½*5*⅓(10√(3))= ⅓(25√(3)) cm²= 14.433756...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3613330

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/3612226

Calculating Area ADCB, a Cylic Quadrilateral.a = ½(45)a = 22.5°a is angle CBD.b = a+45b = 22.5+45b = 67.5°b is angle ABE.sin67.5 = 10/cc = 10.8239220029 units.c is AB = AD.d² = 2c²d² = 2*10.8239220029²d = √(2*10.8239220029²)d = 15.3073372946 units.d is BD, the diameter of the circle.cos22.5 = e/15.3073372946e = 15.3073372946cos22.5e = 14.1421356237 units.e is BC.sin22.5 = f/15.3073372946f = 15.3073372946sin22.5f = 5.85786437626 units.f is CD.It implies, area ADCB, a cylic quadrilateral is;½(10....

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