Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3675194

Calculating Blue Area.It is;Area quarter circle with radius 4 cm - Area triangle with height and base 4 cm respectively.¼*4²π - ½*4²= ¼(16π)-½(16)= 4π-8= 4(π-2) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3675178

Calculating length CH (x).a = 9+6a = 15 unitsa is AH.b = a+xb = (15+x) units.b is AC.Notice.EF = EH = 6 units.6²+c² = 9²c² = 81-36c = √(45)c = 3√(5) units.c is AF.3√(5) ~ 156 ~ dCross Multiply.3√(5)d = 6*15d = 6√(5) units.d is O1H, the radius of the inscribed yellow half circle.e = c+2de = 3√(5)+2*6√(5)e = 15√(5) units.e is AB, the ascribed of the ascribed half circle.It implies, length x is;3√(5) ~ (15+x)9 ~ 15√(5)Cross Multiply.9(15+x) = 3√(5)*15√(5)15+x = 25x = 25-15x = 10 units.x is CH, the...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3674113

Calculating 2R, diameter of the red inscribed circle.a = ½(16)a = 8 units.a is the radius of the inscribed yellow half circle.b = (16-R) units.c²+R² = b²c² = (16-R)²-R²c = √(256-32R+R²-R²)c = √(256-32R) units.d = (8+R) units.e = (8-R) units.It implies, R is;e²+c² = d²(8-R)²+√(256-32R)² = (8+R)²64-16R+R²+256-32R = 64+16R+R²-16R+256-32R = 16R256 = 64RR = 4 units.Again, R is the radius of the inscribed red square.It implies 2R, diameter of the inscribed red circle is; = 2(4)= 8 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3674066

Calculating the area of the semi circle.Let r be the radius of the half circle.a = 2r units.a is the diameter of the half circle.b² = (2r)²+7²-2*7*2rcosxb² = 4r²+49-28rcosx --- (1).b² = 2(15)²+2(15)²cosxb² = 450+450cosx --- (2).cosx = 7/(2r) --- (3).Calculating r.Equating (1) and (2).4r²+49-28rcosx = 450+450cosx4r²-28rcosx = 401+450cosx --- (4).Substituting (3) in (4).4r²-28r(7/(2r)) = 401+450(7/(2r))4r²-98 = 401+(1575/r)4r² = 499+(1575/r)4r³-499r = 1575Observing trial and error approach.r = 12...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3673116

Calculating x.Let a be each of the two equal lengths.b² = 2a²b = √(2a²)b = √(2)a units.c = 2√(2)+5√(2)c = 7√(2) units.It implies;a ~ b7√(2) ~ dTherefore;a ~ √(2)a7√(2) ~ dCross Multiply.ad = 14ad = 14 units.Again.5√(2) ~ e14 ~ 7√(2)Cross Multiply.14e = 70e = 70/14e = 5 units.Therefore, length x is;d+e= 14+5= 19 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3673113

Calculating length CD.a = 360-60-60-90a = 150°a is angle BCD.b = 180-ab = 180-150b = 30°sin30 = c/x½ = c/xc = ½(x) units.d = x+xd = 2x units.d is AB.e = c+x+10e = ½(x)+x+10e = ½(3x+20) units.e = d.It implies;Calculating x.2x = ½(3x+20)4x = 3x+20x = 20 units.Recall.c = ½(x) units.And x = 20 units.c = ½(20)c = 10 units.Therefore, length CD is;(CD)²+c² = x²(CD)² = 20²+10²(CD)² = 400-100CD = √(300)CD = 10√(3) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3672167

Calculating Area Blue, Inscribed Square Area.Notice.2 units is the radius of the inscribed circle.tan60 = 2/a√(3) = 2/aa = ⅓(2√(3)) units.a = 1.15470053838 units.sin60 = 2/b½√(3) = 2/b4 = √(3)bb = ⅓(4√(3)) units.b = 2.30940107676 units.c = a+bc = ⅓(2√(3))+⅓(4√(3))c = ⅓(6√(3))c = 2√(3) units.Let r be the radius of the ascribed quarter circle.d = ½(r) units.e = c+de = (2√(3)+½(r)) units.f = (r-2) units.Calculating r.(r-2)² = 2²+(2√(3)+½(r))²r²-4r+4 = 4+12+2√(3)r+¼(r²)r²-4r = 12+2√(3)r+¼(r²)¼(3r²)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3672162

Calculating length PD, let it be x.Notice.At the inscribed triangle.AB should be 4 cm.While AC should be 6 cm.Resolving with the above corrections.5² = 6²+4²-2*6*4cosa48cosa = 52-25a = acos(27/48)a = 55.7711336722°a is angle BAC.b = ½(a)b = 0.5(55.7711336722)b = 27.8855668361°(5/sin55.7711336722) = (4/sinc)c = 41.4096221093°c is angle ACB.d = 180-a-cd = 180-55.7711336722-41.4096221093d = 82.8192442185°d is angle ABC.e = 180-b-ce = 180-27.8855668361-41.4096221093 e = 110.704811055°e is angle APC...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3672158

Calculating angle QRS.Let it be x.a = 45-32a = 23°a is ang SPQ.b = 90-ab = 90-23b = 67°b is angle PQS.c = b-45c = 67-45c = 22°c is angle RQS.d = 180-45d = 135°d is angle QSR.It implies, thec- required angle QRS (x) is;x = 180-c-dx = 180-22-135x = 23°OrA smart approach is;Angle QPS = Angle QRS = 23°They are angles at the same segment of the circle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3671255

Calculating x/y.Let the side length of the congruent squares be 1 unit each.x² = 1²+1²x² = 2x = √(2) units.a = 1+1+1a = 3 units.b² = 1²+a²b² = 1²+3²b = √(10) units.b is the radius of the circle.Andb = (x+y) units.Calculating y.√(10) = √(2)+yy = √(10)-√(2)y = √(2)(√(5)-1) units.It implies, x/y is;√(2)/(√(2)(√(5)-1))= 1/(√(5)-1)= (1+√(5))/(4)= ¼(1+√(5)) units.= 0.80901699437 units.

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