Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3564546

Sir Mike Ambrose is the author of the question.Area R ÷ Area S to 2 decimal places.Let the base of the inscribed rectangle be 2 units.tan60 = a/(½(2))√(3) = aa = √(3) units.a is the height of the inscribed rectangle.Calculating the single side length of the inscribed square, let it be x.It implies;x+((x)/√(3))+√(3)x=2x=2√(3)/(4+√(3)) units.Therefore;Area R is;x² = (2√(3)/(4+√(3)))²= 12/(19+8√(3)) square units.= 0.36522557677 square units.Area S is;Area triangle with height √(3) units and base 1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3564540

Sir Mike Ambrose is the author of the question.Calculating Area Triangle WArea W to 3 decimal places is;Area triangle with length 5 units and base 7.5sin86.4166783015 units - Area triangle of with length 5 units and 4sin86.4166783015 units - Area triangle with length 2.75 units and 3.5sin34.7719440318 units.= 5.98826978347≈ 5.988 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3564520

Calculating Area Blue.a² = 2(12)²a = 12√(2) units.a is the diagonal of the square.b = ½(a)b = 6√(2) units.tanc = 4/12c = atan(⅓)°d = 45-cd = (45-atan(⅓))°d = 26.5650511771°e² = 12²+4²e = √(12²+4²)e = 12.6491106407 units.sin26.5650511771 = r/(6√(2))r = 3.7947331922 units.r is the radius of the inscribed circle.cos26.5650511771 = f/(6√(2))f = 7.5894663844 units.g = e-fg = 12.6491106407-7.5894663844g = 5.0596442563 units.tanh = (12/4)h = atan(3)°h = 71.5650511771°tanj = r/gj = atan(3.7947331922/5....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3563345

Calculating Length AB.Notice.The ascribed plane shape is a square.Let x be the radius of the inscribed half circle.Let y be the radius of the inscribed circle.a² = 2(12)²a = 12√(2) cm.b = a-12b = (12√(2)-12) cm.Calculating x.It implies;x = bx = (12√(2)-12) cm.x = 4.97056274848 cm.Again, x is the radius of the inscribed half circle.c = 12-xc = 12-(12√(2)-12)c = (24-12√(2))c = 7.02943725152 cm.Calculating y.d = ½(12√(2))-yd = (6√(2)-y) cm.It implies;2y² = (6√(2)-y)²2y² = 72-12√(2)y+y²y²+12√(y)-72...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3562569

Calculating Area of the inscribed circle.a ~ 66 ~ 8Cross Multiply.8a = 6*62a = 9a = ½(9) units.b = a+8b = ½(9)+8b = ½(25) units.b is the diameter of the ascribed half circle.c = ½(b)c = ¼(25) units.c is the radius of the ascribed half circle.d = c-ad = ¼(25)-½(9)d = ¼(7) units.Let e be the radius of the inscribed circle.f = e-df = e-¼(7)f = ¼(4e-7) units.g = c-eg = ¼(25)-eg = ¼(25-4e) units.Calculating e.g² = e²+f²(¼(25-4e))² = e²+(¼(4e-7))²(625-200e+16e²)/16 = e²+(16e²-56e+49)/16625-200e+16e²...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3562568

Calculating Area of the Blue inscribed half circle.Let r be the radius of the inscribed half circle.a²+12² = r²a = √(r²-144) cm.b = r-ab = (r-√(r²-144)) cm.c = 2(12)c = 24 cm.Calculating r.(r-√(r²-144)) ~ 1212 ~ 24 Cross Multiply.24(r-√(r²-144)) = 144r-√(r²-144) = 6(r-6)² = r²-144r²-12r+36 = r²-14412r = 1802r = 30r = 15 cm.Again, r is the radius of the inscribed half circle.Area inscribed half circle is;½(πr²)= ½*15*15π= ½(225π) cm²= 353.429173529 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3560557

Let a be the side length of the regular hexagon.3² = 1²+a²-2acosb2acosb = a²-8cosb = (a²-8)/(2a) --- (1).c²+(a²-8)² = (2a)²c² = 4a²-(a⁴-16a²+64)c = √(-a⁴+20a²-64) units.sinb = c/(2a)sinb = √(-a⁴+20a²-64)/(2a) --- (2).(2√(3))² = 1²+a²-2acos(240-b)11 = a²-2a(cos240cosb+sin240sinb)11 = a²-2a(-½(cosb)-½√(3)sinb)11 = a²+acosb+a√(3)sinb --- (3).Substituting (1) and (2) in (3).11 = a²+a((a²-8)/(2a))+a√(3)(√(-a⁴+20a²-64)/(2a))11 = a²+½(a²-8)+½√(3)√(-a⁴+20a²-64)22 = 2a²+a²-8+√(3)√(-a⁴+20a²-64)30-3a² = √...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3560515

Calculating Area of the inscribed circle.Let r be the radius of the inscribed circle.a = (9-r) units.b = (9+r) units.It implies;c²+a² = b²c² = (9+r)²-(9-r)²c² = 81+18r+r²-(81-18r+r²)c² = 36rc = √(36r)c = 6√(r) units.d = c-9d = (6√(r)-9) units.e = (9-r) units.Calculating r.(9-r)² = r²+(6√(r)-9)²81-18r+r² = r²+36r-108√(r)+81-18r = 36r-108√(r)108√(r) = 54r2√(r) = r4r = r²r = 4 units.Again, r is the radius of the inscribed circle.Therefore, area inscribed circle is;πr²= 4²*π= 16π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3559752

Calculating r, radius of the quarter circle.a = x+16a = (x+16) cm.b = 9+12b = 21 cm.Calculating x.12*(9+21) = 16*(x+(x+16))12*30 = 16*(2x+16)90 = 8x+64x = 26/8x = 13/4x = 3.25 cm.Recall.a = x+16And x = 3.25 cm.a = 3.25+16a = 19.25 cm.Therefore r, radius of the quarter circle is;r² = 9²+19.25²r = √(9²+19.25²)r = 21.25 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3559728

Sir Mike Ambrose is the author of the question.Calculating Area Yellow ÷ Area ABCD exactly as a single fraction.Let the bigger inscribed square be 2 units.BC = 1+2+x = (3+x) units.Calculating x.(3+x)²=1²+8-4√(2)cos1353+x=√(13)x = (√(13)-3) units.Therefore;BC (side of the ascribed square ABCD) is;BC = 3+√(13)-3BC = √(13) units.Therefore;Area ABCD is;√(13)²= 13 square units.Area Yellow is;Area trapezium with parallel sides √(13) units and (√(13)-3) unit respectively, and height (√(13)-2) unit - A...

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Mathematics Question and Solution

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