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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3488183

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Calculating Area Yellow Inscribed Trapezoid.(2√(2)/sin135) = (√(2)/sina)a = 20.70481105464°b = 180-20.70481105464-135b = 24.29518894536°(c/sin24.29518894536) = (2√(2)/sin135)c = 1.64575131106 units.Where c is the height of the yellow inscribed trapezoid.(2√(2)/sin45) = (√(2)/sind)d = 20.70481105464°e = 180-20.70481105464-45e = 114.29518894536°(2√(2)/sin45) = (f/sin114.29518894536)f = 3.64575131106 units.Where f is the base parallel length of the inscribed yellow trapezoid.g = 90-20.70481105464...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3487344

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Calculating Area Yellow.Notice.The ascribed quadrilateral is a square with side 6 units.a² = x²+3²a = √(x²+9) units.b = 2ab = 2√(x²+9) units.c = (6+x) units.Notice.b = cTherefore, calculating x.2√(x²+9) = 6+x4(x²+9) = (6+x)²4x²+36 = 36+12x+x²3x² = 12x3x = 12x = 4 units.d = 6-xd = 6-4d = 2 units.It implies, yellow area is;Area square with side 6 units - Area triangle with height 4 units and base 3 units - Area triangle with height 2 units and base 6 units.= 6²-½(4*3)-½(6*2)= 36-6-6= 36-12= 24 sq...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3487330

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Calculating Area Purple.a = 0.5(90-60)a = 15°cos15 =4/bb = 4.14110472164 cm.Where b is the side length of the inscribed regular triangle purple area.Purple area is;Area triangle with height 4.14110472164 cm and base 4.14110472164sin60 cm= 0.5*4.14110472164*4.14110472164sin60= 7.4256258422 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3486470

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Red Area is;7(area square with side 1 unit) - Area triangle with base 1 unit and height 4 units - Area square with side 1 unit - Area triangle with height and base 1 unit respectively.= 7(1*1) - 0.5(1*4) - (1*1) - 0.5(1*1)= 7 - 2 - 1 - 0.5= 3.5 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3485706

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Sir Mike Ambrose is the author of the question.8 = 4a = (8-a) Cross multiply.64-8a = 4a64 = 12aa = ⅓(16) cm.b = 8-ab = ⅓(8) cm.c = atan(⅔)°d = atan(3/2)°⅔ = 8/ee = 12 cm⅔ = f/8f = ⅓(16) cm.g = 8-fg = ⅓(8) cm.Area Orange exactly in cm² is;Area square units side 8 cm - Area rectangle with length 4 cm and width ⅓(8) cm.= 8² - 4*⅓(8)= 64 - ⅓(32)= ⅓(192-32)= ⅓(160) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3484848

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Sir Mike Ambrose is the author of the question.Calculating Area Shaded.y = 2^(x)P coordinate is;P(1, 2)dy/dx = 2^(x)In2For x = 1dy/dx = 2In2It implies;Q coordinate is;Q(0, (2-2In2))Therefore;Shaded area exactly in square units is;(Area under the curve at x = 0 and x = 1) - (Area trapezoid with parallel sides (2-2In2) unit and 2 unit, and height 1 unit).= (1/(In2)) - (½*1*(2+2-2In2))= (1/(In2)) - (2-In2)= (In2-1)²/In2 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3484033

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Sir Mike Ambrose is the author of the question.a² = 11*11+10*10-2*11*10cos50a = 8.92113926968 cm.(8.92113926968/sin50) = (10/sinb)b = 59.16920318624°c = 180-50-59.16920318624c = 70.83079681376° d = 0.5ad = 4.46056963484 cm.e² = 4.46056963484²+5.5²-2*4.46056963484*5.5cos59.16920318624e = 5 cm.(5/sin59.16920318624) = (5.5/sinf)f = 70.83079681376°g² = 4.46056963484²+5²-2*4.46056963484*5cos70.83079681376g = 5.5 cm.(5.5/sin70.83079681376) = (5/sinh)h = 59.16920318624°i = 180-59.16920318624-70.83079...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3483232

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Area m : Area n is;(√(2)x)² : (½(3x))²= 2x² : ¼(9x²)= 4(2x²) : 4(¼(9x²))= 8x² : 9x²= 8 : 9

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3482410

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Sir Mike Ambrose is the author of the question.Calculating a.½((0, 6√(3)) (-3, 3√(3)) (a, (16a/3)) (0, 6√(3))) = ¼(9+20√(3))Therefore;a = x = ½√(3) unity = 16a/3 = (16/3)(√(3)/2)y = ⅓(8√(3)) unit.Therefore;Area green is;Area triangle with height (√(697-72√(3))/2√(3)) unit and base 6sin(120-atan(16√(3)/(36-3√(3))) unit= ½*(√(697-72√(3))/2√(3))*6sin(120-atan(16√(3)/(36-3√(3)))= ½(√(2091-216√(3)))sin(120-atan(16√(3)/(36-3√(3))) = ½√(2091-216√(3)sin(120-atan((64√(3)+16)/141) square units.= 20.26666...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3481580

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Sir Mike Ambrose is the author of the question.Area green is;½*3*6sin120= ½(9√(3)) cm²Radius, r of the circle is;r = √(63)/√(3)r = √(21) cmTherefore;Area circle is;π(√(21))²= 21π cm²It implies;Area green ÷ Area circle is;½(9√(3)) ÷ 21π= 3√(3)/14π = 3√(3)/44

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