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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

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3480837

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Sir Mike Ambrose is the author of the question.Calculating Exactly, Golden Area ÷ Large Square Area.Let the side of the large square be 2 units.Therefore;Area large square is;2²= 4 square units.Area gold is;Area triangle with height 2 units and base ⅔ units - Area triangle with height 1 unit and base ½(√(2))sin135 units.= (½*⅔*2) - (½*1*½(√(2))sin135)= ⅔ - ¼= 5/12 square units.Therefore;Area Gold ÷ Area Large Square Exactly is;(5/12)÷4= 5/48

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3480462

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Calculating Area Green.Let theta be x.a = 90-39a = 51°x+x+27 = 512x = 51-272x = 24x = 12°b+x+39 = 180b = 180-39-xAnd x = 12°b = 180-39-12b = 129°Let c be the hypotenuse of the ascribed right-angled triangle.Therefore;c/sin129 = d/sin12d = (csin12)/(sin129)d = 0.26753235702c units.Calculating c.½*c*0.26753235702c*sin39 = 260.08418178377c² = 26c² = 308.85541783c = 17.5742828539 units.Again, c is the hypotenuse of right-angled triangle.sin39 = e/ce = csin39e = 17.5742828539sin39e = 11.0598545581 u...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3480099

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Calculating The Red Inscribed Semi Circle Area.Let x be the area of the inscribed half circle.a² = 2x²a = √(2)x units.Calculating x.1² = x²+(√(2)x)²1 = x²+2x²1 = 3x²x = 1/√(3) units.Again, x is the radius of the red inscribed half circle.Area red inscribed half circle is;½*πx²= ½*π(1/√(3))²= ⅙(π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3479279

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Total coloured area is;2(area triangle with height 6.69991708075 cm and base √(65)sin(atan(10)-atan(7/4)) cm= 2*½*6.69991708075*√(65)sin(atan(10)-atan(7/4))= 6.69991708075*3.28362272769= 22 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3478376

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Sir Mike Ambrose is the author of the question.Radius, r of the inscribed circle is;r = (3-√(3)) cmTherefore;Area purple exactly in square cm is;Area trapezium with parallel sides (3-√(3)) cm and 4 cm, and height 3+√(3) cm - 2(area triangle with height (3-√(3)) cm and base (3√(3)-3) cm).= ½*(3+√(3))(4+3-√(3)) - 2(½*(3-√(3))*(3√(3)-3))= (9+2√(3)) - (12√(3)-18)= (27-10√(3)) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3477389

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Sir Mike Ambrose is the author of the question. P(2, (16√(5)/5))Gradient of the curve at the tangent is;= 24/5√(5)Therefore;Gradient of the curve at the normal is;= -5√(5)/24It implies;Coordinate Q is;Q(4, (167√(5)/60))Therefore area R as a single fraction in square units is;(Area under the curve at x = 2 and x = 4) - (Area trapezium with parallel sides 167√(5)/60 units and 16√(5)/5 units, and height 2 units).= (24-⅓(8√(5))) - ½*2((167√(5)/60 )+(16√(5)/5))= 24 - ⅓(8√(5)) - (359√(5)/60)= (1440-5...
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