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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3505933

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Calculating Area Shaded STR (Triangle).Let the square side be a.tan60 = a/bb = a/tan60 cm.Where b is BS.Calculating a.2b+a = 62(a/tan60)+a = 6a((2/tan60)+1) = 6a = 6/((2/tan60)+1)a = 6(2√(3)-3) cm.Area Shaded STR is;½(a²)= ½(6(2√(3)-3))²= 54(7-4√(3)) cm²= 3.87702556513 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3504946

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Calculating Area Green Region.a² = 10²+10²-2*10*10cos150a = 19.31851652578 cm.Where a is AF which is equal AE.(19.31851652578/sin150) = (10/sinb)b = 15°Where b is angle BAF.tan15 = c/10c = 2.67949192431 cm.It implies;Area Green Region is;Area square with side 10 cm - 2(area triangle with height 2.67949192431 cm and base 10 cm.= 10² - ½*2(10*2.67949192431)= 100-26.7949192431= 73.2050807569 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3504120

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Calculating Length PQ.a = atan(2)°b = 180-2ab = (180-2atan(2))°Where b is angle FBP.Notice;CQ = 5 cm.Calculating FP using sine rule.Let FP be c.(c/sin(180-2atan(2))) = (5/sin(atan(2)))c = 4.472135955 cm.Therefore;Length PQ is;CF-CQ-FP= √(10²+5²)-5-4.472135955 = (3√(5)-5) cm.= 1.7082039325 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3502811

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Let the small inscribed square side be a.Let b be the side of the big inscribed square.It implies;b = 10-aTherefore;Calculating a.tan(90-75) = a/(10-a)a(1+tan15) = 10tan15a = (10tan15)/(1+tan15)a = ⅓(15-5√(3)) cm.b = 10-⅓(15-5√(3))b = ⅓(15+5√(3)) cm.Blue area is;a²+b²= (⅓(15-5√(3)))²+(⅓(15+5√(3)))²= ⅓(200) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3501846

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Let the radius of the inscribed circle be r.The square side is 8 cm.b = 8-rc = 2+rTherefore;(2+r)² = 2²+(8-r)²4+4r+r² = 4+64-16r+r²4r = 64-16r20r = 645r = 16r = (16/5) cm.Area circle is;πr²= 256π/25 cm²= 32.16990877276 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3500465

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Calculating Area Rectangle.Let BC be x.a²+x² = 15²a² = 225-x²a = √(225-x²) units.a is CD.15 ~ x5 ~ bCross Multiply.b = x/3 units.15 ~ √(225-x²)5 ~ cCross Multiply.c = ⅓√(225-x²) units.d = x+bd = x+(x/3)d = ⅓(4x) units.e = a+ce = √(225-x²)+⅓√(225-x²)e = ⅓(4√(225-x²)) units.f = 15+5f = 20 units.f is BE.g² = (⅓(4√(225-x²)))²+(x/3)²g² = ⅑(3600-16x²)+(x²/9)g² = 400-⅑(16x²)+(x²/9)g² = 400-⅑(15x²)g² = ⅑(3600-15x²)g = ⅓√(3600-15x²) units.g is CE.It implies;⅓√(3600-15x²) ~ 20√(225-x²) ~ xCross Multiply....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3499443

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Sir Mike Ambrose is the author of the question.Let length AB = 2 units.Therefore;Side length of the square, x is;2x² = 2²x = √(2) units.It implies;Area square is;x²= √(2)²= 2 square units.Area green is;Area triangle with height √(3) units and base 1 unit - Area triangle with height ⅓√(3) units and base 1 unit - Area triangle with height tan15 units and base 1 unit= ½(√(3)*1) - ½(⅓√(3)*1) - ½(tan15*1)= ½√(3) - (√(3)/6) - ½(tan15)= ⅓√(3) - ½(2-√(3))= (5√(3)-6)/6 square units.Therefore;Area Green...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3498160

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Calculating x (AD) the diameter of the half circle.a² = 2²+11²+2*2*11cosya² = 125+44cosy --- (1).a² = x²+7²-2*7*xcosya² = x²+49-14xcosy --- (2).Where a is AC.cosy = 7/x --- (3).Equating (1) and (2).125+44cosy = x²+49-14xcosy76 = x²-14xcosy-44cosy --- (4).Calculating x.Substituting (3) in (4).76 = x²-14x(7/x)-44(7/x)76 = x²-98-(308/x)174 = x²-(308/x)174x = x³-308x³-174x = 308It implies;x = 14 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3497356

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Calculating D, diameter of the ascribed half circle.Let r be the radius of the half circle.a = ½(7)a = 3.5 units.b²+a² = r²b² = r²-3.5²b = √(r²-12.25) units.c = 2bc = 2√(r²-12.25) units.cosx = 7/(2r) --- (1).(2√(r²-12.25))² = 2(3²)+2*3*3cosx4(r²-12.25) = 18+18cosx4r²-49 = 18+18cosx4r²-67 = 18cosxcosx = (4r²-67)/18 --- (2).Equating (1) and (2).7/(2r) = (4r²-67)/187/r = (4r²-67)/9Cross Multiply.63 = 4r³-67rIt implies;r = 4.5 units.Therefore, D, diameter of the half circle is;D = 2rAnd r = 4.5 uni...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3496461

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Sir Mike Ambrose is the author of the question.Let the side let of the square be 2 units.Therefore;Area square is;2²= 4 square units.It implies;Area R is;Area trapezoid with parallel sides 1 unit and 1.44948974279 unit, and height 2 unit - Area square with side 1 unit - Area triangle with base and height 1 unit respectively - Area sector with radius √(2) unit and angle 32.3335306935° - Area triangle with height √(2) unit and base 0.3178372452 unit= (½*2(1+1.44948974279)) - (1*1) - (½*1*1) - (32...

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media video Young Sheldon

Young Sheldon

media video Young Sheldon

Young Sheldon

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Maths at The Mela

media video Solving Simultaneous Linear Equation Using Elimination and Substitution Approach

Solving Simultaneous Linear Equation Using Elimination and Substitution Approach

media video 46 Fun Mathematics Questions with Answers

46 Fun Mathematics Questions with Answers

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