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Mathematics Question and Solution

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

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

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

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

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

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

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

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

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

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

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...