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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1940597

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Let a side length of the three congruent regular pentagon be 1 unit. Single interior angle of the three congruent regular pentagon is; ⅕(180*3) = 108° a² = 1²+0.5²-2*1*0.5cos108 a = 1.24860602048 units. (1.24860602048/sin108) = (1/sinb)  b = 49.6138224408° c² = 2-2cos108 c = 1.61803398875 units. d = c+1+1 d = 3.61803398875 units. e² = 1+3.61803398875²-2*3.61803398875cos108 e = 4.04057395938 units. (4.04057395938/sin108) = (1/sinf) f = 13.6138224408° g = 180-108-13.6138224408 g = 58.386...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1940540

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a² = 8²-2² a = √(60) a = 2√(15) units. Let the radius of the inscribed circle be r. Considering similar triangle ratios, calculating r. r = 2√(15) 1 = 2 Cross Multiply. r = √(15) units. b² = 1+r² b² = 1+√(15)² b = √(16) b = 4 units. c = 2+4+√(15)+x c = (6+√(15)+x) units. d² = (x+√(15))²-√(15)² d² = x²+2√(15)x+15-15 d = √(x²+2√(15)x) Again, considering similar triangle ratios to get x. √(15) = 2√(15) √(x²+2√(15)x) = (6+√(15)+x) Cross Multiply √(15)(6+√(15)+x) = 2√(15)*√(x²+2√(15)x) (6√(1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1938983

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Notice; Angle DCE = -(Angle BCG) Let Angle DCE be x. Calculating K. 4k² = 4k²+k²-4k*kcosx cosx = (¼) --- (1). 144 = 4k²+k²-2k²(-cosx) 144 = 5k²-2k²(-cosx) Substituting (1) in (2). 144 = 5k²-2k²(-¼) 144 = 6k² k² = 24 k = 2√(6) units. It implies; 2k is 4√(6) units. Total square areas is; (4√(6))²+(2√(6))² = 16*6+4*6 = 6(16+4) = 6(20) = 120 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1938157

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Let a be the side of the inscribed square. 24² = a²+(2a)² 576 = a²+4a² 576 = 5a² a = √(576/5) a = 10.733126292 cm. b = ½(24-10.733126292)  b = 6.633436854 cm. b = AM = DR. c = atan(6.633436854/(6+10.733126292)) c = 21.62463392714° c = Angle BPQ. tan21.62463392714 = d/10.733126292 d = 4.25488424946 cm. d = MQ. e = 0.5(10.733126292)+4.25488424946 e = 9.62144739546 cm. e = MO. f = 9.62144739546+12 f = 21.62144739546 cm. f = DM. g = atan(12/21.62144739546) g = 29.03047057929° g = Angle CMD...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1937325

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AB = BC = AC = 6 cm. cos60 = 2/BE BE = 4 cm. (DE)² = 4²-2² DE = 2√(3) cm. (CE)² = (2√(3))²+4² CE = 12+16 CE = 2√(7) cm EF = (2√(7)-4) cm. EF = 1.29150262213 cm. AE = 6-4 AE = 2 cm. Angle AEF = 180-(atan(2/2√(2√(3)))+atan(4/2√(3))) Angle AEF = 100.89339464913° Therefore; Area Shaded, Triangle AEF is; 0.5*2*1.29150262213sin100.89339464913 = 1.26823039846 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1936683

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Let CE be 2 units. Let the square side be a. b =√(2)a unit. c = (√(2)a-2) units. Calculating a  2²+2² = (√(2)a-2)² 8 = 2a²-4√(2)a+4 2a²-4√(2)a-4 = 0 a²-2√(2)a-2 = 0 (a-√(2))² = 2+2 a = √(2)±2 Therefore; a ≠ (√(2)-2) units. a = (√(2)+2) units. a is AD. tand = (2/(√(2)+2)) d = atan(2/(√(2)+2))° d = 30.36119340482° d is angle ABF. (BG)² = 1+3.41421356237²-2*1*3.41421356237cos45 BG = 2.79793265193 units. (2.79793265193/sin45) = (1/sine) e = 14.63880659519° e is angle CBG. Calculating re...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1935913

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a² = (2r)²+r² a = √(5r²) a = √(5)r units. b = a-r b = (√(5)r-r) units. Considering similar triangle ratios. Calculating r.  (√(5)r-r) = 2r √(2) = r Cross Multiply.  (√(5)r-r) = 2√(2) r(√(5)-1) = 2√(2) r = (2√(2))/(√(5)-1) r = ½(√(2)+√(10)) units. r = 2.28824561127 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1935892

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Let the base of the composite triangle be 1 unit. a = 180-3-6 a = 171° (1/sin171) = (b/sin3) b = 0.33455515209 units. c = 180-30-87 c = 180-117 c = 63° (d/sin30) = (1/sin63) d = 0.56116311882 units. e² = 0.56116311882²+0.33455515209²-2*0.56116311880.334555152092*0.33455515209cos81 e = 0.60670683106 units. Calculating x. (0.60670683106/sin81) = (0.33455515209/sinx) x = 33°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1935400

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i = k  And i = 180-101-39 = 40° Therefore; k = 40° While j = 101° Or Calculating i. i = 90-11-39 i = 90+50 i = 40° a = 180-2(11) a = 180-22 a = 158° b = 360-a b = 360-158 b = 202° Calculating j. j = ½(b) j = ½(202) j = 101° Calculating k. k+j+39 = 180 k+101+39 = 180 k = 180-140 k = 40°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1935378

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Considering similar triangle ratios. 3 = (3+x) y = (4+y) Where x and y are the width and length of the inscribed rectangle ascribing the red region respectively. Cross Multiply. 12+3y = 3y+xy xy = 12 square units. Therefore; Area Red Region is; ½(xy) = ½(12) = 6 square units.
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