Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2927091

Let the inscribed purple regular heptagon side be 1 unit. a = ⅐(180*5) a = ⅐(900)° b = 180-⅐(900)) b = ⅐(360)° c = 180-2(⅐(360)) c = ⅐(540)° (1/sin⅐(540)) = (d/sin⅐(360)) d = 0.8019377358 unit. e = 2(0.8019377358)+1 e = 2.60387547161 units. f = 90-½(⅐(540)) f = ⅐(360)° cos⅐(360) = g/2.60387547161 g = 1.62348980186 units. h = 2g h = 3.24697960372 units. Where h is the base of the ascribed quadrilateral. cos(180/7) = i/0.8019377358 i = 0.72252093395 unit....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2925743

Let the side of the regular pentagon be 1 unit. Notice; The regular pentagon side is equal the regular heptagon side. a = ⅐(180*5) a = ⅐(900)°, the single interior angle of the regular heptagon. b = 2a b = ⅐(450)° tan(⅐(450)) = c/0.5 c = 1.03826069829 units. cos(⅐(450)) = 0.5/d d = 1.15238243548 units. Where d is the radius of the ascribed circle. e = 180-⅐(900) e = ⅐(360)° tan18 = 0.5/f f = 1.53884176859 units. g² = 1.15238243548²-0.5² g = 1.03826069828 u...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2923831

a² = 11*11+10*10-2*11*10cos50 a = 8.92113926968 cm. (8.92113926968/sin50) = (10/sinb) b = 59.16920318624° c = 180-50-59.16920318624 c = 70.83079681376° d = 0.5a d = 4.46056963484 cm. e² = 4.46056963484²+5.5²-2*4.46056963484*5.5cos59.16920318624 e = 5 cm. (5/sin59.16920318624) = (5.5/sinf) f = 70.83079681376° g² = 4.46056963484²+5²-2*4.46056963484*5cos70.83079681376 g = 5.5 cm. (5.5/sin70.83079681376) = (5/sinh) h = 59.16920318624° i = 180-59.16920318624-70.83...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2923822

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/2922210

Area of coloured region is; Area triangle with height 53.3479501083 m and base 60sin(102.042575143) m + Area triangle with height 100.027406452 m and base √(5924)sin(12.5245961777) m = (½*53.3479501083*60sin(102.042575143)) + (½*100.027406452*√(5924)sin(12.5245961777)) = 1565.21739131 + 834.782608691 = 2400 m²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2922194

Sir Mike Ambrose is the author of the question. Let beta be x. x = (½*atan(5/4)) Let alpha be y. y = (½*atan(4/5))° Therefore; Area blue exactly in cm² decimal is; Area triangle with height (10/cos(0.5atan(4/5))) cm and base (8/cos(0.5atan(5/4)))sin45 = ½* (10/cos(0.5atan(4/5)))*(8/cos(0.5atan(5/4)))sin45 = √(2)(5/cos(0.5atan(4/5)))(4/cos(0.5atan(5/4))) = 33.2562362737 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2920536

a = 360-140-80 a = 140° cos20 = 1/b b = 1.06417777248 units. (1.06417777248/sin30) = (c/sin70) c = 2 units. d² = 4+1.064177772482²-4*1.06417777248cos140 d = 2.89711998506 units. (2.89711998506/sin140) = (2/sine) e = 26.34291130662° f = 40-e f = 13.65708869338° g = 20+26.34291130662 g = 46.34291130662° tan46.34291130662 = h/2 h = 2.09602104183 units. Area ABC is; 0.5*2*2.09602104183 = 2.09602104183 square units. Calculating Area Blue. i = 60-26.34291130662 i = 33.65708869338= (1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2920515

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) units. y = 16a/3 = (16/3)(√(3)/2) y = ⅓(8√(3)) units. Therefore; Area green is; Area triangle with height (√(697-72√(3))/2√(3)) units and base 6sin(120-atan(16√(3)/(36-3√(3))) units. = 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...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2918704

Let the radius of the ascribed semi circle be 2 units. Let the radius of the inscribed circle be r. Calculating r. (2-r)² = r² + 1 Therefore; r = ¾ units. It implies; OH = r = ¾ units. Notice; AH = AK = 3 units. Therefore; AH ÷ OH is; 3 ÷ (¾) = 4 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2918689

Shaded area is; Area triangle with height 4√(3) cm and base 8 cm + 2(area sector with radius 8 cm and angle 60° - area triangle with height 4√(3) cm and base 8 cm) - Area circle with radius 2 cm - 2(area sector with radius 4 cm and angle 60°) - 2(area sector with radius 4 cm and angle 60° + area triangle with height 2√(3) cm and base 4 cm). = 16√(3) + 2(⅓(32π)-16√(3)) - 4π - 2(⅓(8π)) - 2(⅓(8π)+4√(3)) = 16√(3) + ⅓(64π) - 32√(3) - 4π - ⅓(16π) - ⅓(16π) + 8√(3) = ⅓(32π)-4π-8√(3) =...

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