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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2832555

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Calculating green inscribed triangle area. a² = 2(3√(2))² a² = 2*18 a = √(36) a = 6 units. a is AB. b = ½(360-90) b = ½(270) b = 135° b is angle ABC. c² = 6²+(4√(2))²-2*6*4√(2)cos135 c = 2√(29) units. c = 10.7703296143 units. c is AC. 2d² = (2√(29))² d² = 2*29 d = √(58) units. d = 7.6157731059 units. d is AO = CO the radius of the ascribed quarter circle. sine = (0.5*6)/7.6157731059 e = 23.1985905135° f = 2e f = 46.3971810271° g = 90-f g = 43.6028189729°...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2825213

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a² = R²+6² a = √(R²+36) units. a is OB. b = a-4 b = (√(R²+36)-4) units. c = 6+4 c = 10 units. It implies; 10² = R²+(√(R²+36)-4)² 100 = R²+R²+36-8√(R²+36)+16 48 = 2R²-8√(R²+36) 8√(R²+36) = 2R²-48 4√(R²+36) = R²-24 16(R²+36) = R⁴-48R²+576 16R²+576 = R⁴-48R²+576 16R² = R⁴-48R² R² = 64 R = 8 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2824411

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Let a be the big inscribed circle radius. Let b be the bigger inscribed circle radius. Let c be the biggest inscribed circle radius. It implies; a+b = 12 --- (1). a+c = 22 --- (2). b+c = 26 --- (3). Therefore, resolving (1), (2) and (3) simultaneously to get a, b and c. Subtracting (2) from (3). (b+c = 26)-(a+c = 22) It implies; b-a = 26-22 b-a = 4 a = b-4 --- (4). Substituting (4) in (1) to get b. a+b = 12 And at (4), a = b-4 b-4+b = 12 2b = 12+4 b...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2811317

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Let the two equal lengths be 1 unit. (a) Calculating Theta exactly. Let theta be x. Calculating x. x = ½(90-54) x = ½(36) x = 18° sin18 = a/1 a = 0.30901699437 units. b = 90-18 b = 72° sin72 = c/1 c = 0.9510565163 units. tan36 = 0.9510565163/d d = 1.30901699437 units. Area Blue is; 0.5(1.30901699437*0.9510565163) = 0.62247457122 square units. e = 270-54-96 e = 120° f = 360-120-72 f = 168° g² = 2-2cos168 g = 1.98904379074 units. h = ½(180-168...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2805007

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tana = 5/10 a = atan(½)° b = 2a b = 2atan(½)° c = b-45 c = 8.1301023542° d² =10²+10²-2*10*10cos8.1301023542 d = 1.4177804018 units. e = ½(180-8.1301023542) e = 85.9349488229° f² = 2*5² f = 5√(2) units. g = 10-f g = 2.9289321881 units. tanh = 10/5 h = atan(2)° j = 2h j = 2atan(2)° k = j-90 k = 36.8698976458° It implies, area red is; Area triangle with height 1.4177804018 units and base 2.9289321881sin85.9349488229 units + Area sector with radius 10 un...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2802130

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Sir Mike Ambrose is the author of the question. Let the side of the inscribed square be 2 units. Therefore; Area S is; Area triangle with two side 1.03528 units and 0.53589838486 units, and angle 45° = 0.5*1.03528*0.53589838486sin45 = 0.1961531464 space units. Area R is; Area triangle with two side (2√(2)-2) units and 0.65961372961 units, and angle 72.367664609° = 0.5*(2√(2)-2)*0.65961372961sin72.367664609 = 0.26038499708 space units. Area T is; Area sector with radi...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2798968

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Sir Mike Ambrose is the author of the question. Let the small inscribed circle's radius be x. x = 6 cm. Let the big inscribed circle's radius be y. y = 8 cm. Let the ascribed semi circle's radius be z.  = 18 cm. a² = 10²-8² a = 6 cm. b² = 12²-6² b = 6√(3) cm c = atan(z/b)° c = atan(18/6√(3))° c = atan√(3)° c = 60° d = atan(z/a)° d = atan(18/6) d = atan(3)° (a) Angle theta exactly is; Let it be e. e = 180-c-d e = 180-60-atan3 e = (120-atan3)° e = (atan(⅓)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2797465

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a = 4+6 a = 20 cm. a is the radius of the quarter circle. tanb = 10/4 b = atan(5/2)° c = b+60 c = (60+atan(5/2))° c = 128.1985905136° (10/sin128.1985905136) = (4/sind) d = 18.3215084855° e = 180-d-c e = 180-18.3215084855-128.1985905136 e = 33.4799010009° (10/sin128.1985905136) = (f/sin33.4799010009) f = 7.0195191472 cm. g = 180-atan(5/2)-60 g = (120-atan(5/2))° g = 51.8014094864° Therefore, are blue is; Area quarter circle with radius 10 cm - Area triangle w...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2793964

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a = 180-54-30-18 a = 180-102 a = 78° a is angle ABC. b = 78+54 b = 132° b is angle BAD. c = 180-b-18 c = 180-150 c = 30° c is angle ADB. Let AD = 1 unit. (1/sin18) = (d/sin30) d = 1.6180339887 units. d is AB. (1.6180339887/sin54) = (e/sin(18+30)) e = 1.486289651 units. e is AC. f² = 1²+1.486289651²-2*1*1.486289651cos54 f = 1.2090569266 units. f is CD. Therefore, the required angle, ? is; Let it be g. (1.2090569266/sin54) = (1/sing) g = 42° Again, g i...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2791220

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Let a be the radius of the circle. 10² = 2a²+2a²cosd 50 = a²+a²cosd --- (1). 10² = (2a)²+e² e = √(100-4a²) units. √(100-4a²)² = 2a²-2a²cosd 100-2a² = 2a²-2a²cosd 2a²cosd = 4a²-100 cosd = (2a²-50)/a² --- (2). Substituting (2) in (1). 50 = a²+a²(2a²-50)/a² 100 = 3a² a² = ⅓(100) a = ⅓(10√(3)) units. a = 5.7735026919 units. Again, a is the radius of the circle and also the width of the blue area. cose = 5/a cose = 5/⅓(10√(3)) cose = 15/(10√(3)) cose = 3/(2√(3)) cos...
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