Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2500086

Let the side length of the regular pentagon be 1 unit. a² = 2-2cos108 108° is the single interior angle of the regular pentagon. a = 1.6180339887 units. Area Blue is; 0.5*1*1.6180339887sin72 = 0.7694208843 square units. Calculating Yellow Area. b = ½(360-2(108)) b = ½(360-216) b = ½(144) b = 72° cos72 = c/1 c = 0.3090169944 units. d² = 0.3090169944²+1²-2*0.3090169944*1cos36 d = 0.7716809592 units. (0.7716809592/cos36) = (0.3090169944/sine) e = 13.6138224422°...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2500064

It implies; 2a² = √(2)² Where a is the radius of the half circle. Therefore; 2a² = 2 a² = 1 a = 1 unit. Again, a is the radius of the half circle. x (diameter of the half circle) is; x = 2a And a = 1 unit. x = 2(1) x = 2 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2497970

b = (180(12-2))/12 b = 1800/12 b = 150° b is the single interior angle of the regular dodecagon. 2c+2(150) = 360 2c = 60 c = 30° d = 150-15-30 d = 105° 2e+3(150) = 180(5-2) 2e = 540-450 2e = 90 e = 45° f = e-15 f = 30° g = 180-d-f g = 180-105-30 g = 45° It implies; (h/sin105) (2√(2)/sin30) h = 5.4641016151 units. j = 150-15-45 j = 90° k = 180-105-30 k = 45° (l/sin45) = (2√(2)/sin105) l = 2.0705523608 units. l is the side length of the regular dodecagon. m² = 2(2.0705523608)²-2(2.070...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2497946

Let the inscribed square side be 1 unit. sin30 = a/1 a = ½ units. a = 0.5 units. a is the side length of one of the four congruent isosceles inscribed shaded triangle. sin60 = b/1 b = ½√(3) units. c = b-a c = ½√(3)-½ c = ½(√(3)-1) units. c = 0.3660254038 units. d = a+a+c d = 0.5+0.5+½(√(3)-1) d = 1+½(√(3)-1) d = ½(1+√(3)) units. d = 1.3660254038 units. d is the side length of the ascribed square. Shaded Area is; Area square with side 1 unit + 2(area square with side ½ units). = 1²+2(0.5...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2497237

Let a be half AB. b = ½(3+7+2) b = 6 units. It implies; 6² = a²+c² c² = 36-a² c = √(36-a²) units. d = ½(7) d = 3.5 units. e = 4-3.5 e = 0.5 units. Therefore; c² = 0.5²+f² f² = 36-a²-0.25 f² = 35.75-a² f = √(35.75-a²) units. Calculating a, radius of the red half circle (half AB). a² = f²+3.5² a² = √(35.75-a²)²+3.5² 2a² = 35.75+12.25 2a² = 48 a² = 24 a = 2√(6) units. Again, a is the radius of the red half circle (half AB). It implies; AB = 2a AB =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2496487

a² = 12²+5² a = √(144+25) a = 13 cm. a is AD. 12b+5b+13b = 12*5 30b = 60 b = 2 cm. b is the radius of the inscribed circle. tanc = 12/5 c = atan(12/5)° tan(0.5c) = 2/d ⅔ = 2/d d = 3 cm. d is AB. e = 12-2 e = 10 cm. Let CD = f f+3f = 13 4f = 13 f = ¼(13) cm. Again, f is CD. g = 3*¼(13) g = ¼(39) cm. g is AC. h = g-3 h = ¼(39-12) h = ¼(27) cm. h is BC. j = (180-2atan(12/5))° k² = 2(¼(39))²-2(¼(39))²cos(180-2atan(12/5)) k = ½(15) cm. Area green...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2495869

tana = 1/2 a = atan(1/2)° b = 180-2a b = (180-2atan(½))° Notice! Radius of the inscribed circle is 1 unit. Area yellow is; Area triangle with height 2 units and base 1 unit - Area square with side 1 unit + Area quarter circle with radius 1 unit - Area sector with radius 1 unit and angle (180-2atan(½))° + Area triangle with height 1 unit and base (180-2atan(½)) units. = (½*2*1)-(1*1)+(0.25π*1*1)-((180-2atan(½))π*1*1/360)+(0.5*1*1sin(180-2atan(½))) = 1-1+0.7853981634-1.107148...

Mathematics Question and Solution.

https://onlineedumath.com/update/mathematics-question-and-solution/2495838

Calculating the area of the inscribed square. a = ½(45) a = 22.5° b = ½(180-45) b = ½(135) b = 67.5° cos22.5 = 6/c c = 6.4943532018 units. Let d be the side length of the inscribed square. e² =2d² e = √(2)d unit. e = 1.4142135624d units. (√(2)d/sin45) = (f/sin67.5) f = 1.847759065d units. (g/sin45) = (d/sin67.5) g = 0.7653668647d units. d is; c = f+g 1.847759065d+0.7653668647d = 6.4943532018 2.6131259297d = 6.4943532018 d = 2.4852813743 units. Again, d i...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2493984

sin30 = a/6 a = 3 units. b = 6+2a b = 12 units. Let the side length of the inscribed regular pentagon be c. Calculating c. d² = 2c²-2c²cos108 108° is the single interior angle of the regular pentagon. d = 1.6180339887c units. e²+(0.5c)² = (1.6180339887c)² e = 1.5388417685c units. tan60 = 0.5c/f f = 0.2886751346c units. It implies; e+f = b 1.5388417685c+0.2886751346c = 12 1.8275169031c = 12 c = 6.5662867357 units. Again, c is the side length of the inscribed r...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2493882

Radius of the half circle is 5 units. 5² = 4²+a² a = 3 cm. tanb = 4/3 b = atan(4/3)° c = 2b c = 2atan(4/3)° 5² = 3²+d² d = √(25-9) d = 4 cm. tane = 3/4 e = atan(3/4)° f = 2e f = 2atan(3/4)° Shaded Area is; Area sector with radius 5 cm and angle 2atan(4/3)° - Area triangle with height 5 cm and base 5sin(2atan(4/3)) cm - Area sector with radius 5 cm and angle 2atan(3/4)° + Area triangle with height 5 cm and base 5sin(2atan(3/4)) cm. = (2atan(4/3)π*5*5/360)-(0.5...

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