Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2782863

Sir Mike Ambrose is the author of the question. Let the side length of the large square be x. Calculating x. 25=x²+64-(16xcos20) Therefore; x = 11.70244 cm. It implies; Area large square is; 11.70244² = 136.947101954 cm² Area orange is; Area triangle with height and base 8 cm each - Area triangle with two side 2.91176187413 cm and 2.27177411013 cm, and angle 70° = ½(8)² - ½(2.91176187413*2.27177411013)sin70 = 32 - 3.107970027 = 28.892029973 cm² Therefore; Are...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2782853

Sir Mike Ambrose is the author of the question. Let the side length of the regular pentagon be 2 units. a² = 2²-1² a = √(3) units. b² = 5-4cos108 b = 2.49721204096 units. (2.49721204096/sin108) = (2/sinc) c = 49.61382244056° d = 90-c d = 40.38617755944° e² = 3+2.49721204096²-2√(3)*2.49721204096cos40.38617755944 e = 1.62694657231 units. (1.62694657231/sin40.38617755944) = (√(3)/sinf) f = 43.61382244075° g = 180-e-f g = 96° Let the radius of the inscribed circle b...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2782725

Calculating Area Red. 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 s...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2782384

Let EB be a. Let AE be b. It implies; a/b = √(2) --- (1). a+b = 1 --- (2). From (1). a = √(2)b --- (3). Substituting (3) in (2) to get b. √(2)b+b = 1 b(√(2)+1) = 1 b = (√(2)-1)/(2-1) b = √(2)-1 b = 0.4142135624 units. b is AE. At (2). a+b = 1 And b = 0.4142135624 units. a = 1-0.4142135624 a = 0.5857864376 units. a is EB. tanc = 1/0.4142135624 c = 67.5° c is angle AED. d =½(c) d = 33.75° Let x be the small inscribed circle radius. e = (0.4...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2781334

Let the radius of the circle be 2 units. a = (2-b) units. Calculating b. 2² = 2(2-b)² 4 = 2(4-4b+b²) b²-4b+2 = 0 (b-2)² = -2+(-2)² (b-2)² = 2 b = 2±√(2) It implies; b ≠ (2+√(2)) b = (2-√(2)) units. b = 0.5857864376 units. c = 2-b c = 2-0.5857864376 c = 1.4142135624 units. c = √(2) units. Therefore; The required angle, theta is; Let it be d. d = 45-atan(√(2)/2)) d = 45-35.2643896828 d = 9.7356103172° Therefore; tan(theta) is; tand tan(9.735610317...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2781317

Let the regular pentagon side length be 1 unit. sin72 = a/1 a = 0.9510565163 unit. Let the side length of the two inscribed congruent squares be b. c = ½(b) unit. b+½(b) = 0.9510565163 b = 0.63403767753 unit. Area Yellow is; (0.63403767753)² = 0.40200377653 square units. Calculating Area Blue. d = ½(1-0.63403767753) d = 0.18298116124 unit. e² = 2-2cos108 e = 1.61803398875 units. f = ½(1.61803398875-0.63403767753) f = 0.49199815561 unit. Area Blue is;...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2779791

Let a be the radius of the circle. b = 2a units. b is the diameter of the circle. Calculating a observing similar plane shape (right-angled triangle) ratios. It implies; 2a - 5 9 - a Cross Multiply. 2a² = 45 a = √(45/2) a = 3√(5)/√(2) a = ½(3√(10)) units. a = 4.7434164903 units. Again, a is the radius of the circle. Calculating the side length of the inscribed yellow square. Let it be c. 2a² = c² c² = 2(½(3√(10)))² c² = ½(90) c = √(45) c = 3√(5) units. c...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2779762

Let the radius of the ascribed half circle be a. Let the radius of the inscribed half circle be b. It implies; a² = b²+(0.5*12)² a²-b² = 6² a²-b² = 36 square units. Where (a²-b²) is the blue area.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2779739

Sir Mike Ambrose is the author of the question. Let the radius of the semi circle be 1 unit. Therefore; Area lilac is; Area triangle with height 0.26794919243 units and base 1 unit = ½(0.26794919243) = 0.13397459622 space units. Area large square is; Area square with side length 1 unit = 1² = 1 square unit. Area Lilac ÷ Area Large Square in its simplest decimal form is; 0.13397459622 ÷ 1 = 0.13397459622 ≈ 0.134 to 3 decimal places.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2779183

Let AB be 5 units. Radius, r of the inscribed circle is; 5*(2/5) r = 2 units. a = ⅛(180*6) a = 135° 2b² = 25 b = ½(5√(2)) units. c = 180-0.5(135)-45 c = 67.5° sin22.5 = ½(5√(2))/d d = 9.23879532511 units. Where d is the side length of the regular triangle (purple area). Area purple is; 0.5*9.23879532511*9.23879532511sin60 = 36.95994598698 square units. Calculating Area Shaded. e = 360-135-45-67.5-60 e = 52.5° f² = 50-50cos135 f = 9.23879532511 units...

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