Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2695220

Let the side length of the inscribed blue square be a. b²+a² = 6² b = √(36-a²) units. c = a+b c = (a+√(36-a²)) units. It implies; (a+√(36-a²)) - 6 10 - (a+√(36-a²)) Cross Multiply. 60 = (a+√(36-a²))² √(60) = a+√(36-a²) (2√(15)-a)² = 36-a² 60-4√(15)a+a² = 36-a² 2a²-4√(15)a+24 = 0 a²-2√(15)+12 = 0 (a-√(15))² = -12+√(15)² a = √(15)±√(3) It implies; a = √(15)-√(3) a = √(3)(√(5)-1) units. a = 2.1409325386 units. Again, a is the side length of the inscribed blue...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2695207

Let the base of the triangle be 2 units. a = 180-30-135 a = 15° (1/sin15) = (b/sin135) b = 2.7320508076 units. c² = 2²+2.7320508076²-2*2*2.7320508076cos30 c = 1.4142135624 units. Therefore, the required angle, x is; (1.4142135624/sin45) = (1/sinx) x = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2694752

Sir Mike Ambrose is the author of the question. Area largest square is; (3√(5))² = 45 cm² Area purple is; Area triangle with height equal base equal 3√(5) cm - area triangle with side 2.23606797751 cm and 2.37170824513 cm, and angle 45° - area triangle with side 1.49071198498 cm and 1.72487872371 cm, and angle 45° = ½(3√(5))² - ½(2.23606797751*2.37170824513)sin45 - ½(1.49071198498*1.72487872371)sin45 = (45/2) - (15/8) - (10/11) = (1735/88) cm² Therefore; Area purple ÷ Ar...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2693903

Let AB = CD = 1 unit. a = 180-40-30 a = 110° a is angle CAD. (1/sin110) = (b/sin40) b = 0.6840402867 units. b is AC. Therefore, the required angle, x is; (0.6840402867/sinx) = (1/sin30) x = 20°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2693601

Calculating area shaded blue. a² = (2+√(3))²+(3+2√(3))² a² = 4+4√(3)+3+9+12√(3)+12 a² = 28+16√(3) a = √(28+16√(3)) units. a = 7.4641016151 units. a is the hypotenuse of the ascribed right-angled triangle. Let b be the radius of the inscribed half circle. c = (2+√(3))-b c = (3.7320508076-b) units. d = √(28+16√(3))-(3+2√(3)) d = 7.4641016151-6.4641016151 d = 1 unit. Calculating b. (3.7320508076-b)² = 1²+b² 13.9282032303-7.4641016151b = 1 12.9282032303 = 7.4641016151b...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2693506

Let the side length of the regular octagon be 1 unit. a = ⅛*180(8-2) a = ⅛(180*6) a = 135° a is the single interior angle of the regular octagon. 2b² = 1² b² = ½ b = ½√(2) units. c = 1+b c = (1+½√(2)) units. c is the side length of the square. Therefore, the required angle, x is; tanx = (½√(2))/(1+½√(2)) x = 22.5°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2693501

Let the side length of the inscribed regular hexagon which is equal the side length of the ascribed regular nonagon be 1 unit. a = ⅙*180(6-2) a = 120° a is the single interior angle of the inscribed regular hexagon. b = ⅑*180(9-2) b = 140° b is the single interior angle of the ascribed regular nonagon. c = 140-120 c = 20° d = 180-20-60 d = 100° (1/sin100) = (e/sin20) e = 0.3472963553 units. (1/sin100) = (f/sin60) f = 0.8793852416 units. g = 1-f g = 0.1206147584 u...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2693497

a = 180-45-75 a = 60° b = 180-75-60 b = 45° c = a-b c = 15° d = 180-60-45-c d = 60° e = 180-45-d e = 75° Therefore, the required angle, alpha, let it be f, is; f = 180-d-e f = 180-60-75 f = 180-135 f = 45°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2692696

a = 1+1.5 a = 2.5 units. b²+1.5² = 2.5² b² = 6.25-2.25 b = √(4) b = 2 units. Let c be the radius of the ascribed semi circle. d = 1+2 d = 3 units. e = (3-c) units. f = 1.5+e f = 1.5+3-c f = (4.5-c) units. Calculating c. c² = (4.5-c)²+1.5² c² = 20.25-9c+c²+2.25 9c = 22.5 c = 2.5 units. Again, c is the radius of the ascribed half circle. Therefore, the required length, ?, diameter of the ascribed half circle is; ? = 2c ? = 2*2.5 ? = 5 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2692646

Let the base of the inscribed green triangle be a. b = (8-a) units. c² = a²+8² c = √(a²+64) units. It implies; (8-a)+8 = √(a²+64) (16-a)² = a²+64 256-32a+a² = a²+64 32a = 192 4a = 24 a = 6 units. Therefore, area green inscribed triangle is; ½*a*8 = ½*6*8 = 24 square units.

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