Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2757249

a² = 3²+4² a = √(25) a= 5 units. a is the radius of the yellow semi circle. 4² = 2*5²-2*5²cosb 16 = 50-50cosb 50cosb = 50-16 cosb = 34/50 b = acos(17/25) b = 47.1563569564° c = 2b c = 94.3127139128° Therefore, length AB is; AB = √(5²+5²-2*5*5cos94.3127139128) AB = 7.3321211119 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2757245

Calculating radius of the inscribed circle. a² = 4²+4²-2*4*4cos(90-30) a² = 32-32cos60 a² = 16 a = 4 units. Let b be the radius of the inscribed circle. c = (4-b) units. d = 60-45 d = 15° e² = 2b² e = √(2)b units. Calculating b. It implies; cos15 = (0.5*4)/e cos15 = 2/√(2)b b = 2/(√(2)cos15) b = 1.4641016151 units. Again, b is the radius of the inscribed circle. Therefore, area inscribed circle is; πb² = π(1.4641016151)² = 6.7342977158 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2756232

Let the two equal lengths be 1 unit each. a² = 2(1²) a = √(2) units. b = 180-60-45 b = 75° (√(2)/sin45) = (c/sin75) c = 1.9318516526 units. d = c-a d = 0.5176380902 units. e = 180-45 e = 135° It implies; f² = 0.5176380902²+1.9318516526-2*0.5176380902*1.9318516526cos135 f = √(2) units . Calculating the required angle, x. (√(2)/sin135) = (1/sinx) x = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2756094

Let a = 1 unit. b = ½(2+1) b = 1.5 units. b is the radius of the semi circle. c = (1.5-d) units. d is the radius of the inscribed circle. e = 2-1.5 e = 0.5 units. It implies; (1.5-d)² = d²+0.5² 2.25-3d+d² = d²+0.25 3d = 2 d = ⅔ units. Again, d is the radius of the inscribed circle. tanf = (2/3)/2 f = atan(⅓)° Therefore, the required angle, x is; x = 2f x = 2atan(⅓)° x = 36.8698976458°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2755522

Sir Mike Ambrose is the author of the question. Let the side length of the square be 10 units. Area square is; 10² = 100 square units. Therefore; Area R is; Area semi circle with radius 5 units - Area sector with radius 5 units and angle (180-2atan(2))° - Area kite with diagonal lengths √(10²+5²) units and √(200-200cos(2atan(½))) units + Area sector with radius 10 units and angle 2atan(½)° - 2(area quarter circle with radius 5 units - area triangle with height and base 5 unit...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2755513

Calculating Area Orange. Let the side length of the inscribed regular triangle be a. It implies; 2a = 15 a = 7.5 units. Area sector with radius 15 units and angle 60° - Area regular triangle with side length 7.5 units. = (60π*15*15/360)-(½*7.5*7.5sin60) = ½(75π)-¼(56.25√(3)) = 93.4527600282 square.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2755502

Calculating inscribed shaded area green. a = ⅛*180(8-2) a = ¼*180*3 a = 135° a is the single interior angle of the regular octagon. 2b² = 4² b² = 8 b = 2√(2) units. c = 2b c = 4√(2)units. Therefore, green inscribed area is; Area triangle with height 4√(2) units and base 4√(2)sin45 units - Area triangle with height and base 2√(2) units respectively. = (½*4√(2)*4√(2)sin45)-(½*2√(2)*2√(2)) = 8√(2)-4 = 4(2√(2)-1) square units. = 7.313708499 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2755494

Let the side length of the square be a. b²+(0.5a)² = 5² b = √(25-¼(a²)) units. c = a-b c = (a-√(25-¼(a²))) units. It implies, calculate a. 8² = (0.5a)²+(a-√(25-¼(a²)))² 64 = ¼(a²)+a²-2a√(25-¼(a²))+25-¼(a²) 39 = a²-2a√(25-¼(a²)) 2a√(25-¼(a²)) = a²-39 4a²(25-¼(a²)) = a⁴-78a²+1521 100a²-a⁴ = a⁴-78a²+1521 2a⁴-178a²+1521 = 0 Let p = a². 2p²-178p+1521 = 0 p²-89p = -(1521/2) (p-½(89))² = -(1521/2)+(-0.5*89)² (p-44.5)² = 1219.75 p = 44.5±√(1219.75) p ≠ 44.5-√(1219.75) p =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2755491

a = ½(360-2*50) a = ½(260) a = 130° b² = 50²+50²-2*50*50cos130 b = 90.6307787037 units. sin(0.5*50) = r/c c = 2.3662015832r units. sin(0.5*50) = R/d d = 2.3662015832R units. It implies; 2.3662015832r+r+R+2.3662015832R = 90.6307787037 3.3662015832r+3.3662015832R = 90.6307787037 3.3662015832(r+R) = 90.6307787037 r+R = 90.6307787037/3.3662015832 r+R = 26.9237526225 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2755109

a = (6-b) cm. b is the radius of the two congruent inscribed yellow half circles. 6 cm is the radius of the ascribed half circle and also, the diameter of the inscribed circle. c = ½(6) c = 3 cm. c is the radius of the inscribed yellow circle. d = (3+b) cm. It implies, calculating b. (6-b)²+3² = (3+b)² 36-12b+b²+9 = 9+6b+b² 36 = 18b b = 2 cm. Again, b is the radius of the inscribed two equal half circles. Therefore, area blue is; Area half circle with radius 6 cm -...

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