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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2726665

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½*a*b = 10 a = (20/b) cm. a is CD. b is OD. b² - 30 (20/b)² - 10 Cross Multiply. 10b² = 12000/b² 10b⁴ = 12000 b⁴ = 1200 b⁴ = 1200 b² = √(1200) b² = 20√(3) b = √(20√(3)) b = 5.8856619128 cm. Again, b is OD. Recall. a = (20/b) And b = 5.8856619128 cm. a = 20/5.8856619128 a = 3.3980884897 cm. Again, a is CD. Therefore, the required red angle is; Let it be c. tanc = b/a c = atan(5.8856619128/3.3980884897) c = atan(1.7320508076) c = atan(√(3))° c = 60° A...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2721532

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Calculating shaded red area. Let a be the diameter of the inscribed big half circle. Therefore, 2a is the diameter of the bigger half circle or a is the radius of the bigger half circle. a² = 6²+b² b = √(a²-36) cm. 6 - c a - √(a²-36) Cross Multiply. c = (6√(a²-36))/a cm. Again. a - √(a²-36) √(a²-36) - d Cross Multiply. d = (a²-36)/a cm. e = a+d e = (2a²-36)/a cm. Calculating a. 10² = ((2a²-36)/a)²+((6√(a²-36))/a)² 100 = ((4a⁴-144a²+1296)/a²)+((36a²-1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2720381

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Calculating area of the ascribed square and area of the inscribed blue triangle. a²+8² = 10² a = 6 units. 6 - 8 b - x Cross Multiply. 8b = 6x b = ¼(3x) units  c²+x² = 8² c = √(64-x²) units. It implies; b+c = x ¼(3x)+√(64-x²) = x √(64-x²) = ¼(x) 64-x² = x²/16 (64*16)-16x² = x² 17x² = (64*16) x = √((64*16)/17) x = 32/√(17) x = (32√(17))/17 units. x = 7.7611400012 units. x is the side length of the ascribed square. Therefore, area of the ascribed square (x²)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2720039

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Let AB be 1 unit. a = 180-18-36 a = 126° a is angle APB. (1/sin126) = (b/sin18) b = 0.3819660113 units  b is BP. c = 180-(12+18)-(36+36) c = 180-102 c = 78° c is angle ACB. (1/sin78) = (d/sin30) d = 0.5111702974 units. d is BC. e² = 0.5111702974²+0.3819660113²-2*0.5111702974*0.3819660113cos36 e = 0.3021133709 units. e is CP. Therefore, the required angle, x is; (0.3021133709/sin36) = (0.3819660113/sinx) x = asin(0.7431448256) x = 48°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2718323

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Calculating area of the bigger square. Let the bigger square side length be a. b² = 4 b = 2 cm. b is the side length of the big square. c = (a+2) cm. d = (a-2) cm. Calculating a², area of the bigger square. (a-2)²+(a+2)² = 10² a²-4a+4+a²+4a+4 = 100 2a²+8 = 100 2a² = 92 a² = ½(92) a² = 46 cm² Again, a² is the area of the bigger square.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2716941

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a² = 10²+5² a = √(125) a = 5√(5) units. a is the side length of the square. b² = 2a² b² = 2(5√(5))² b = √(250) b = 5√(10) units. b is the diagonal of the square. c = 90+atan(10/5) c = (90+atan(2))° d = 90-atan(5/10) d = (90-atan(0.5))° d = atan(2)° cos(atan(2)) = e/10 e = 4.472135955 units. f = 5√(5)-e f = 6.7082039325 units. 2 = g/4.472135955 g = 8.94427191 units. h = 90-atan(8.94427191/6.7082039325) h = atan(4/3)° j = atan(3/4)° k = 180-45-j k = 98....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2713706

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Let the length (base) of the ascribed rectangle be 4 units. Therefore its width/breadth/height is 2 units. tana = 4/2 a = atan(2)° b = atan(½)° sin(atan(½)) = c/4 c = 3.577708764 units. d² = 4²+2² d = 4.472135955 units. e = d-c e = 0.894427191 units. f = 180-2atan(½) f = 126.8698976458° g² = 2-2cos126.8698976458 g = 1.788854382 units. h = g-e h = 0.894427191 units. j = 90-atan(2) j = 26.5650511771° k = 2j k = 53.1301023542° cos53.1301023542 = l/1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2710974

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Calculating area of the inscribed red square. Let a be the side length of the inscribed red square. 2b² = a² b = √(a²/2) b = ½√(2)a units. 10² = a²+(½√(2)a)²-2a*½√(2)acos(45+90) 100 = ½(3a²)+a² 100 = ½(5a²) 5a² = 200 a² = 40 square units. Where a² is the area of the inscribed red square.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2706404

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6²+6²-2*6*6cosa = 12²+3²+2*12*3cosa 72-72cosa = 153+72cosa 144cosa = 81 a = acos(9/16) a = 55.7711336722° b = 180-a b = 124.2288663278° 6²+12²-2*6*12cosc = 6²+3²+2*6*3cosc 180-144cosc = 45+36cosc 180cosc = 135 c = acos(3/4) c = 41.4096221093° d = 180-c d = 138.5903778907° e = 180-a-c e = 82.8192442185° (12/sin82.8192442185) = (f/sin55.7711336722) f = 10 units. g = 180-b-c g = 14.3615115629° (6/sin14.3615115629) = (h/sin124.2288663278) h = 20 units. Theref...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2691653

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Let the side length of the three congruent regular hexagon be 1 unit each. Notice. The inscribed green triangle is equilateral. a² = 2-2cos120 120° is the single interior angle of one of the congruent regular hexagon. a = √(3) units. b = 2a b = 2√(3) units. c² = (2√(3))²+1² c² = 13 c = √(13) units. tand = 2√(3)/1 d = atan(2√(3))° e = atan(1/(2√(3)))° f = 120-d f = 46.102113752° It implies; (1/sin60) = (g/sin(atan(2√(3))) g = 1.1094003925 units. (1/sin6...
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