Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2580424

a² = 6²+4² a = √(36+16) a = √(52) a = 2√(13) units. a is AD = CD. tanb = 6/4 b = atan(3/2)° b is angle ADB. c = 180-b c = (180-atan(3/2))° c is angle BDC. Calculating x, length BC. x² = 4²+(2√(13))²-2*4*2√(13)cos(180-atan(3/2)) x = 10 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2580404

Calculating Area Red. It is; Area sector with radius 5 m and angle 60° - Area triangle with height 5 units and base 5sin60 m + 2(area triangle with height 2.5 m and base 2.5 m) = (60π*5*5/360)-(0.5*5*5sin60)+2(0.5*2.5*2.5sin60) = ⅙(25π)-¼(25√(3))+⅛(25√(3)) = (100π-150√(3)+75√(3))/24 = (100π-75√(3))/24 m² = 7.6773106163 m²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2580130

Calculating the total shaded area with respect to d. a = (180(12-2))/12 a = 150° a is the single interior angle of the regular dodecagon. b = 105-75 b = 30° sin30 = c/d c = ½(d) units. sin60 = e/d e = ½√(3)d units. f = e-c f = ½√(3)d-½(d) f = ½(√(3)-1)d units. g = ½(f) g = ¼(√(3)-1)d units. It implies, total shaded green area in terms of d is; 2(area trapezoid with parallel side lengths ½(d) units and ½√(3)d units respectively and height ¼(√(3)-1)d units. =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2580097

Let the two equal lengths be 1 unit each. a = 180-45 a = 135° b = 180-135-15 b = 30° (1/sin15) = (c/sin30) c = 1.9318516526 units. d² = 1.9318516526²+1²-2*1.9318516526cos45 d = 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/2578254

Calculating Area Triangle ABC. a = 7-5 a = 2 units. b = 180-60-45 b = 75° sin75 = 7/c c = 7.2469332629 units. c is the side length of the triangle ABC, equilateral triangle. Area triangle ABC is; 0.5*7.2469332629²sin60 = 22.7409791417 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2578241

Let the radius of the ascribed semi circle be a. b = ½(a) units. c = b+a c = ½(3a) units. It implies; (½(3a))² = (½a)²+6² ¼(9a²) = ¼(a²)+36 ¼(8a²) = 36 8a² = 36*4 a² = 18 a = 3√(2) units. Again, a is the radius of the ascribed semi circle. b = ½(a) b = ½(3√(2)) units. b is the radius of the inscribed yellow half circle. Area inscribed yellow circle is; ½(½(3√(2)))²π = ⅛(18)π = ¼(9π) square units. Area ascribed semi circle is; ½(3√(2))²π = 9π square units...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2578138

a = 40+10 a = 50° a is angle ACB. b = 180-10-50 b = 120° b is angle ADC. (6/sin120) = (c/sin10) c = 1.2030698654 units. c is CD. (6/sin120) = (d/sin50) d = 5.3073115854 units. d is AD. e = 180-120 e = 60° e is angle ADB. f = 180-50-50 f = 80° f is angle ABD. The required length, x (BD) is; (5.3073115854/sin80) = (x/sin40) x = 3.4641016151 units. Again, x is the required length, BD.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2577656

Let the side length of the inscribed two blue squares be 1 unit. a² = 2(1)² a = √(2) units. a is the radius of the green half circle. b = a-1 b = (√(2)-1) units. tanc = 1/(√(2)-1) c = atan(1/(√(2)-1))° c = 67.5° d = 90-c d = 22.5° Notice! The side length of the green square is equal the radius of the red half circle. Therefore; e = a = √(2) units. e is the side length of the green square. f² = 2e² f² = 2√(2)² f = √(4) f = 2 units. f is the black diagonal...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2577209

Let the ascribed square side be be 4 units. Area ascribed square is; 4² = 16 square units. Calculating the inscribed red rectangle. It implies; a² = 2²+1² a = √(5) units. b² = 1²+0.5² b = √(5/4) b = ½√(5) units. Observing similar plane shape (right-angled) side length ratios. 2 - ½√(5) c - ½ Cross Multiply. √(5)c = 2 c = ⅕(2√(5)) units. d = c+a d= ⅕(2√(5))+√(5) d = ⅕(7√(5)) units. Therefore, area inscribed red rectangle is; a*d = √(5)*⅕(7√(5)) = 7 square units. It implies; Shaded...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2576729

Let 1 be R. Calculating r. a = (1-r) units. tanb = 1/2 b = atan(½) c = 180-2b c = (180-2atan(½))° d = 180-c d = 2atan(½)° It implies, calculate r. sin(2atan(½)) = r/(1-r) ⅘ = r/(1-r) 5r = 4-4r 9r = 4 r = (4/9) units. Therefore; R/r is; 1÷(4/9) = 9/4 = 2¼ = 2.25

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