Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3132789

Calculating angle x. a = 3-1 a = 2 units. a is CD. tanb = 3/1 b = atan(3)° b is angle BCF equal angle ACD. It implies; ((⅕(3√(10)))/sin(atan(3))) = (2/sinc) c = asin(1) c = 90° c is angle CAD. 2² = d²+(⅕(3√(10)))² d² = 4-⅕(18) d = √(⅕(2)) I d = ⅕√(10) units. d = 0.63245553203 units. d is AC. e = 180-b e = 180-atan(3) e = 108.434948823° e is angle ACB. f² = 1²+0.63245553203²-2*1*0.63245553203cos(180-atan(3)) f = 1.3416407865 units. f is AB. Therefore;...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3132782

Let a be 1 unit. e = ½(180-45) e = 67.5° tane = 1/f f = 1/tan(67.5) f = 0.41421356237 units g = 1-f g = 1-0.41421356237 g = 0.58578643763 units. 2h² = a² 2h² = 1 h = ½√(2) units. h = 0.70710678119 units. j = 45-(90-67.5) j = 45-22.5 j = 22.5° Therefore, calculating lengths b, c and d. Length b is; b = √(1²+0.58578643763²) b = 1.15894165104 units. Length c is; c = √(1²+0.41421356237²) c = 1.08239220029 units. Length d is; d² = 1.08239220029²+0.7...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3132062

Let the ascribed square side length be 1 units. Calculating area red inscribed circle. a = (1-b) units. Where b is the radius of the red inscribed circle sin30 = b/(1-b) 2b = 1-b 3b = 1 b = ⅓ units. Area red inscribed circle is; πb² = π(⅓)² = ⅑(π) square units. Calculating area blue ascribed circle. Let c be the radius of the blue inscribed circle. cos30 = 1/d ½√(3) = 1/d d = 2/√(3) d = ⅓(2√(3)) units. e = 1-d e = ⅓(2√(3))-1 e = ⅓(2√(3)-3) units. Ther...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3132058

Let a be the radius of the inscribed pink circle. b = (a+7) units. c = (7-a) units. d = (a-2) units. Calculating a. It implies; a ~ (a+7) (a-2) ~ (7-a) Cross Multiply. a(7-a) = (a-2)(a+7) 7a-a² = a²+7a-2a-14 2a²-2a-14 = 0 a²-a-7 = 0 (a-½)² = (7)+(-½)² (a-½)² = ¼(29) a = ½±√(¼(29)) a = ½±½√(29) It implies; a ≠ ½(1-√(29)) a = ½(1+√(29)) a = 3.19258240357 units. Again, a is the radius of the inscribed pink circle. Recall. b = (a+7) units. And a = 3.19258240357 units. b = 3.19258240357+7...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3131092

Let 2 units be the side length of the ascribed square. Therefore; a = 1 units. a is the radius of the inscribed circle. Calculating angle x. b² = 2(1)² b = √(2) units. c = ½(b) c = ½√(2) units. d= a+c d = 1+½√(2) d = ½(2+√(2)) units. Therefore, angle x is; tanx = d/c x = atan((½(2+√(2)))/(½√(2))) x = atan((2+√(2))/√(2)) x = atan(½(2√(2)+2)) x = atan(1+√(2)) x = 67.5°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3130418

Calculating shaded area. tana = 10/5 a = atan(2)° b = 2a b = 2atan(2)° c = 180-b c = (180-2atan(2))° d = ½(c) d = (90-atan(2)) d = atan(0.5)° It implies; tan(atan(0.5)) = e/5 ½ = e/5 e = 2.5 units. f = 10-e f = 7.5 units. g² = 2(2.5)²-2(2.5)²cos(2atan(2)) g = 4.472135955 units. Shaded Area is; ½(10*10)-½(10*4.472135955sin(atan(0.5)) = 50-10 = 40 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3130416

a² = r²+(½(r))² a = ½√(5)r units. 2r ~ ½√(5)r b ~ ½(r) Cross Multiply. ½√(5)rb = r² √(5)rb = 2r² b = 2r/√(5) b = ⅕(2√(5))r units. b is the height of the inscribed triangle. c²+b² = (2r)² c² = (2r)²-(⅕(2√(5))r)² c² = 4r²-⅕(4r²) c² = ⅕(16r²) c = ⅕(4√(5))r units. d = c-a d = ⅕(4√(5))r-½√(5)r d = (8√(5)r-5√(5)r)/10 d = ⅒(3√(5))r units. d is the base of the inscribed triangle. Area inscribed triangle is; ½*b*d = ½*⅕(2√(5))r*⅒(3√(5))r = ⅒(3)r² square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3129465

Let a be the radius of the circ. Calculating a. b = ½(12+2) b = 7 units. c = ½(4+6) c = 5 units. d = b-2 d = 7-2 d = 5 units. It implies; a² = c²+d² a = √(5²+5²) a = √(50) a = 5√(2) units. Again, a is the radius of the circle. Calculating blue area. e² = 6²+12² e = 6√(5) units. sinf = (0.5*6√(5))/(5√(2)) f = 71.5650511771° g = 2f g = 143.130102354° h² = 4²+2² h = 2√(5) units. sinj = (0.5*2√(5))/(5√(2)) j = 18.4349488229° k = 2j k = 36.86989...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3129461

Let a be the radius of the white inscribed circle. b = (10-a) units. Calculating a. b² = a²+5² (10-a)² = a²+25 100-20a+a² = a²+25 75 = 20a a = ¼(15) units. a = 3.75 units. c = ½(10) c = 5 units. c is the radius of the bigger circle ascribing the white circle. Therefore, area yellow is; πc²-πa² = π(5²-(¼(15))²) = π(25-(225/16)) = π((175/16) = 10.9375π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3128875

x+a = y+2a 4+a = 3+2a a = 1 unit. It could be; a = √(4²+3²)-4 a = 5-4 a = 1 unit. Or 2a = √(4²+3²)-3 2a = 5-3 a = 1 unit. Which makes; b = y+a b = 4+a And a = 1 unit. b = 5 units. Or b = y+2a b = 3+2a And a = 1 unit. b = 3+2(1) b = 5 units. c = x-y c = 4-3 c = 1 unit. Let PQ = d. Calculating d b² = c²+d² 5² = 1²+d² d² = 25-1 d = √(24) d = 2√(6) units. Again, d is the required length, PQ.

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