Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2608493

Calculating the radius of each of the inscribed circles. The biggest inscribed circle radius, R is; 3.3R+2.73R+1.84R=(2.73x1.84) 7.87R = (2.73x1.84) R = 0.63827191868 m. The bigger inscribed circle radius, r is; Sin(0.5tan–¹(1.84/2.73))=(R-r)/(R+r) And R = 0.63827191868 m ≈ 0.638272 m. Therefore; 0.3=(0.638272-r)/(0.638272+r) 1.3r=0.438272 r = 0.33713230769 m. The big inscribed circle radius, q is; Sin(0.5tan–¹(1.84/2.73))=(r-q)/(r+q) Therefore; 0.3=(0.3371...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2607022

Let the side length of the square be 1 unit. Calculating angle x. a = 180-60-70 a = 50° (1/sin50) = (b/sin60) b = 1.1305158748 units. c² = 2(1)² c = √(2) units. c is the diagonal of the square. d = ½(c) d = 0.5√(2) units. e = 180-20-45 e = 115° (1/sin115) = (f/sin45) f = 0.7802060087 units. g = b-f g = 0.3503098661 units. (1/sin115) = (h/sin20) h = 0.377377474 units. j = d-h j = 0.3297293072 units. k² = 0.3297293072²+0.3503098661²-2*0.3297293072*0.35...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2606879

Let to theta be x. Let beta be y. a = 90+60 a = 150° b = ½(180-150) b = 15° c = 90-15 c = 75° Therefore, x is; x = c x = 75° (vertically opposite angles are equal). Calculating y. Let the side length of the square be 1 unit. cos15 = 1/d d = 1.0352761804 units. e² = 1+1.0352761804²-2*1.0352761804cos75 e = 1.2393136749 units. Angle y is; (1.2393136749/sin75) = (1/siny) y = 51.2060231134° It implies; tanx ÷ tany is; = tan(75)÷tan(51.2060231134) =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2606868

Notice! The angles are exterior. The sum of exterior angles is 360° The exterior angles are arranged in order (clockwise). Calculating x. It implies; x+2x+86+70 = 360 3x = 360-156 3x = 204 x = 68°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2606862

Notice! Radius of the inscribed half circle is 4 units. tana = 4/8 a = atan(½)° b = 2a b = 2atan(½)° c = 90-b c = (90-2atan(½))° tan(90-2atan(½)) = d/8 d = 6 units. Therefore, area green inscribed triangle is; ½*8d = ½*8*6 = 24 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2606855

Calculating x. a²+x² = 1 a = √(1-x²) units. b = (1+x) units. Observing similar plane shape (right-angled) side length ratios. 1 - √(1-x²) (1+x) - x Cross Multiply. x = √(1-x²)(1+x) x² = (1-x²)(1+2x+x²) x² = 1+2x+x²-x²-2x³-x⁴ x⁴+2x³+x²-2x-1 = 0 It implies; x = 0.8832035059 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2604719

Let AB, the side length of the square be 1 unit. a² = 1²+1² a = √(2) units. b²+1² = c² c = √(b²+1) units. b is the radius of the inscribed circle. d² = 2b² d = √(2)b units. It implies; c+d = a √(b²+1)+√(2)b = √(2) b²+1 = (√(2)-√(2)b)² b²+1 = 2-4b+2b² b²-4b+1 = 0 Resolving the above quadratic equation via completing the square approach to get b, radius of the circle. (b-2)² = -1+(-2)² (b-2)² = 3 b = 2±√(3) It implies; b ≠ 2+√(3) b = (2-√(3)) units. b = 0.2679...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2604706

Calculating Area of the inscribed Square. tana = 18/9 a = atan(2)° Let the side length of the square be b. sin(atan(2)) = b/c c = (b/0.894427191) units. cos(atan(2)) = d/b d = 0.4472135955b units. It implies; c+d = 9 (b/0.894427191)+0.4472135955b = 9 1.5652475842b = 9 b = 5.749889085 units. Therefore, area inscribed square is; b² = (5.749889085)² = 33.0612244898 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2602525

Length NP is; √(20²+5²) - (9+3) = √(425) - 13 = (5√(17)-13) cm. = 7.6155281281 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2602515

Red area is; ½(Area rectangle of length 2r and width r - area semi circle of radius r) - area triangle of height 2r and base r + area sector of radius r and angle (2tan–¹(1/2))° + area triangle of two equal length r, and angle (180-2tan–¹(1/2))°, the angle they form. Therefore; ½(2r²-½(πr²))-r²+((2tan–¹(1/2))*(πr²)÷360)+½(r²sin(180-2tan–¹(1/2)) = r²-¼(πr²)-r²+0.14758361765πr²+0.4r² = 0.4r²+0.14758361765πr²-0.25πr² = 0.4r²−0.1024163824πr² = r²(0.4−0.3217505546) = (0.0782494454)...

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