Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2502222

Let the inscribed blue equilateral triangle side length be 1 unit. Area Blue is; 0.5*1*1sin60 = ¼√(3) square units. = 0.4330127019 square units. 0.5² = 2-2cosa 2cosa = 2-0.25 cosa = (1.75/2) a = 28.9550243719° b = ½(180-28.9550243719) b = 75.5224878141° c = 90-b c = 14.4775121859° sin60 = d/1 d = ½√(3) units. d = 0.8660254038 units. e² = 1²+0.8660254038²-2*0.8660254038cos14.4775121859 e = 0.2700907567 units. (0.2700907567/sin14.4775121859) = (1/sinf) f = 67.7...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2502207

Let the radius of the ascribed circle be 1 unit. Area ascribed circle is; πr² = π(1)² = π square units. Calculating area inscribed red circle. Let it's radius be r. sin30 = a/1 a = ½ units. a = 0.5 units. It implies; (1-r)² = r²+(r+0.5)² 1-2r+r² = r²+r²+r+0.25 r²+3r-0.75 = 0 4r²+12r-3 = 0 Resolving the above quadratic equation via completing the square approach to get r, radius of the inscribed red circle. r²+3r = ¾ (r+(3/2))² = ¾+(3/2)² (r+(3/2))² = 3 r =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2502178

Method A. 15² = (0.5*24)²+a² a² = 225-144 a = √(81) a = 9 units. a is the side length of the two equal lengths each. tanb = 9/12 b = atan(¾)° c = 90-b c = atan(4/3)° Therefore, r, radius of the circle is; cos(atan(4/3)) = (0.5*15)/r r = 7.5/cos(atan(4/3)) r = 12.5 units. Method B. 15² = (0.5*24)²+d² d² = 225-144 d = √(81) d = 9 units. d is the side length of the two equal lengths each. Let r be the radius of the circle. e = r-d e = (r-9) units. It im...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2502142

a*x = 15*15 ax = 225 a = 225/x --- (1). (30+a)*x = (30+15)*(30+15) ax+30x = 2025 --- (2). Substituting (1) in (2). (225/x)x+30x = 2025 225+30x = 2025 30x = 2025-225 30x = 1800 x = 1800/30 x = 60 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2501577

Let R=the bigger inscribed circle radius. Let r=the smaller inscribed circle radius. It implies; R+r=6√(3)-6=4.4 R+r=4.4 -------(1). And the radius of the ascribed semi circle is 6 unit. Calculating the R and r considering the radius of the ascribed semi circle. R=√((6-R)²-R²) Therefore; R=√(36-12R) ----- (2). r=√((6-R)²-R²) Therefore; r=√(36-12r) ----- (3). Adding (2) and (3). R+r=√(36-12R)+√(36-12r)-----(4) Therefore equating (1) and (4). √(36-12R)+√(3...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2500596

a² = 2-2cos120 a = √(3) units. (2/sin30) = (√(3)/sinb) b = 25.6589062733° c = 180-30-b c = 124.3410937267° d² = √(3)²+2²-2*2√(3)cos124.3410937267 d = 3.3027756377 units. e² = 3.3027756377²+1-2*3.3027756377cos60 e = 2.9335219916 units. (2.9335219916/sin60) = (1/sinf) f = 17.1705468636° Therefore, the required angle, Beta is; b+f = 25.6589062733+17.1705468636 = 42.8294531369°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2500544

Let a be the side length of the regular hexagon. Calculating a. (14/sin60) = (16/sinb) b = 81.7867892983° c = 180-60-b c = 38.2132107017° It implies; (a/sin38.2132107017) = (14/sin60) a = 10 units. Again, a is the side length of the regular hexagon. sin30 = d/a d = 10sin30 d = 5 units. e = 2d+a e = 2(5)+10 e = 20 units. f = e-16 f = 4 units. It implies; (4/sin81.7867892983) = (g/sin38.2132107017) g = 2.5 units. Therefore the required area, area pink...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2500406

Observing similar plane shape (scalene triangle) side length ratios. Therefore; y - 8 20 - 22 It implies; (y/20) = (8/22) (y/20) = (4/11) Cross Multiply. 11y = 80 y = 80/11 units. y = 7.2727272727 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2500296

3² = 2c²-2c²cos108 108° is the single interior angle of the regular pentagon. It implies; 9 = 2.6180339887c² c² = 3.4376941013 c = 1.8541019663 units. c is the side length of the regular pentagon. sin36 = d/c d = 1.8541019663sin36 d = 1.089813792 units. cos36 = e/c e = 1.8541019663cos36 e = 1.5 units. f = 1.5-1 f = 0.5 units. tang = f/d tang = 0.5/1.089813792 g = atan(0.5/1.089813792) g = 24.6453719079° h = 54-g h = 29.3546280921° j = 180-108-h j = 42.6453719...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2500105

Let the side length of the regular pentagon be 1 unit. a² = 2-2cos108 108° is the single interior angle of the regular pentagon. a = 1.6180339887 units. sin72 = b/1 b = 0.9510565163 units. cos72 = c/1 c = 0.3090169944 units. Area Blue is; 0.5*1*1sin108+0.5*0.3090169944*0.9510565163 = 0.6224745712 square units. d = a-c d = 1.6180339887-0.3090169944 d = 1.3090169943 units. Area Yellow is; 0.5*1.3090169943*0.9510565163 = 0.6224745712 square units. Therefore; Blue Area ÷ Yellow Area is;...

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