Mathematics Question and Solution.

https://onlineedumath.com/update/mathematics-question-and-solution/2495838

Calculating the area of the inscribed square. a = ½(45) a = 22.5° b = ½(180-45) b = ½(135) b = 67.5° cos22.5 = 6/c c = 6.4943532018 units. Let d be the side length of the inscribed square. e² =2d² e = √(2)d unit. e = 1.4142135624d units. (√(2)d/sin45) = (f/sin67.5) f = 1.847759065d units. (g/sin45) = (d/sin67.5) g = 0.7653668647d units. d is; c = f+g 1.847759065d+0.7653668647d = 6.4943532018 2.6131259297d = 6.4943532018 d = 2.4852813743 units. Again, d i...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2493984

sin30 = a/6 a = 3 units. b = 6+2a b = 12 units. Let the side length of the inscribed regular pentagon be c. Calculating c. d² = 2c²-2c²cos108 108° is the single interior angle of the regular pentagon. d = 1.6180339887c units. e²+(0.5c)² = (1.6180339887c)² e = 1.5388417685c units. tan60 = 0.5c/f f = 0.2886751346c units. It implies; e+f = b 1.5388417685c+0.2886751346c = 12 1.8275169031c = 12 c = 6.5662867357 units. Again, c is the side length of the inscribed r...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2493882

Radius of the half circle is 5 units. 5² = 4²+a² a = 3 cm. tanb = 4/3 b = atan(4/3)° c = 2b c = 2atan(4/3)° 5² = 3²+d² d = √(25-9) d = 4 cm. tane = 3/4 e = atan(3/4)° f = 2e f = 2atan(3/4)° Shaded Area is; Area sector with radius 5 cm and angle 2atan(4/3)° - Area triangle with height 5 cm and base 5sin(2atan(4/3)) cm - Area sector with radius 5 cm and angle 2atan(3/4)° + Area triangle with height 5 cm and base 5sin(2atan(3/4)) cm. = (2atan(4/3)π*5*5/360)-(0.5...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2493854

a = 180-2(70) a = 40° a is angle ACB. b = 180-2a b = 180-80 b = 100° b is angle AOC. Let r be the radius of the circle. Calculating r. Observing Cosine Rule. 9² = 2r²-2r²cos100 81 = 2.3472963553r² r² = 81/2.3472963553 r = √(34.5077858686) r = 5.874332802 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2493834

Calculating r, radius of the circle. a = (12-r) units. It implies; r² = 5²+(12-r)² r² = 25+144-24r+r² 24r = 169 r = (169/24) units. r = 7.0416666667 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2492951

Calculating length AB. Let a be the radius of the inscribed semi circle. b = 180-45 b = 135° c² = 12²+5²-2*5*12cos45 c = 9.1731775442 units. d = c+a d = (9.1731775442+a) units. e = 12+5 e = 17 units. Calculating a. 17² = a²+(9.1731775442+a)² 289 = a²+84.1471862574+18.3463550884a+a² 2a²+18.3463550884a-204.8528137426 = 0 Resolving the quadratic equation. a = 6.52481 units. f = 2a+c f = 2(6.52481)+9.1731775442 f = 22.2227975442 units f is CD = AB. AB = 22...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2492054

Calculating equation of the curve. (y-12) = a(x-6)² At x = 0, y = 0 Calculating a. -12 = a*36 a = -⅓ It implies; y-12 = -⅓(x-6)² y = -⅓(x²-12x+36)+12 y = ⅓(-x²+12x-36+36) y = ⅓(12x-x²) Where y = ⅓(12x-x²) is the equation of the curve. Calculating x when y = (32/3). (32/3) = = ⅓(12x-x²) 32 = 12x-x² x²-12x+32 = 0 Calculating x via completing the square approach. (x-6)² = -32+36 x = 6±√(4) x ≠ 6-2 x = 6+2 x = 8 units. Calculating x when y = 0. 0 = ⅓(12x-x²) x² =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2491932

Let AB = BC = 1 unit. tan30 = a/1 a = ⅓√(3) units. a = 0.5773502692 units. a is CD. cos30 = 1/b b = ⅓(2√(3)) units. b = 1.1547005384 units. b is BD. c = 90-30 c = 60° c is angle BDC. 3(90)+90+60+d = 180(5-2) 270+150+d = 540 540° is the sum of the interior angles of the irregular pentagon ABCDE. d = 540-270-150 d = 540-420 d = 120° d is angle BAE. e = 90-30 e = 60° e is angle ABD. f² = 1.1547005384²+1-2*1.1547005384cos60 f = 1.0856485596 units. f is AD. (...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2490228

Let R be radius of the quarter circle. Let r be radius of the inscribed circle. R will be; √(3²+4²) R=√(25) R= 5 unit. Calculating r, radius of the inscribed circle. Therefore r will be; (5-r)²=(3+r)²+r² 25-10r+r²=9+6r+r²+r² r²+16r-16=0 Therefore resolving the quadratic equation via completing the square side to get r, radius of the inscribed circle. (r+8)² = 16+8² (r+8)² = 80 r = -8±√(80) r = -8±4√(5) It implies; r ≠ -8-4√(5) r = -8+4√(5) r = 4(√(5)-2) units....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2490219

Let the base of the equilateral be 1 unit. a = 180-22-16-38 a = 180-76 a = 104° (1/sin104) = (b/sin38) b = 0.6345091075 units. c = 180-16-44-38 c = 180-98 c = 82° (1/sin82) = (d/sin(38+44)) d = 1 unit. e² = 0.6345091075²+1-2*0.6345091075cos22 e = 0.4753825906 units. (0.4753825906/sin22) = (0.6345091075/sinf) f = 30° f is the required angle. ? = 30°

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