Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3372005

Calculating the inscribed orange triangle area. Notice. a = ½(3) a = 1.5 units. a is the radius of the inscribed circle. tanb = 3/4 b = atan(3/4)° tan(atan(¾)) = c/1.5 ¾ = ⅓(2c) c = 9/8 units. d = (3/2)-(9/8) d = 3/8 units. e = (180-atan(4/3))° e = 126.869897646° (1.5/sin(180-atan(4/3))) = ((3/8)/sinf) f = 11.5369590328° g = 180-2f g = 156.926081934° h² = 2*1.5²-2*1.5²cos156.926081934 h = 2.93938769134 units. j = 180-(180-atan(4/3))-11.5369590328 j =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3368641

Calculating k. k^(k) = 3^(k)*3⁹ k^(k) = 3^(k)*19683 (k/3)^(k) = 19683 It implies; k = 9 Checking Accuracy. (9/3)⁹ = 3⁹ = 19683

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3367656

Calculating x, the side length of the square ABCD. 3² = x²+4²-2*4xcosy 9 = x²+16-8xcosy 8xcosy = x²+7 cosy = (x²+7)/(8x) --- (1). 5² = x²+4²-2*x*4cos(90-y) 25 = x²+16-8xcos(90-y) 8x(cos90cosy-sin90siny) = x²-9 -8xsiny = x²-9 8xsiny = 9-x² siny = (9-x²)/(8x) --- (2). At (2). (9-x²) is opposite. (8x) is hypotenuse. Calculating adjacent. Let it be a. a²+(9-x²)² = (8x)² a² = 64x²-(81-18x²+x⁴) a² = 64x²-81+18x²-x⁴ a = √(82x²-81-x⁴) units. Again, a is adjacent. It...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3366743

Calculating Shaded Area. a²+10² = 12² a = √(144-100) a = √(44) a = 2√(10) units. b = 10-a b = (10-2√(10)) units. b = 3.67544467966 units. cosc = 10/12 c = acos(⅚)° d = 90-45-c d = (45-acos(⅚))° tan(45-acos(⅚)) = e/10 e = 2.02410864032 units. f = 10-e f = 10-2.02410864032 f = 7.97589135968 units. Therefore, shaded area is; ½*d*f = 0.5*3.67544467966*7.97589135968 = 14.6574737317 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3365919

Calculating Yellow Area. Area yellow is; Area triangle with two side 8√(2) cm and 2 cm, and angle 45° - Area triangle with height (20/3) cm and base ⅓(20tan(45-2atan(⅓))) cm = ½*2*8√(2)sin45 - ½*(20/3)*⅓(20tan(45-2atan(⅓))) = 8 - (200tan(45-2atan(⅓)))/9 = (72 - 200tan(45-2atan(⅓)))/9 cm² = (2666664 - 1058200)/333333 cm² = 1608464/333333 cm² = (1608464÷5291)/(333333÷5291) = 304/63 cm² = 4.8253968254 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3364971

Calculating Blue Area. Let a be the congruent inscribed side lengths each of the inscribed right-angled isosceles triangle. b = ½(2) b = 1 unit. c²+1² = a² c = √(a²-1) units. d = ½(180-45) d = ½(135) d = 67.5° tan67.5 = e/1 e = 2.4142135624 units. e = (1+√(2)) units. f = 1+e f = (2+√(2)) units. f is the radius of the ascribed quarter circle. g = f-2 g = √(2) units. Therefore, blue area is; ½*(fg) = ½*(2+√(2))*√(2) = ½(2+2√(2)) = (1+√(2)) square units. =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3364068

Calculating Area Purple ÷ Area Orange Exactly. Area purple is; Area trapezoid with two parallel sides 4 cm and (50/9) cm, and height (7/2) cm - Area triangle with height 4 cm and base 3 cm. = ½(4+(50/9))*(7/2) - ½(4*3) = (602/36) - 6 = 193/18 cm² Area orange is; Area trapezoid with two parallel sides 6 cm and (40/9) cm, and height (7/2) cm - Area triangle with height (40/9) cm and base (40/13) cm. = ½(6+(40/9))*(7/2) - ½((40/9)*(40/13)) = (329/18) - (800/117) = 4...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3363181

Calculating x. a = (x+1) units. b² = 1²+1² b² = 2 b = √(2) units. c = b+r c = (1+√(2)) units. Therefore, x is; (x+1)² = x²+(1+√(2))² x²+2x+1 = x²+1+2√(2)+2 2x = 2√(2)+2 2x = 2(√(2)+1) x = (1+√(2)) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3362313

Calculating angle alpha, let it be x. Let R = 2 units. a = (2-r) units. Calculating r. It implies; r²+r² = (2-r)² 2r² = 4-4r+r² r²+4r-4 = 0 (r+2)² = 4+(2)² r = -2±√(8) r = -2±2√(2) It implies; r ≠ -2-2√(2) units. r = -2+2√(2) units. r = (2√(2)-2) units. b = R+r b = 2+2√(2)-2 b = 2√(2) units. It implies, angle alpha, x is; tanx = r/b x = atan((2√(2)-2)/(2√(2))) x = atan(1-(1/√(2))) x = atan(½(2-√(2)) x = 16.3249499369° Again,...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3361473

Calculating the required angle, let it be x. Let 1 unit be the side length of the both congruent inscribed equilateral triangles. a = ⅑*180(9-2) a = 140° a is the single interior angle of the regular nonagon. b = ½(a) b = 70° c = 180-2b c = 40° d² = 2-2cos40 d = 0.68404028665 units. d is the side length of the regular ascribed nonagon. e = 60+70 e = 130° f² = 1²+0.68404028665²-2*1*0.68404028665cos130° f = 1.53208888624 units. (1.53208888624/sin130) = (1/sing)...

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