Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2682636

3² = a²+2² a² = 9-4 a = √(5) units. a is PQ. cosb = 2/3 b = acos(2/3)° b is angle BAC. tan(acos(2/3)) = (5√(5))/c c = 10 units. c is AB. d = 10-3 d = 7 units. d is PB. e² = 7²+(5√(5))² e² = 49+125 e = √(174) units. e is CP, diameter of the circle. f = ½(e) f = ½√(174) f is radius of the circle. Area circle is; πf² = π(½√(174))² = ¼(174)π = 43.5π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2682632

a = (x+60)° b = (180-7x)° Notice. Blue triangle is equilateral. Yellow triangle is isosceles. L1 is parallel to L2 Therefore; a = b (x+60) = (180-7x) 8x = 120 2x = 30 x = 15°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2681412

Sir Mike Ambrose is the author of the question. Area green, exactly is; Area square with length 8 cm - area trapezoid with two parallel length 8 cm and 2 cm, and height 3 cm - area trapezoid with two parallel length 7 cm and 2cm, and height 5 cm - area triangle with height 3√(5) cm and base ½(3√(5)) cm. = 64-½(30)-½(45)-¼(3√(5))² = ¼(256-195) = 61/4 = 15¼ cm² = 15.25 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2681406

Calculating area inscribed yellow rectangle. tana = 20/15 a = atan(4/3) b = 90+a b = (90+atan(4/3))° (20/sin(90+atan(4/3))) = (15/sinc) c = 26.7436839504° d = 180-(90+atan(4/3))-26.7436839504 d = 10.1262136954° (e/sin10.1262136954) = (20/sin(90+atan(4/3))) e = 5.8605710995 units. e is the width of the inscribed yellow rectangle. f = 90-26.7436839504 f = 63.2563160496° h = 180-2(63.2563160496) h = 53.4873679008° It implies; j² = 20²+20²-2*20*20cos53.487367900...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2681057

1. The Right-angled Triangle. Calculating length x. a = 9+6 a = 15 cm. a is the height of the ascribed right-angled triangle. 17 cm is the hypotenuse of the ascribed right-angled triangle. b²+15² = 17² b² = 17²-15² b² = 289-225 b = √(64) b = 8 cm. b is the base of the ascribed right-angled triangle. It implies; Calculating s. s² = 6²+8² s² = 36+64 s² = 100 s = √(100) s = 10 cm. 2. The Half Circle. Calculating A. Notice. 4 cm is the radius of the h...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2680117

Let the radius of the quarter circle be a. b = 2a units. c = 4+5 c = 9 units. It implies; 3²+9² = (2a)² 4a² = 90 a² = ½(45) a = ½(3√(10)) units. a = 4.7434164903 units. Again, a is the radius of the quarter circle. d² = 5²-(½(3√(10)))² d² = 25-½(45) d = √(½(5)) d = ½√(10) units. e = a-d e = ½(3√(10))-½√(10) e = √(10) units. f = 180-atan(3)-atan(3/4) f = 71.5650511771° It implies; The required length, x is; x² = 4²+10-2*4√(10)cos71.5650511771 x = 4.24...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2680115

Let a be the radius of the circle. tanb = (0.5a)/a b = atan(0.5)° cos(atan(0.5)) = c/6 c = 5.366563146 units. sin(atan(0.5)) = d/6 d = 2.683281573 units. cosatan(0.5) = 3/a a = 3.3541019662 units. Again, a is the radius of the circle. It implies; Area of the rectangle is; ac = 3.3541019662*5.366563146 = 18 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2678846

Calculating r, radius of the quarter circle Notice. AE = DE = 1 unit. CE = 6+1 CE = 7 units. AC = √(2r²) AC = √(2)r units. It implies; (√(2)r)² = 7²+1² 2r² = 50 r = √(25) r = 5 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2678829

Calculating R, radius of the inscribed circle. a² = 2*3² a = 3√(2) units. a is the diagonal of the square. b² = 2R² b = √(2)R units. c = a-b c = (3√(2)-√(2)R) units. It implies; (3√(2)-√(2)R)² = R²+3² 18-12R+2R² = R²+9 R²-12R+9 = 0 Calculating R, resolving the above quadratic equation via completing the square approach. (R-6)² = -9+(-6)² (R-6)² = 27 R = 6±√(27) R = 6±3√(3) It implies; R ≠ 6+3√(3) units. R = 6-3√(3) units. R = 3(2-√(3)) units. R = 0.80384...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2678823

The single side length of the square, ABCD is; 2√(5) cm Area Blue is; Area square with length 2√(5) cm - area triangle with height 2√(5) cm and base √(5) cm - area triangle with height 4cm and base 2 cm = (2√(5))² - ½(2√(5)*√(5)) - ½(4*2) = 20 - 5 - 4 = 11 cm²

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