Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2836358

Let the side of the two congruent sides be a units. The small green square side will be; ⅔(a) units. The big green square side is; 2a - ⅔(a) = ⅓(4a) units. b = a-⅔(a) b = ⅓(a) units. Calculating a. 9² = a²+(⅓(a))² 81 = a²+(⅑(a²)) 81 = 10a²/9 a = 27/√(10) a = ⅒(27√(10)) units ⅔(a) = ⅕(9√(10)) units. ⅓(4a) = ⅕(18√(10)) units. Area Green is; (⅕(9√(10)))² + (⅕(18√(10)))² = ⅕(162) + ⅕(648) = ⅕(810) = 162 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2836347

Sir Mike Ambrose is the author of the question. Shaded area exactly in cm² is; Area triangle with height 84 cm and base 13 cm - Area triangle with height 10 cm and base 24 cm + Area triangle with height (65/12) cm and base 13 cm. = (½*13*84) - (½*10*24) + (½*13*(65/12)) = 546 - 120 + (845/24) = (13104 - 2880 + 845)/24 = 11069/24 cm² = 461.208333333 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2834884

Radius of the circle, r is; r² = (8+(3/2))²+4² r² = (19/2)²+16 r² = ¼(425) r = ½(5√(17)) cm Therefore; Area shaded, exactly in cm² decimal is; Area sector with radius ½(5√(17)) cm and angle (180-atan(8/19)-atan(16/13))° - Area triangle with height 8 cm and base 6.5 cm - Area triangle with height 8 cm and base 3.36842105263 cm -Area trapezoid with two parallel side 7.36842105263 cm and 8 cm, and height 1.5 cm = ((180-atan(8/19)-atan(16/13))*π(2.5√(17))²÷360) - (0.5*8*6.5) - (...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2834872

Calculating r, radius of the circle. (r/tan40)+(r/tan30) = 10 r = 3.42020143326 cm. (a/sin60) = (10/sin40) a = 13.47296355334 cm. tan40 = 3.42020143326/b b = 4.07603734548 (4.07603734548/sin50) = (c/sin80) c = 5.24005260459 cm. d = 13.47296355334-4.07603734548 d = 9.39692620786 cm. e² = 9.39692620786²+5.24005260459²-2*9.39692620786*5.24005260459cos130 e = 13.38142726359 cm. (13.38142726359/sin130) = (5.24005260459/sinf) f = 17.45618925878° g = 100-17.4561892587...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2833470

Sir Mike Ambrose is the author of the question. Let the small inscribed square side length be x. The big inscribed square side length will be 2x. Calculating x. ⅓(√(3)x+2√(3)x+3x+6x) = 18+6√(3) (9+3√(3))x = 3(18+6√(3)) (3+√(3))x = 6(3+√(3)) x = 6 cm. Therefore: Area Blue exactly in decimal is; Area triangle with height 12sin(atan(12/(12+4√(3))))÷sin(180-atan(12/(12+4√(3)))-atan(⅔)) cm and base 12sin(atan(⅔)) cm. = ½*12sin(atan(12/(12+4√(3))))÷sin(180-atan(12/(12...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2833435

Notice; AB = 25 units. cos30 = 8√(3)/a a = 16 units. tan30 = b/(8√(3)) b = 8 units. Let the side of the regular hexagon be c. c = ¼(AC) Calculating AC. ((AC)/sin46) = (25/sin74) AC = 18.70821971213 units. c = ¼(18.70821971213) c = 4.67705492803 units. Calculating r, radius of the circle. (r/sin16) = (25/sin148) r = 13.00374294827 units. Calculating Shaded Area. cos30 = d/4.67705492803 d = 4.05044838257 units. e = 2d e = 8.10089676514 units. tan16...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2832555

Calculating green inscribed triangle area. a² = 2(3√(2))² a² = 2*18 a = √(36) a = 6 units. a is AB. b = ½(360-90) b = ½(270) b = 135° b is angle ABC. c² = 6²+(4√(2))²-2*6*4√(2)cos135 c = 2√(29) units. c = 10.7703296143 units. c is AC. 2d² = (2√(29))² d² = 2*29 d = √(58) units. d = 7.6157731059 units. d is AO = CO the radius of the ascribed quarter circle. sine = (0.5*6)/7.6157731059 e = 23.1985905135° f = 2e f = 46.3971810271° g = 90-f g = 43.6028189729°...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2832383

Calculating the required angle, alpha. Let it be a. Let R = 2 units, radius of the ascribed semi circle. b = (2-r) units. It implies; Calculating r. (2-r)² = 2r² 2-r = √(2r²) 2-r = √(2)r 2 = √(2)r+r (1+√(2))r = 2 r = 2/(1+√(2)) r = 2(1-√(2))/(-1) r = 2(√(2)-1) units. r = 0.8284271247 units. r is the radius of the inscribed circle. c = 2+r c = 2+0.8284271247 c = 2.8284271247 units. Therefore, the required angle, alpha (a) is; tana = r/c tana = 0.828427124...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2831022

9^(x)+9*9^(x) = 270 Let 9^(x) be p. p+9p = 270 10p = 270 p = 27 And p = 9^(x) 9^(x) = 27 2x = 3 x = 3/2 x = 1.5

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2831015

7² = 3²+8²-2*3*8cosa 48cosa = 9+64-49 a = acos((9+64-49)/48) a = 60° Calculating b, radius of the inscribed circle. 3b+8b+7xb = 3*8sin60 18b = 12√(3) 3b = 2√(3) b = ⅔√(3) units. b = 1.1547005384 units. tan30 = b/c c = 1.1547005384/tan30 c = 2 units. Therefore, length x is? x² = 2²+2²-2*2*2cos60 x² = 8-8cos60 x² = 8-4 x = √(4) x = 2 units.

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