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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2460071

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r = √(7²+6²) r = √(85) cm. r is the radius of the semi circle. x² = √(85)²+6² x² = 85+36 x = √(121) x = 11 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2459663

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r² = x²+(6+8)² --- (1). r² = (10+x)²+6² --- (2). Where r is the radius of the ascribed quarter circle. Equating (1) and (2) to get x. r² = r² x²+(6+8)² = (10+x)²+6² x²+196 = 100+20x+x²+36 196-136 = 20x x = 60/20 x = 3 units. Calculating r, radius of the ascribed quarter circle. At (1). r² = x²+(6+8)² And x = 3 units. r² = 3²+14² r² = 9+196 r = √(205) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2459286

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AB is; 4+2tan(½(180-2atan(2))) = 4+1 = 5 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2459118

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Calculating angle x. Let the side length of the regular pentagon be 1 unit. Calculating r, radius of the inscribed two equal circles. a = ⅕*180(5-2) a = 108° b = ½(a)  b = ½*108 b = 54° sin54 = r/c c = r/(sin54) units. c = 1.2360679775r units. d = 108-½(180-108) d = 108-36 d = 72° tan72 = e/0.5 e = 1.5388417686 units. It implies; 3r+c = e 3r+1.2360679775r = 1.5388417686 4.2360679775r = 1.5388417686 r = 0.363271264 units. Again, r is the radius of each of the...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2459080

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a = 2+2+R+R a = (4+2R) a is the diameter of the ascribed circle. b = ½(a) b = ½(4+2R) b = (2+R) units. b is the radius of the ascribed circle. c = b-R c = (2+R)-R c = 2 units. d = b-2 d = (2+R)-2 d = R units. It implies; R = 4 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2458438

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a² = 13²-5² a² = 169-25 a = √(144) a = 12 units. a is AE. tanb = 5/12 b = atan(5/12)° b is angle BAE. c = x+a c = (x+12) units. c is AG. Notice! Triangle ABG is isosceles. AB = AG Therefore; 13 = x+12 x = 13-12 x = 1 unit.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2457841

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Calculating R, radius of the inscribed circle. Observing similar plane shape (right-angled triangle) side length ratios. (12+x) - 6          6 - x Cross Multiply. 36 = 12x+x² x²+12x-36 = 0 (x+6)² = 36+(+6)2 x = -6±√(72) x = -6±6√(2) x ≠ -6-6√(2) x = 6√(2)-6 x = 6(√(2)-1) units. x = 2.4852813742 units. a = x+12 a = 6(√(2)-1)+12 a = 2.4852813742+12 a = 14.4852813742 units. a is the diameter of the ascribed semi circle. b = ½(a) b = 0.5*14.4852813742 b = 7.2426406871 units. b is the radius of...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2457614

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a² = 16+9-2*3*4cos150 a = 6.7664325675 cm.  a is the side length of the regular triangle. (6.7664325675/sin150) = (4/sinb) b = 17.192123734° c = 60-b c = 60-17.192123734 c = 42.807876266° Calculating x. x² = 3²+6.7664325675²-2*3*6.7664325675cos42.807876266 x = 5 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2457589

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Calculating lengths x plus y. Calculating y. y² = 2(10)² y = √(2(10)²) y = 10√(2) cm. y = 14.1421356237 cm. Calculating x. sina = 12/20 a = asin(3/5)° b = ½(a) b = (0.5asin(3/5))° Therefore, length x is; sin(0.5asin(3/5)) = x/20 x = 20sin(0.5asin(3/5)) x = 6.3245553203 cm. It implies; x+y is; = 6.3245553203+14.1421356237 = 20.466690944 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2457550

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Let R be 1 unit (radius of the inscribed big circle). Let a be the ascribed square side length. Calculating a. (a-1)² = 1²+(0.5a)² a²-2a+1 = 1+¼(a²) 4a²-a² = 8a 3a = 8 a = ⅓(8) units. Again, a is the side length of the ascribed square. sin60 = b/a ½√(3) = b/(⅓(8)) ⅓(8√(3)) = 2b b = ⅙(8√(3)) units. b = ⅓(4√(3)) units. c = a-r c = (⅓(8)-r) units. d = (⅓(8)+r) units. Calculating r, radius of the small inscribed circle. d² = c²+r² (⅓(8)+r)² = (⅓(8)-r)²+(½*⅓(8))² ⅑(64)+⅓(16r)+r² = ⅑(64)-⅓(16...
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