Mathematics Question and Solution

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

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...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2456459

(x³)² = (x²)²+x² x⁶ = x⁴+x² Dividing through by x² x⁴ = x²+1 Let p be x² p² = p+1 p²-p-1 = 0 Resolving the above quadratic equation via completing the square approach to get p. (p-½)² = 1+(-½)² (p-½)² = (5/4) p = ½±√(5/4) p = ½±½√(5) It implies; p ≠ ½(1-√(5)) p = ½(1+√(5)) units. And p = x² Therefore; x² = ½(1+√(5)) x = √(0.5(1+√(5))) units. x = 1.2720196495 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2456209

Calculating angle EHB. Let the side length of the regular pentagon be 1 unit. Therefore; AB = BC = CD = DE = AE = 1 unit. a² = 2-2cos108 a = 1.6180339887 units. a is BE, the square side length. b = 108-½(180-108) b = 108-½(72) b =108-36 b = 72° b is angle BED c = 90-b c = 90-72 c = 18° c is angle DEG. sin18 = d/1 d = 1sin18 d = 0.3090169944 units. cos18 = e/1 e = 1cos18 e = 0.9510565163 units. f = a-e f = 1.6180339887-0.9510565163 f = 0.6669774724 unit...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2456135

Let the radius of the circle be r. Calculating r. 2(½*12r)+(r*r)+2(½*8r) = ½(12+r)(8+r) (12r)+r²+(8r) = ½(12+r)(8+r) 24r+2r²+16r = 96+20r+r² 40r+2r² = 96+20r+r² r²+20r-96 = 0 (r+10)² = 96+100 r = -10±√(196) r = -10±14 It implies; r ≠ -24 r = 4 cm. Therefore, the height (h) and base (b) of the ascribed right-angled triangle is; h = 12+4 = 16 cm. b = 8+4 = 12 cm. Green painted area is; Area triangle with height 16 cm and base 12 cm - Area circle with radius 4 cm. = ½(16*12)-π(4)² = 96-16π...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2455106

Calculating length PQ. tana = 12/5 a = atan(12/5) a = 67.380135052° b = 180-a b = 180-67.380135052 b = 112.619864948° c = ½(b) c = ½(112.619864948) c = 56.309932474° tan56.309932474 = r/d d = r/(1.5) tan45 = r/e e = r It implies; d+e = 5 r/(1.5)+r = 5 r+1.5r = 7.5 2.5r = 7.5 r = 3 units. r is the radius of the circle. f = 90+a f = 90+67.380135052 f = 157.380135052° f is the angle subtending PQ and the radius of the circle. It implies; (PQ)² = 2(3)²-2(3)²cos157.380135052 PQ = 5.883484...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2454585

Therefore; a = 224 b = 9 c = 79 d = 54

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2454558

Sir Mike Ambrose is the author of the question. Let AC = 2 units. Therefore; AB = ⅖ unit. BC = (8/5) unit. It implies; Area Blue is; Area triangle with height ⅖ unit and base 2sin120 units + Area triangle with height 1 unit and base 0.4√(31)sin111.05172443535 units = (½*⅖*2sin120) + (½*1*0.4√(31)sin111.05172443535) = √(3)/5 + 1.03923048454 = 1.38564064606 square unit. Area Green is; Area triangle with height √(7) units and base 3√(21)/5 units = ½*√(7)*3√(21)/5 = 21√(3)/10 square un...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2453932

(7/sin60) = (8/sina) a = 81.7867892983° b = 120-a b = 120-81.7867892983 b = 38.2132107017° (7/sin60) = (c/sin38.2132107017) c = 5 units. c is the side length of the regular hexagon. cos60 = c/d ½ = 5/d d = 10 units. e = d-8 e = 10-8 e = 2 units. Observing similar plane shape (scalene triangle) side length ratios. 5 - f 8 - 2 Cross Multiply. 8f = 10 f = (5/4) units. Therefore, the required length, ? Is; c - f = 5-(5/4) = ¼(20-5) = ¼(15) units. = 3.75 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2453179

Observing similar plane shape (right-angled triangle) side length ratios. 3 - 5 a - (4-a) Cross Multiply. 5a = 3(4-a) 5a = 12-3a 8a = 12 a = (3/2) units. b = 4-a b = 4-(3/2) b = (5/2) units. c = √(3²+(3/2)²)+√(5²+(5/2)²) c = √(¼(45))+√(¼(125)) c = ½(3√(5))+½(5√(5)) c = ½(8√(5)) units. c = 4√(5) units. It implies; Calculating the required length, x. c² = 3²+x² (4√(5))²-3² = x² x² = 80-9 x = √(71) units. x = 8.4261497732 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2452768

a² = 2²+4² a = √(20) cm. a is radius OA = OB. tanb = 4/2 b = atan(2)° c = atan(½)° d = 180-b-c d = 180-atan(2)-atan(½) d = 90° d is the angle subtended by chord AB and the radius (OA and OB) of the sector. Area yellow is; (90π*√(20)²/360)-(½*√(20)²*sin90) = 5π-10 = 5(π-2) cm² = 5.7079632679 cm²

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