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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3172117

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Calculating Shaded Area. Notice. a = ½(4) a = 2 cm. a is the radius of the inscribed circle. tanb = 4/2 b = atan(2)° c = 90-b c = atan(½)° d² = 2²+4² d = √(20) cm. d = 2√(5) cm. Therefore, shaded area is; ¼(2*2π)+2(½*√(20)*2sin(atan(½))) = π+4 = (π+4) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3172109

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Calculating x. a = 23+x+3 a = (26+x) units. b = 23-3 b = 20 units. Length x is; 21²+20² = (26+x)² 441+400 = 676+52x+x² x²+52x-165 = 0 (x+26)² = 165+26² (x+26)² = 676+165 (x+26)² = 841 x = -26±√(841) x = -26±29 It implies; x ≠ -26-29 x = -26+29 x = 3 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3170573

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Area Pink is; ¼(π*3²)-½(3²)+3((¼π)-½) = ¼(9π)-½(9)+¼(3π)-½(3) = ¼(12π)-½(12) = 3π-6 = 3(π-2) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3170568

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Calculating x. Notice. AC = BC  Let a be alpha. It implies; sina = 0.5x/(1+4) sina = ⅒(x) --- (1). Again. sina = 1/x --- (2). Equating (1) and (2). ⅒(x) = 1/x x² = 10 x = √(10) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3170567

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Calculating the area of the ascribed half circle. Let a be the radius of the circle. b² = a²+(3√(3))² b = √(a²+27) units. c = ½(3+b) c = ½(3+√(a²+27)) units. d = c-3 d = ½(3+√(a²+27))-3 d = ½(√(a²+27)-3) units. e² = (3√(3))²-(½(√(a²+27)-3))² e² = 27-¼(a²+27-6√(a²+27)+9) e² = ¼(108-a²-27+6√(a²+27)-9) e² = ¼(72-a²+6√(a²+27)). It implies; Calculating a. e²+c² = a² ¼(72-a²+6√(a²+27))+(½(3+√(a²+27)))² = a² ¼(72-a²+6√(a²+27))+¼(9+6√(a²+27)+a²+27) = a² 72-a²+6√(a²+27)+36+6√(a²+27)+a² = 4a²...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3168936

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Calculating red Area. Notice. 3x = 4-x 4x = 4 x = 1 unit. Where x is the radius of the red quarter circle and the inscribed white half circle. y = 4-x And x = 1 units. y = 4-1 y = 3 units. y is the radius of the bigger quarter circle. Area red is; ¼(πy²)-¼(πx²) = ¼*π(3²-1²) = ¼*π(9-1) = ¼*π*8 = 2π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3168930

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Calculating Length FE. tana = 10/20 a = atan(½)° b = 2a b = 2atan(½)° c = 180-2b c = 180-4atan(½) d² = 2(20)²-2(20)²cos(180-4atan(½))° d = 24 units. d is DE. Notice. DF = 20 units. Therefore, the required length, FE is; FE = DE-DF  FE = 24-20 FE = 4 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3167213

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Calculating Shaded Area. a = 180-40-30 a =110° a is angle BAC. (6/sin40) = (b/sin30) b = 4.66717148058 cm. b is AB. c = a-90 c = 20° cos20 = (0.5*4.66717148058)/d d = 2.48335007498 cm. d is the radius of the circle. e = 180-2c e = 140° Therefore Shaded Area is; (140π*2.48335007498²/360)-(0.5*2.48335007498²sin140) = 7.53444556147-1.9820444633 = 5.55240109817 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3167208

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Calculating Shaded Area. Radius of the inscribed circle is 3 cm. tana = 6/3 a = atan(2)° b = a-45 b = (atan(2)-45)° c = 90-a c = atan(½)° d = 180-2c d = 180-2atan(½) d = 2atan(2)° e² = 2(3)² e = 3√(2) cm. f² = 2(3)²-2(3)²cos(2atan(2)) f = 5.366563146 cm. Therefore shaded area is; 0.5*3√(2)*5.366563146sin(atan(2)-45) = 3.6 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3165466

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Let the side length of the square be a. tanb = (0.5a)/a b = atan(½)° c = 90-b c = atan(2)° sin(atan(2)) = d/a d = 0.894427191a units. Calculating a. (3√(2))² = (0.894427191a)²+a²-2*0.894427191a*acos(atan(2)) 18 = 0.8a²+a²-0.8a² a² = 18 square units. Where a² is the area of the square.
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