Mathematics Question and Solution

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

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

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

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

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.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3165459

Calculating Length MA. Let a be the AB, side length of the regular hexagon. b = (a-3) units. b is AM. c = ½(a) units. c is AK = FK. Let d be CM = KM. It implies; d² = a²+3²-2*3acos120 d² = a²+9+3a --- (1). Again. d² = (0.5a)²+(a-3)²-2*½(a)(a-3)cos120 d² = ¼(a²)+a²-6a+9+½(a²-3a) d² = ¼(a²+4a²-24a+36+2a²-6a) d² = ¼(7a²-30a+36) --- (2). Equating (1) and (2). a²+9+3a = ¼(7a²-30a+36) 4a²+12a+36 = 7a²-30a+36 42a = 3a² a = ⅓(42) a = 14 units. Again a is AB, the side length of the regular...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3164095

Calculating Area Triangle ABC. Let a be the height of the triangle. b = √(a²+4) units. b is AC. c = √(a²+9) units. c is BC. Calculating a. ½(2+3)a = ½√(a²+4)*√(a²+9)sin45 10a = √(2(a²+4)(a²+9)) 100a² = 2a⁴+26a²+72 2a⁴-74a²+72 = 0 a⁴-37a²+36 = 0 a⁴-a²-36a²+36 = 0 a²(a²-1)-36(a²-1) = 0 (a²-1)(a²-36) = 0 It implies; a² = 1 or a² = 36 Therefore; a ≠ √(1) units. a = √(36) a = 6 units. Area Triangle ABC is; ½*a*(AB) = ½*6*(2+3) = 3(5) = 1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3164092

Let AD be x. a² = x²+4²-2*4*xcos30 a = √(x²+16-4√(3)x) units. a is BD. b² = x²+3²-2*3*xcos30 b = √(x²+9-3√(3)x) units. b is CD. c² = 4²+3²-2*3*4cos60 c² = 25-12 c = √(13) units. c is BC. Calculating length AD (x). It implies; a+b = c √(x²+16-4√(3)x)+√(x²+9-3√(3)x) = √(13) x²+9-3√(3)x = (√(13)-√(x²+16-4√(3)x))² x²+9-3√(3)x = 13-2√(13)√(x²+16-4√(3)x)+x²+16-4√(3)x 20-√(3)x = 2√(13)√(x²+16-4√(3)x) (20-√(3)x)² = 52(x²+16-4√(3)x) 400-40√(3)x+3x² = 52x²-208√(3)x+832 49x²-168√(3)x+432 = 0...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3162934

Calculating R, radius of the circle. a²+4² = 6² a = √(36-16) a = √(20) a = 2√(5) units. b²+4² = 10² b = √(100-16) b = √(84) b = 2√(21) units. Therefore; 2√(5)*2√(21) = 4c c = √(105) units. d = 4+c d = (4+√(105)) units. d = 14.246950766 units. e = ½(d) e = 7.123475383 units. f = ½(a+b) f = ½(2√(5)+2√(21)) f = (√(5)+√(21)) units. g = f-a g = (√(5)+√(21))-2√(5) g = (√(21)-√(5)) units. g = 2.34650771746 units. Therefore R, radius of the circle is; R² =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3162932

Let the two equal lengths be 1 unit each. (1/sin20) = (a/sin40) a = 1.87938524157 units. b = a-1 b = 0.87938524157 units. (1/sin20) = (c/sin(180-40-20)) c = 2.53208888624 units. d² = 2(2.53208888624)²-2(2.53208888624)cos40 d = √(3) units. d = 1.73205080757 units. e² = 1.73205080757²+0.87938524157²-2*1.73205080757*0.87938524157cos50 e = 1.34729635533 units. (1.34729635533/sin50) = (√(3)/sinf) f = 80° Where f is the required angle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3161756

Green Area is; Area square with length 4 units - Area quarter circle with radius 2 units - Area square with length 2 units - 3(Area square with length 1 unit) - Area square with length 3 units + Area quarter circle with radius 3 units - Area square with length 2 units + Area quarter circle with radius 2 units - Area quarter circle with radius 1 unit. = 16 - ¼(4π) - 4 - 3 - 9 + ¼(9π) - 4 + ¼(4π) - ¼(π) = 16 - 20 + ¼(9π - π) = ¼(8π) - 4 = 2π - 4 = 2(π-2) square units.

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