Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2611880

Calculating r, radius of the inscribed half circle. a = (r-3) units. b = (r-6) units. It implies; r² = a²+b² r² = (r-3)²+(r-6)² r² = r²-6r+9+r²-12r+36 r²-18r+45 = 0 Resolving the above quadratic equation via factorization approach to get r, radius of the inscribed half circle. r²-3r-15r+45 = 0 r(r-3)-15(r-3) = 0 (r-3)(r-15) = 0 It implies; r ≠ 3 r = 15 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2611871

Notice! Diagonal of the inscribed rectangle is also the radius of the ascribed circle. Let it be a. b = (a-2) units. c = (a-1) units. It implies; a² = (a-1)²+(a-2)² a² = a²-2a+1+a²-4a+4 a²-6a+5 = 0 Resolving the above quadratic equation via factorization approach to get a, the diagonal of the inscribed rectangle. a²-5a-a+5 = 0 a(a-5)-1(a-5) = 0 (a-5)(a-1) = 0 It implies; a ≠ 1 unit. a = 5 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2610461

Sir Mike Ambrose is the author of the question. Let the single side length of the square be 2 unit. Therefore; Area blue is; Area triangle of height (12+2√(3))/9 unit and base 2 unit - Area triangle of height 1 unit and base 2 unit. = ½(2x(12+2√(3))/9) - ½(2x1) = (12+2√(3))/9) - 1 = ⅑(3+2√(3)) square unit. Area red is; Area triangle with height (√(3)/3) unit and base (5+√(3))/22 unit. = ½(√(3)/3)((√(3)+5)/22) = (3+5√(3))/132 square unit. Therefore; Area Red/Ar...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2610436

Sir Mike Ambrose is the author of the question. Area A exactly as a single fraction is; Area sector with radius 2√(2) units and angle 60° - Area equilateral triangle with side 2√(2) units+ Area rectangle with length √(2) units and width √(2)tan15 unit. = (60π*(2√(2))²/360) - (½(2√(2))²*sin60) + (√(2)*√(2) tan15) = ⅓(4π) - 2√(3) + 2tan15 = ⅓(4π-6√(3)+6tan15) square units. = ⅓*4(3-3√(3)+π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2610156

Let a, the side length of the two equal squares be 1 unit. tanb = 1/2 b = atan(½)° c = 90-b c = (90-atan(½))° c = atan(2)° d = ½(atan2)° It implies; tan(0.5atan(2)) = x/1 0.6180339887 = x/1 x = 0.6180339887 units. Therefore; a ÷ x is; 1/0.6180339887 = 1.6180339887 = ½(1+√(5))

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2610143

Calculating r, radius of the circle. a²+r² = 17² a = √(289-r²) units. It implies; r² = √(289-r²)²+7² 2r² = 289+49 2r² = 338 r² = 169 r = 13 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2610136

Let BC be 1 unit. a = 180-70-52 a = 58° a is angle BDC. (1/sin58) = (b/sin52) b = 0.9292052623 units. b is BD. c = 189-14-54 c = 112° c is angle BAC. (1/sin112) = (d/sin54) d = 0.8725529358 units. d is AB. e = 70+14 e = 84° e is angle ABD. It implies; f² = 0.8725529358²+0.9292052623²-2*0.8725529358*0.9292052623cos84 f = 1.2063464293 units. f is AD. Therefore, the required angle, x is; (1.2063464293/sin84) = (0.8725529358/sinx) x = 46°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2610124

a = 2+3 a = 5 units. 5² = 3²+b² b = √(25-9) b = 4 units. c = 2+4 c = 6 units. Let d be the radius of the ascribed semi circle. e = c-d e = (6-d) units. d² = f²+3² f = √(d²-9) units. Calculating d. √(d²-9)² = (6-d)²+3² d²-9 = 36-12d+d²+9 12d = 54 2d = 9 d = ½(9) units. Again, d is the radius of the half circle. Therefore, diameter, g of the half circle is; g = 2d g = 2*½(9) g = 9 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2608549

Sir Mike Ambrose is the author of the question. Let the two equal length of the inscribed isosceles triangle (red area) be 2 unit. Therefore; Area blue will be; Area triangle with lengths 1 unit and ⅔ unit, and angle 60°, the angle they form. = ½(1*⅔sin60) = (√(3)/6) square unit. Area red will be; 2(area triangle with height 1 unit and base √(3) unit). = 2x½(1x√(3)) = √(3) square unit. Area green will be; 2(area triangle with height 2 unit and base ⅔(√(3)) unit...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2608527

Calculating Yellow Shaded Area. Radius, r of the inscribed circle is; Cos30=r/4 r = 2√(3) in. Length AD = DB = 2 in. Or Length AD = DB is; = √(4²-(2√(3))²) = √(16-12) = √(4) = 2 in. Therefore yellow area will be; 6(area triangle of height 2√(3) in and base 2 in - area sector of with radius 2√(3) in and angle 30°) = 6(½(2√(3)x2) - (30π(2√(3))²/360) = 6(2√(3) - π) in² = 1.9350537693 in²

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