Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2710974

Calculating area of the inscribed red square. Let a be the side length of the inscribed red square. 2b² = a² b = √(a²/2) b = ½√(2)a units. 10² = a²+(½√(2)a)²-2a*½√(2)acos(45+90) 100 = ½(3a²)+a² 100 = ½(5a²) 5a² = 200 a² = 40 square units. Where a² is the area of the inscribed red square.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2710269

Let the base of the quadrilateral be 1 unit. a = 180-58-28-38 a = 180-124 a = 56° (1/sin56) = (b/sin(58+28)) b = 1.2032796622 units. (1/sin56) = (c/sin38) c = 0.7426219217 units. d = 180-72-46 d = 180-118 d = 62° (1.2032796622/sin62) = (e/sin46) e = 0.9803152067 units. f = 180-28-38-46 f = 180-112 f = 68° (1/sin68) = (g/sin(38+46)) g = 1.0726264165 units. h = 180-78-58 h = 44° (1.0726264165/sin44) = (j/sin78) j = 1.5103639435 units. k = j-c k = 0.767...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2709806

Let the ascribed quarter circle radius be 1 unit. 2a² = 1² a² = ½ a = ½√(2) units. a = 0.7071067812 units. a is the radius of the blue inscribed quarter circle. Area blue inscribed quarter circle (S2) is; ¼*πa² = ¼(½√(2))²π = ⅛π square units. c = (1-b) units. b is the radius of the yellow inscribed semi circle. It implies, calculate b. (1-b)² = b²+(½√(2))² 1-2b+b² = b²+½ 2b = ½ b = ¼ units. b = 0.25 units. Again, b is the radius of the yellow semi circle. Area...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2709796

Sir Mike Ambrose is the author of the question. Let the single side length of the small square be 1 unit. Therefore; Area large square is; ((1/tan15)-1)² = 7.46410161514 square units. Area orange is; Area large square - area triangle with height 2.73205080757 units and base 0.73205080757 units - area triangle with height 2 units and base 0.53589838486 units - area triangle with height 2.19615242271 units and base 0.58845726812 units. = 7.46410161514 - ½(2.73205080757*0.732...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2709788

Sir Mike Ambrose is the author of the question. The width of the rectangle is; (5/2)+(11/2) = 8 cm Area Rectangle is; 12*8 = 96 cm² Area Green is; ½(7*12)sin((90-atan(5√(3)/11))) = 33 cm² Area green ÷ Area rectangle is; 33 ÷ 96 = 11/32 = 0.34375 = 0.34 to 2 decimal places.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2709264

Let the square side length be 1 unit. tan15 = a/1 a = 0.2679491924 units. b = 1+a b = 1.2679491924 units. c = 1-a c = 0.7320508076 units. d² = 1²+1² d = √(2) units. e = 90-15 e = 75° f² = √(2)²+1.2679491924²-2√(2)*1.2679491924cos15 f = 0.378937382 units. g² = 0.378937382²+0.7320508076²-2*0.7320508076*0.378937382cos75 g = 0.7320508076 units. Therefore; Since c = g, the required angle, x is; x = 180-2e x = 180-2*75 x = 180-150 x = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2708447

Let the length of the two congruent rectangle be 1 units. Let the width be x. Therefore; 1 ~ (1+x) x ~ 1 It implies; 1/x = (x+1)/1 x²+x-1 = 0 (x+½)²=1+¼ x = -½±½√(5) Therefore; x ≠ -½(1+√(5)) x = ½(√(5)-1) units. Let alpha be y. Therefore; Tany = (1+x)/1 Tany = (1+x) And x = ½(√(5)-1) units. Tany = ½(√(5)-1)+1 Tany = ½√(5)-½+1 Tany = ½√(5)+½ Tany = ½(1+√(5))

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2707965

Let the side length of the blue inscribed equilateral triangle be 1 unit. Area blue equilateral triangle is, 0.5*1²*sin60 = ¼√(3) square units. = 0.4330127019 square units. It implies; ⅓(¼√(3)) = ½*a²*sin60 ⅙ = ½*a² a² = ⅓ a = √(⅓) a = ⅓√(3) units. a is the side length of each of the three congruent red regular triangles. Therefore; b = 2a+1 b = 2*⅓√(3)+1 b = ⅓(2√(3)+3) units. c² = (⅓(2√(3)+3))²+(⅓√(3))²-2*⅓√(3)*⅓(2√(3)+3)cos120 c² = ⅑(12+12√(3)+9)+⅓+⅑(6+3√(3))...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2707956

Sir Mike Ambrose is the author of the question. The equation of the curve is; Using the points; (4, 4), (2, 0), (12, 0) And, y =ax²+bx+c Therefore; 4=16a+4b+c 0=4a+2b+c 0=144a+12b+c Solving the three equations via matrix determinant approach. It implies; a = -¼, b = ½(7) and c = -6 Substituting a, b and c values in the standard quadratic equation. The equation of the curve is; y = -¼(x²) + ½(7x) - 6 The maximum turning point of the curve is; At y = 6¼...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2707948

Sir Mike Ambrose is the author of the question. Area BDE is; Area triangle with side 2 units and 3 units, and angle 120° = ¼(6√(3)) = ½(3√(3)) square units. = 2.59807621135 square units. Area yellow is; Area triangle with side 2 units and 3 units, and angle 120° - Area triangle with side 2 units and 1.74355957742 units, and angle (60-atan(3√(3)/7))° + Area triangle with side 1.30766968306 units and 3 units, and angle atan(3√(3)/7)° - Area triangle with side 1 unit and 0.4175747...

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