Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3218392

Calculating The Required Shaded Area. a = ¼(2R) a = ¼(6) a = 1.5 units. b² = 2(3)² b = 3√(2) units. c² = 1.5²+3²-2*3*1.5cos150 c = ½(3√(7)) units. c = 4.36396936676 units. d² = (3√(2))²+1.5²-2*1.5*3√(2)cos45 d = ½(3√(5)) units. d = 3.35410196625 units. 3² = 3.35410196625²+4.36396936676²-2*4.36396936676*3.35410196625cose 29.2743964674cose = 21.294228634 e = acos(21.294228634/29.2743964674) e = 43.331039462° Required Shaded Area is; (0.5*4.36396936676*3.35410196...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3197241

Calculating Area Ascribed Rectangle. Let x be the width of the ascribed rectangle. Let y be the length of the ascribed rectangle. a = 1+2+1 a = 4 units. b = (x-2) units. c = (y-2) units. 4² = (x-2)²+(y-2)² 16 = x²-4x+4+y²-4y+4 8 = x²+y²-4x-4y x²+y² = 8+4(x+y) --- (1). d²+1² = 2² d = √(3) units. e = (x-1) units. f = 2d+2e f = (2x+2√(3)-2) units. f = (2x+1.46410161514) units. f is the diagonal of the ascribed rectangle. It implies; f² = x²+y² (2x+1.464...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3189442

Calculating Green Area. Notice. The composite plane shape is not drawn to scale. Let x be the width of the ascribed rectangle. Therefore; x = the diameter of the inscribed small half circle, and also the radius of the inscribed quarter circle. b = (1+x) units. b is the length of the ascribed rectangle, and also the diameter of the big inscribed half circle. c = ½(b) c = ½(1+x) units. c is the radius of the big inscribed half circle. d = ½(x) units. d is the radius of the small inscribe...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3183735

Calculating Area Blue. Let a be the side length of the ascribed quarter circle. tan30 = 2/b b = 2√(3) units. c = (a-2) units. d = (2√(3)+½(a)) d = ½(4√(3)+a) units. Calculating a. (a-2)² = (½(4√(3)+a))²+2² a²-4a+4 = ¼(48+8√(3)a+a²)+4 4a²-16a = 48+8√(3)a+a² 3a²-(16+8√(3))a-48 = 0 3a²-29.8564064606a-48 = 0 Therefore; a = 11.3605 units. Again, a is the radius of the ascribed quarter circle. (11.3605)² = e²+f² f = √(129.06096025-e²) e is the side length of the inscribed blue square....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3158612

Sir Mike Ambrose is the author of the question. Calculating Exactly Green Area. Let x be the side length of each of the three inscribed congruent squares. a² = 2x² a = √(2)x a is the diagonal of the squares. b = 45-30 b = 15° cos15 = c/(√(2)x) c = ½(√(3)+1)x units. c = 1.36602540378x units. d = c+x d = ½(1+√(3))x+x d = ½(3+√(3))x units. d is the height of the ascribed regular hexagon. e² = 2²+2²-2*2*2cos120 e = √(8+4) e = 2√(3) units. e = d. Calculating x....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3155706

Let the three equal lengths be x. a = x+x+x+8+8 a = (16+3x) units. a is the diameter of the ascribed half circle. b = ½(a) b = ½(16+3x) units. b is the radius of the ascribed half circle. c² = x²+8² c = √(64+x²) units. It implies; Calculating x. √(64+x²) ~ (16+3x) x ~ √(64+x²) Cross Multiply. 16x+3x² = 64+x² 2x²+16x-64 = 0 x²+8x-32 = 0 (x+4)² = 32+(4²) (x+4)² = 48 x = -4±√(48) x = -4±4√(3) It implies; x ≠ -4-4√(3) units. x =4(√(3)-1) units. x = 2.9282032302...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3133538

b = √(a) b = √(48) b = √(16*3) b = 4√(3) units. b is the radius of the ascribed quarter circle. c = ½(b) c = 2√(3) units. c is the radius of the inscribed half circle. tan30 = d/c ⅓√(3) = d/(2√(3)) d = 2 units. e = b-d e = (4√(3)-2) units. e = 4.92820323028 units. e is the side length of the inscribed regular triangle. f = ½(e) f = (2√(3)-1) units. tan60 = g/f g = √(3)(2√(3)-1) g = (6-√(3)) units. g = 4.26794919243 units. g shares the inscribed regular triangle to two equal right-angled t...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3122933

Let a be the radius of the red inscribed circle. b = (4-a) units. tan30 = a/b ⅓√(3) = a/(4-a) 3a = 4√(3)-√(3)a (3+√(3))a = 4√(3) a = 4√(3)/(3+√(3)) a = ⅙(12√(3)-12) a = (2√(3)-2) units. a = 1.46410161514 units. Therefore, area red inscribed circle is; πa² = π(2√(3)-2)² = π(12-8√(3)+4) = (16-8√(3))π = 8(2-√(3))π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3096012

a = 180-30-45 a = 105° (1/sin105) = (b/sin30) b = 0.51763809021 units. c = √(2)-b c = 0.89657547217 units. (0.89657547217/sin105) = (d/sin30) d = 0.46410161514 units. It implies, area green is; 0.5*0.46410161514*0.89657547217sin45 = 0.14711431703 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3081841

Let x be a side length of the largest possible inscribed square of the scalene triangle with sides 13 cm, 14 cm and 15 cm respectively. It implies; x = 14÷((1/tan(53.13))+1+(1/tan(67.38))) x = 14÷(2.16667222427) x = 6.46152188743 cm. Therefore, area largest possible inscribed square A is; A = x² A = 6.46152188743² A = 41.7512651017 cm²

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