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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2808389

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Sir Mike Ambrose is the author of the question. The side length of the congruent Hexagons is 2 units. Area hexagon is; (6*2²)/(4tan30) = 6√(3)) square units. a² = 8-8cos30 a = √(8-4√(3)) units. Area ABC is; ½(√(8-4√(3)))²sin60 = (2√(3)-3) square units. (d) Area Hexagon ÷ Area ABC exactly is; (6√(3)) ÷ (2√(3)-3) = 6(2+√(3)) (c) Calculating Area Green to 2 decimal places. b = a + √(2²+2²) b = √(8-4√(3))+2√(2) b = (√(2)+√(6)) units. b = 3.86370330516 units. ...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2806638

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Let the side length of the regular pentagon be 4 units. a = ⅕*180(5-2) a = 108° a is the single interior angle of the regular pentagon. b = ½(180-a) b = ½(180-108) b = 36° c = 2 units. d = 90-36 d = 54° tan54 = e/2 e = 2.7527638409 units. Area yellow is; 0.5*2*2.7527638409 = 2.7527638409 square units. Calculating area regular pentagon. f² = 4²+4²-2*4*4cos108 f = 6.472135955 units. g = 108-90 g = 18° sin18 = h/4 h = 1.2360679775 units. cos18 = j/4...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2806591

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Notice. 7 square units is the are of the small pink half circle. ½*a²π = 7 And π = 22/7 ½*a²(22/7) = 7 11a² = 49 √(11)a = 7 a = 7√(11)/11 units. a = 2.110579412 units. a is the radius of the small pink half circle. b = 2a b = 14√(11)/11 units. b = 4.2211588241 units. b is the diameter of the small pink half circle. Let c be the side length of the small square. Calculating c. d = ⅛*180(8-2) d = ⅛*180*6 d = ½*45*6 d = 135° d is the single interior angle of a regu...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2805677

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a² = 4²+3² a = √(25) a = 5 units. a is MN, the side length of the inscribed square. b² = 2(5²) b = 5√(2) units. b is MK, the diagonal of the inscribed square. tanc = 4/3 c = atan(4/3)° c is angle ALK. d = atan(3/4)° d is angle CLM. e = ½(90)+d e = (45+atan(3/4))° e is angle CLN. f = 180-45-e f = 135-(45+atan(3/4)) f = 90-atan(3/4) f = atan(4/3)° f is angle CNL. Calculating x (CL). (x/sin(atan(4/3))) = (5√(2)/sin45) x = 8 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2805415

mathematics-question-and-solution_1730103268CkZlfw.jpeg
Notice. 7 square units is the are of the small pink half circle. ½*a²π = 7 And π = 22/7 ½*a²(22/7) = 7 11a² = 49 √(11)a = 7 a = 7√(11)/11 units. a = 2.110579412 units. a is the radius of the small pink half circle. b = 2a b = 14√(11)/11 units. b = 4.2211588241 units. b is the diameter of the small pink half circle. Let c be the side length of the small square. Calculating c. d = ⅛*180(8-2) d = ⅛*180*6 d = ½*45*6 d = 135° d is the single interior angle of a regu...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2805007

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tana = 5/10 a = atan(½)° b = 2a b = 2atan(½)° c = b-45 c = 8.1301023542° d² =10²+10²-2*10*10cos8.1301023542 d = 1.4177804018 units. e = ½(180-8.1301023542) e = 85.9349488229° f² = 2*5² f = 5√(2) units. g = 10-f g = 2.9289321881 units. tanh = 10/5 h = atan(2)° j = 2h j = 2atan(2)° k = j-90 k = 36.8698976458° It implies, area red is; Area triangle with height 1.4177804018 units and base 2.9289321881sin85.9349488229 units + Area sector with radius 10 un...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2804990

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Sir Mike Ambrose is the author of the question. Let length AB be 3 units. Therefore; AC = 20 units. It implies; Area large square is; 20² = 40 square units. Area orange is; Area triangle with two side 17 units and 32.0319709782 units, and angle (180-atan(10/7))° + Area triangle with two side 20 units and 44.0408623507 units, and angle 53.4269678346° - Area square with side 20 units + Area triangle with height 3 units and base 4.2857142857 units. = (0.5*32.0319709782...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2804295

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Let EB be a. Let AE be b. It implies; a/b = √(2) --- (1). a+b = 1 --- (2). From (1). a = √(2)b --- (3). Substituting (3) in (2) to get b. √(2)b+b = 1 b(√(2)+1) = 1 b = (√(2)-1)/(2-1) b = √(2)-1 b = 0.4142135624 units. b is AE. At (2). a+b = 1 And b = 0.4142135624 units. a = 1-0.4142135624 a = 0.5857864376 units. a is EB. tanc = 1/0.4142135624 c = 67.5° c is angle AED. d =½(c) d = 33.75° Let x be the small inscribed circle radius. e = (0.4...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2803507

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Sir Mike Ambrose is the author of the question. Let the side of the square be 2 units. Therefore; Area square is; 2² = 4 square units. Area S is; Area triangle with two side 0.98921899145 units and 1.02554900641 units, and angle 14.318078177° - Area sector with radius 2 units and angle 28.6361563539° + Area triangle with two side 2 units each, and angle 28.6361563539° = (0.5*0.98921899145*1.02554900641sin14.318078177) - (28.6361563539*2²π÷360) + (0.5*2²sin28.6361563539)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2803491

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Let the square side be 2 units. tan30 = a/2 a = ⅓(2√(3)) unit. sin60 = 2/b b = ⅓(4√(3)) units. c = 2-a c = ⅓(6-2√(3)) units. It implies; Area Shaded is; 2(½*(⅓(6-2√(3)))²sin60) + 2*⅓(6-2√(3)) = ⅓(4√(3)) square units. Area of the composite shape is; Area Shaded + 2(½*(⅓(4√(3)))²sin60) = ⅓(4√(3)) + ⅓(8√(3)) = 4√(3) square units. Therefore; Fraction Shaded is; ⅓(4√(3)) ÷ 4√(3) = ⅓
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