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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2782863

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Sir Mike Ambrose is the author of the question. Let the side length of the large square be x. Calculating x. 25=x²+64-(16xcos20) Therefore; x = 11.70244 cm. It implies; Area large square is; 11.70244² = 136.947101954 cm² Area orange is; Area triangle with height and base 8 cm each - Area triangle with two side 2.91176187413 cm and 2.27177411013 cm, and angle 70° = ½(8)² - ½(2.91176187413*2.27177411013)sin70 = 32 - 3.107970027 = 28.892029973 cm² Therefore; Are...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2764554

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Let the side length of the hexagon be 2 units. a = ⅙*180(6-2) a = 120° b = ½(360-(30+90)) b = ½(240) b = 120° (2/sin45) = (c/sin15) c = 0.7320508076 units. Calculating Area Blue. It is; 8(area triangle with height 2 units and base 0.7320508076sin120 units). = 8(0.5*2*0.7320508076sin120) = 5.0717967699 square units. d = 2+2(2sin30) d = 4 units. d is the radius of the ascribed circle. e = 2+2c e = 3.4641016152 units. f = 2-c f = 1.2679491924 units. It i...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2759076

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Calculating green area. a² = 2²+2² a = √(8) a = 2√(2) units. sin30 = b/a ½ = b/(2√(2)) b = √(2) units. c²+b² = a² c²+√(2)² = (2√(2))² c² = 8-2 c = √(6) units. Area green is; (½*2*2)+(½*√(2)*√(6)) = (2+√(3)) square units. = 3.7320508076 square units. Calculating yellow area. d = -45+60 d = 15° tan15 = b/e e = (√(2)/tan15) units. e = 5.2779168675 units. sin15 = √(2)/f f = (√(2)/sin15) units. f = 5.4641016151 units. 2g² = f² 2g² = 29.8564064601 g =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2757245

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Calculating radius of the inscribed circle. a² = 4²+4²-2*4*4cos(90-30) a² = 32-32cos60 a² = 16 a = 4 units. Let b be the radius of the inscribed circle. c = (4-b) units. d = 60-45 d = 15° e² = 2b² e = √(2)b units. Calculating b. It implies; cos15 = (0.5*4)/e cos15 = 2/√(2)b b = 2/(√(2)cos15) b = 1.4641016151 units. Again, b is the radius of the inscribed circle. Therefore, area inscribed circle is; πb² = π(1.4641016151)² = 6.7342977158 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2752584

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Sir Mike Ambrose is the author of the question. Let the side length of the regular hexagon be 1 unit. Therefore; Area green is; Area triangle with height 1 unit and base ⅓√(3) units. = ½(⅓√(3)*1) = √(3)/6 square units. Area yellow is; Area triangle with two length 0.40929915222 units and 1 unit, and angle (90-atan(⅓(2√(3))))° = ½(0.40929915222*1)sin(90-atan(⅓(2√(3)))) = 0.13397459621 square units. It implies; Area yellow ÷ Area green in its exact decimal is; 0.1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2749327

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Let the side length of the regular hexagon be 1 unit. It implies; x+y = 1 a = ⅙*180(6-2) a = 120° a is the single interior angle of the regular hexagon. b = 360-2a-45 b = 360-285 b = 75° c² = 2-2cos120 c = √(3) units. d = 45-30 d = 15° e = 180-15-75 e = 90° Calculating y. tan75 = √(3)/y y = 0.4641016151 units. Calculating x. Recall. x+y = 1 And y = 0.4641016151 units. x = 1-0.4641016151 x = 0.5358983849 units. Therefore; x : y is; 0.535898...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2735876

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Calculating area shaded blue. a² = (2+√(3))²+(3+2√(3))² a² = 4+4√(3)+3+9+12√(3)+12 a² = 28+16√(3) a = √(28+16√(3)) units. a = 7.4641016151 units. a is the hypotenuse of the ascribed right-angled triangle. Let b be the radius of the inscribed half circle. c = (2+√(3))-b c = (3.7320508076-b) units. d = √(28+16√(3))-(3+2√(3)) d = 7.4641016151-6.4641016151 d = 1 unit. Calculating b. (3.7320508076-b)² = 1²+b² 13.9282032303-7.4641016151b = 1 12.9282032303 = 7.4641016151b...

Mathematics Question and Solution

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

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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/2702955

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Area of the ascribed regular Pentagon is; 5/(4tan(180/5)) = 1.72047740059 square units. Single side length of the formed small inscribed regular Pentagon is; Let it be x. x² = (1/(1-cos108)) - (1/(1-cos108)cos36) x = √(0.14589803375) = 0.38196601125 units. Therefore area of the small inscribed regular Pentagon is; 5(0.38196601125)²/(4tan(180/5)) = 0.25101426986 square units. It implies; Ratios small inscribed regular pentagon to the ascribed big regular pentagon is...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2693601

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Calculating area shaded blue. a² = (2+√(3))²+(3+2√(3))² a² = 4+4√(3)+3+9+12√(3)+12 a² = 28+16√(3) a = √(28+16√(3)) units. a = 7.4641016151 units. a is the hypotenuse of the ascribed right-angled triangle. Let b be the radius of the inscribed half circle. c = (2+√(3))-b c = (3.7320508076-b) units. d = √(28+16√(3))-(3+2√(3)) d = 7.4641016151-6.4641016151 d = 1 unit. Calculating b. (3.7320508076-b)² = 1²+b² 13.9282032303-7.4641016151b = 1 12.9282032303 = 7.4641016151b...
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