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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2891801

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Let x the the side length of the inscribed green equilateral triangle. Calculating area green. a = 180-60 a = 120° b = 45-30 b = 15° c = 180-120-15 c = 180-135 c = 45° d = (4+√(6)) units. It implies; (x/sin45) = ((4+√(6))/sin120) x = 5.2659863237109 units. Again, x is the side length of the inscribed green equilateral triangle. Area green inscribed equilateral triangle is; ½(x²)sin60 = 0.5*5.2659863237109²*sin60 = 12.0077072105782 square units. = 12 square...

Mathematics Question and Solution.

https://onlineedumath.com/update/mathematics-question-and-solution/2891685

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Sir Mike Ambrose is the author of the question. Let alpha be x. Therefore; (12/sin2x)=(8/sinx) And; Sin2x = 2sinxcosx Therefore; (12/2sinxcos)=(8/sinx) 6 = 8cosx x = acos(6/8)° 2x = 2acos(6/8)° Let y be the third interior angle. y = (180-3acos(6/8))° It implies; Area of the triangle in cm² is; Area triangle with height 12 cm and base 8sin(180-3acos(6/8)) cm. = ½*12*8sin(180-3acos(6/8)) = 48sin(180-3acos(6/8)) cm² = 39.686269666 cm².

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2890296

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Sir Mike Ambrose is the author of the question. Let the side length of the green regular hexagon be 1 unit. Therefore the angle theta to 2 d. p. is; (atan(0.5÷(tan72-sin60)) + atan(0.5÷(tan72+sin60)))° = (12.7389799507+7.22564879327)° = 19.964628744° ≈ 19.96°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2890287

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Sir Mike Ambrose is the author of the question. Area green exactly in cm² is; Area ABCD - Area trapezoid with parallel sides (9-2√(3)) cm and (24√(3)-39) cm, and height 6 cm. = (54-6√(3)) - ½*6((9-2√(3))+(24√(3)-39)) = (54-6√(3)) - 3(22√(3)-30) = 54-6√(3)-66√(3)+90 = 144-72√(3) = 72(2-√(3)) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2888719

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Area green is; Area triangle with height 8 units and base 8√(3) units - Area sector with radius 8 units and angle 30° + Area sector with radius 8 units and angle 60° - 2(area triangle with height 8 units and base 8sin60 units). = ½(8*8√(3)) - (30π*64/360) + (60π*64/360) - 2(½*8*8sin60) = 32√(3)-⅓(16π)+⅓(32π)-32√(3) = ⅓(16π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2888702

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Length AB is; 2(65/8) = ¼(65) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2887086

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Sir Mike Ambrose is the author of the question. Area blue total, exactly in cm² decimal is; Area trapezoid with parallel sides (80/7) cm and ((80/7)+21.4285714284) cm, and height 20 cm - Area trapezoid with parallel sides 20 cm and ((80/7)+12.2448979591) cm, and height (60/7) cm - Area triangle with height (80/7) cm and base 12.2448979591 cm. = (½*20((80/7)+((80/7)+21.4285714284))) - (½*(60/7)(20+((80/7)+12.2448979591))) - (½*(80/7)*12.2448979591) = 442.857142855 - 187.172011662 -...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2887075

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(5-r)² = r² + (3+r)² Where r is the radius of the blue inscribed circle. 25-10r+r²=r²+9+6r+r² 25-10r=9+6r+r² 9+6r+r²+10r-25=0 r²+16r-16=0 (r+8)²-16-64=0 (r+8)² = 80 r = -8±4√(5) Therefore; r ≠ -8-4√(5) r = -8+4√(5) r = (4√(5)-8) units. It implies; Area of blue inscribed circle is; π(4√(5)-8)² = (80-64√(5)+64)π = (144-64√(5))π = 16π(9-4√(5)) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2886119

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Sir Mike Ambrose is the author of the question. Side length of the square is; Let it be x. x = √((16*34)-144) x = √(400) x = 20 cm. Therefore; Area Orange exactly in decimal cm² is; Half area square with side 20 cm - Area triangle with height 7.88405797101 cm and base ⅓(32sin(2atan(3/5))) cm - Area triangle with height 8.74642784227 cm and base (12sin(atan(5/3))) cm = ½(20²) - ½(7.88405797101*⅓(32sin(2atan(3/5)))) - ½(8.74642784227*(12sin(atan(5/3)))) = 200 - 37.10144927...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2885693

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Let the radius of the blue inscribed circle be r.   Calculating r. (5+r)² = (5-r)² + (5+√(25-10r))² It implies; 20r = 25+10√(25-10r)+25-10r 30r-50 = 10√(25-10r) (3r-5)² = 25-10r 9r²-30r+25 = 25-10r 9r² = 30r-10r 9r² = 20r It implies; r ≠ 0 r = (20/9) units. Therefore; Area of the blue inscribed circle is; π(20/9)² = 400π/81 square units.
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