Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2417679

Sir Mike Ambrose is the author of the question. P coordinate is; P(3, 9) Q coordinate is; Q(2, 5), Q(3, 5) It implies; Shaded area exactly in its square units simplest form is; (Area under the curve at x = 2 and x = 3) - (Area trapezoid with parallel sides ⅓(23) units and 9 units, and height 1 unit). = (8+ln(81/16)) - ½((23/3)+9) = (8+ln(81/16))-½(50/3) = (8+ln(81/16))-⅓(25) = (ln(81/16)-⅓) square units. = 1.2885270991 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2406151

Sir Mike Ambrose is the author of the question. Let the side of the regular Pentagon be 2 units. Therefore; Area orange is; Area triangle with height 1.7013016167 unit and base 1.8345763534sin(18) units = ½*1.7013016167*1.8345763534sin(18) = 0.48224693327 square unit. Area blue is; Area trapezoid with parallel sides 0.21654236465 unit and 1 unit, and height 1.9021130326 unit ½*1.9021130326(0.21654236465+1) = 1.15700054326 square unit. It implies; Area Orange ÷ Area Blue to 2 decimal place...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2400085

Sir Mike Ambrose is the author of the question. Area Green exactly is; Area trapezoid with parallel sides 3 cm and (6-2√(3)) cm, and height (6-3√(3)) cm. = ½(6-3√(3))(3+(6-2√(3))) = ½(6-3√(3))(9-2√(3)) = ½(72-39√(3)) cm² = 3(24-13√(3))/2 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2397396

Notice, radius of the three congruent circles is 7 cm each. sin60 = a/14 a = 7√(3) cm. a = 12.124355653 cm. b = a+7+7 b = (7√(3)+14) cm. b = 26.124355653 cm. b is the length of the rectangle. tan30 = c/7 c = ⅓(7√(3)) cm. c = 4.0414518843 cm. d = 7+7+2c d = (14+⅓(14√(3)) cm. d = 22.0829037687 cm. d is the width of the rectangle. tane = ⅓(7√(3))/7 e = atan(⅓√(3))° e = 30° cosf = ⅓(7√(3))/7 f = acos(⅓√(3))° f = 54.7356103172° g = 90-e g = 90-atan(⅓√(3)) g = atan(√(3)) g = 60° h = 180-60-5...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2396165

Let r be the radius of the inscribed circle. a = (2-r) units. (2-r)² = r²+b² b² = 4-4r+r²-r² b = √(4-4r) units. c = (1+r) units. d = (1+√(4-4r)) units. It implies, observing Pythagoras Rule on a right-angled triangle with adjacent sides r units and (1+√(4-4r)) units, and hypotenuse (1+r) units to get r, radius of the inscribed circle. (1+r)² = r²+(1+√(4-4r))² 1+2r+r² = r²+1+2√(4-4r)+4-4r 6r-4 = 2√(4-4r) 3r-2 = √(4-4r) (3r-2)² = 4-4r 9r²-12r+4 = 4-4r 9r² = 8r 9r = 8 r = (8/9) units. Agai...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2385722

Calculating area red. a = 4+6 = 10 units. a is the side length of the two congruent regular composite plane shape. b = ½(10) b = 5 units. c = 5-1 c = 4 units. tans = 4/6 d = atan(2/3)° e² = 6²+4² e = √(36+16) e = √(52) e = 2√(13) units. f = 90-d f = 90-atan(2/3) f = atan(3/2)° tanf = g/4 tan(atan(3/2)) = g/4 3/2 = g/4 g = 6 units. It implies, h = e h = 2√(13) units. i = 10-6-1 i = 3 units. Therefore, area red is; Area trapezoid with parallel sides 3 units and 5 units respectively,...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2379840

Observing similar plane shape side lengths ratio. 3 - 5 a - (4-a) Cross Multiply. 3(4-a) = 5a 12-3a = 5a 8a = 12 a = 3/2 units. a is the base of the right-angled triangle with height 3 units. b = 4-a b = 4-(3/2) b = 5/2 units. b is the base of the right-angled triangle with height 5 units. c² = 3²+(3/2)² c² = 9+(9/4) c² = ¼(45) c = ½√(45) units. d² = 5²+(5/2)² d² = 25+(25/4) d² = ¼(125) d = ½√(125) units. e = c+d e = ½√(45)+½√(125) e = (3/2)√(5)+(5/2)√(5) e =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2373881

Sir Mike Ambrose is the author of the question. Area green exactly in decimal is; Notice. Square side is (16/√(17)) units. Area trapezoid with parallel sides (16/√(17)) units and (4/√(17)) units, and height (16/√(17)) units - Area triangle with height (16/√(17)) units and base (4/√(17)) units - Area triangle with height (16/√(17)) units and base 3.1067961165sin(atan(13/16)) units. = ½*(16/√(17))((16/√(17))+(4/√(17))) - ½*(16/√(17))(4/√(17)) - ½*(16/√(17))*3.1067961165sin(atan(13/16)) = (160/...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2370714

1. The proof that triangle EGH is equilateral. Let the side length of the regular hexagon be 1 unit. a = ⅙(180(6-2)) a = 30*4 a = 120° a is single interior angle of the regular hexagon. b² = 1²+0.5²-2*0.5cos120 b = 1.3228756555 units. b is EG. (1.3228756555/sin120) = (0.5/sinc) c = 19.1066053509° c is angle DEG. d² = 2-2cos120 d = 1.7320508076 units. d is BF. e = ½(d) e = ½(1.7320508076) e = 0.8660254038 units. e is FH. g² = 1²+0.8660254038² g = 1.3228756555 units. g is EH. Notice! EH...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2344174

Sir Mike Ambrose is the author of the question. Let the side let of the square be 2 units. Therefore; Area square is; 2² = 4 square units. It implies; Area R is; Area trapezoid with parallel sides 1 unit and 1.44948974279 unit, and height 2 unit - Area square with side 1 unit - Area triangle with base and height 1 unit respectively - Area sector with radius √(2) unit and angle 32.3335306935° - Area triangle with height √(2) unit and base 0.3178372452 unit = (½*2(1+1.44948974279)) - (1*1)...

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