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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2920515

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Sir Mike Ambrose is the author of the question. Calculating a. ½((0, 6√(3)) (-3, 3√(3)) (a, (16a/3)) (0, 6√(3))) = ¼(9+20√(3)) Therefore; a = x = ½√(3) units. y = 16a/3 = (16/3)(√(3)/2) y = ⅓(8√(3)) units. Therefore; Area green is; Area triangle with height (√(697-72√(3))/2√(3)) units and base 6sin(120-atan(16√(3)/(36-3√(3))) units. = ½*(√(697-72√(3))/2√(3))*6sin(120-atan(16√(3)/(36-3√(3))) = ½(√(2091-216√(3)))sin(120-atan(16√(3)/(36-3√(3)))  = ½√(2091-216√(3)sin...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2918704

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Let the radius of the ascribed semi circle be 2 units. Let the radius of the inscribed circle be r. Calculating r. (2-r)² = r² + 1 Therefore; r = ¾ units. It implies; OH = r = ¾ units. Notice; AH = AK = 3 units. Therefore; AH ÷ OH is; 3 ÷ (¾) = 4 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2918689

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Shaded area is; Area triangle with height 4√(3) cm and base 8 cm + 2(area sector with radius 8 cm and angle 60° - area triangle with height 4√(3) cm and base 8 cm) - Area circle with radius 2 cm - 2(area sector with radius 4 cm and angle 60°) - 2(area sector with radius 4 cm and angle 60° + area triangle with height 2√(3) cm and base 4 cm). = 16√(3) + 2(⅓(32π)-16√(3)) - 4π - 2(⅓(8π)) - 2(⅓(8π)+4√(3)) = 16√(3) + ⅓(64π) - 32√(3) - 4π - ⅓(16π) - ⅓(16π) + 8√(3) = ⅓(32π)-4π-8√(3) =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2917430

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Sir Mike Ambrose is the author of the question.  Area green is; ½*3*6sin120 = ½(9√(3)) cm² Radius, r of the circle is; r = √(63)/√(3) r = √(21) cm. Therefore; Area circle is; π(√(21))² = 21π cm² It implies; Area green ÷ Area circle is; ½(9√(3)) ÷ 21π = 3√(3)/14π  = 3√(3)/44

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2917410

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Sir Mike Ambrose is the author of the question. Let the side of the large square be 2 units. Therefore; Area large square is; 2² = 4 square units. Area gold is; Area triangle with height 2 units and base ⅔ units - Area triangle with height 1 unit and base ½(√(2))sin135 units. = (½*⅔*2) - (½*1*½(√(2))sin135) = ⅔ - ¼ = 5/12 square units. Area Gold ÷ Area Large Square is; (5/12)÷4 = 5/48

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2915395

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Total coloured area is; 2(area triangle with height 6.69991708075 cm and base √(65)sin(atan(10)-atan(7/4)) cm. = 2*½*6.69991708075*√(65)sin(atan(10)-atan(7/4)) = 6.69991708075*3.28362272769 = 22 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2915386

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Sir Mike Ambrose is the author of the question.  x = 16 The sequence is; 16, 12, 9, (27/4), ... a = 16 r = ¾  Calculating n. (2187/16383) = (¾)^(n) n = 7. Therefore; The exact last term for the set of values is; L = 16*(¾)^(6) L = 729/256

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2913587

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Shaded area is; Area trapezoid with parallel sides 58 cm and 28 cm, and height 30 cm - Area trapezoid with parallel sides 9 cm and 30 cm, and height 28 cm - Area triangle with height 12 cm and base 12 cm. = ½*30(28+58) - ½*28(9+30) - ½*12*12 = (15*86) - (14*39) - 72 = 1290 - 546 - 72 = 1290 - 618 = 672 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2913582

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Sir Mike Ambrose is the author of the question. Radius, r of the inscribed circle is; r = (3-√(3)) cm. Therefore; Area purple exactly in square cm is; Area trapezium with parallel sides (3-√(3)) cm and 4 cm, and height 3+√(3) cm - 2(area triangle with height (3-√(3)) cm and base (3√(3)-3) cm). = ½*(3+√(3))(4+3-√(3)) - 2(½*(3-√(3))*(3√(3)-3)) = (9+2√(3)) - (12√(3)-18) = (27-10√(3)) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2911913

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Sir Mike Ambrose is the author of the question.  P(2, (16√(5)/5)) Gradient of the curve at the tangent is; = 24/5√(5) Therefore; Gradient of the curve at the normal is; = -5√(5)/24 It implies; Coordinate Q is; Q(4, (167√(5)/60)) Therefore area R as a single fraction in square units is; (Area under the curve at x = 2 and x = 4) - (Area trapezium with parallel sides 167√(5)/60 units and 16√(5)/5 units, and height 2 units). = (24-⅓(8√(5))) - ½*2((167√(5)/60 )+...
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