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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2419488

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Calculating x, the diameter of the semi circle. Notice, we have a cyclic quadrilateral in the complete circle. Therefore; x² =11²+y² y² = x²-121 --- (1). cosa = (11/x) --- (2). y² = 7²+2²+2*2*7cosa --- (3). Therefore, substituting (2) and (1) in (3) to get x, the diameter of the semi circle. x²-121 = 7²+2²+2*2*7(11/x) x²-121 = 53+28(11/x) x³-121x = 53x+308 x³-174x-308 = 0 Resolving the above polynomial equation to get x, the diameter of the half circle. x³-14x²+14x²-174x-308 = 0 It imp...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2418395

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Notice. Area Blue is an inscribed half circle. Let r be the radius of area blue. r² = (r-1)²+(r-2)² r² = r²-2r+1+r²-4r+4 0 = r²-6r+5 Resolving the above quadratic equation via completing the square approach to get r, radius of the inscribed half circle. (r-3)² = -5+(-3)² (r-3)² = 4 r-3 = ±√(4) r = 3±2 Therefore; r ≠ 3-2 = 1 cm. r = 3+2 r = 5 cm. It implies, area blue inscribed semi circle is, ½(πr²/ = 0.5*π(5)² = ½(25π) cm² = 12.5π cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2418181

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Calculating r, radius of the two congruent circles. r² = 2²+a² a = √(r²-4) units. r² = 1²+b² b = √(r²-1) units. c = 2+1+1 c = 4 units. d = b-a d = (√(r²-1)-√(r²-4)) units. Therefore, r, radius of the two congruent circles is; (2r)² = c²+d² (2r)² = 4²+(√(r²-1)-√(r²-4))² 4r² = 16+r²-1-2√((r²-1)(r²-4))+r²-4 2r²-11 = -2√((r²-1)(r²-4)) 4((r²-1)(r²-4)) = (11-2r²)² 4(r⁴-5r²+4) = 121-44r²+4r⁴ -20r²+16 = 121-44r² 24r² = 105 8r² = 35 r = √(35)/√(8) r = ¼√(70) units. r = 2.0916500663 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2418135

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Let x be 4 units. Calculating total blue length of arc. It implies; ½(circumference of a circle with radius 2 units + circumference of a circle with radius 1 unit + circumference of a circle with radius 0.5 units). = ½(2*2π+1*2π+0.5*2π) = ½(4π+2π+π) = ½(7π) units. Calculating green length of arc. It's radius is; ½((x/4)+(x/2)+x) And x is 4 units. ½(1+2+4) = ½(7) units. Therefore green length of arc is; ½(2π(½(7)) = ½(7π) units. It implies; Blue length of arc (½(7π)) = Green length of...

Mathematics Question and Solution

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

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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/2417124

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Sir Mike Ambrose is the author of the question. Calculating the circle's radius, r. The radius of the circle is; Area triangle with base 21 cm and height r cm + Area triangle with base 15 cm and height r cm + Area triangle with base 12√(2) cm and height r cm = Area triangle with base 21 cm and height 12√(2)sin45 cm ½(21r) + ½(15)r + ½(12√(2)r) =  ½*21*12√(2)sin45 21r + 15r + 12√(2)r = 21*12√(2)sin45 (36+12√(2))r = 252r r = 252/(36+12√(2)) cm. r = 3(3-√(2)) cm. It implies; The exac...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2416872

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3²+4²+2*3*4cosa = 5²+6²-2*5*6cosa 25+24cosa = 61-60cosa 84cosa = 36 cosa = (36/84) cosa = (3/7) a = acos(3/7) a = 64.6230664748° b = 2a b = 129.2461329497° c² = 5²+6²-2*5*6cos64.6230664748 c = 5.9401779675 units. Therefore, R, radius of the circle is; sin64.6230664748 = (0.5*5.9401779675)/R R = (0.5*5.9401779675)/(sin64.6230664748) R = 3.2872861147 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2416776

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Calculating the shaded area. a² = 4²+6²-2*4*6cos30 a = 3.2296719057 units. sin15 = b/6 b = 1.5529142706 units. c = 90-15 c = 75° Or cosc = 1.5529142706/6 c = acos(1.5529142706/6) c = 75° (3.2296719057/sin30) = (4/sind) d = 38.2619661994° e = c-d e = 75-38.2619661994 e = 36.7380338006° tan36.7380338006 = f/1.5529142706 f = 1.1591109915 units. Therefore, area shaded is; Area triangle with height 1.5529142706 units and base 3.2296719057sin36.7380338006 units - Area triangle with height...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2415939

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2r² = 2² r = √(2) units. r is the radius of the circle. Area circle is; πr² = π√(2)² = 2π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2414073

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Sir Mike Ambrose is the author of the question. Let the radius of the ascribed semicircle be 2 units. Therefore; Radius of the inscribed circle is 1 unit. It implies; Area R is; Area sector with radius 2 units and angle 45° - Area triangle with height and base √(2) unit each. = (45*2*2π÷360)-½(√(2))² = ½(π)-1 = ½(π-2) square unit. Area S is; Area triangle with height and base √(2) unit each - Area sector with radius 1 unit and angle 90° + Area triangle with height and base 1 unit respect...
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