Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2864602

Sir Mike Ambrose is the author of the question. Area Gold exactly in square cm decimal form is; Area trapezium with two parallel side 6√(13) cm and 14.9769052981 cm, and height 11.6487041207 cm - Area triangle with height 9.6896279025 cm and base 6√(13)sin(atan(1.75)) cm = (½*11.6487041207(6√(13)+14.9769052981)) - (½*9.6896279025*6√(13)sin(atan(1.75))) = 213.23076923 - 91 = 122.23076923 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2864071

Let the two equal inscribed right-angled triangle side lengths be a each. Calculating a. 0.5a² = 50 a² = 100 a = 10 units. a is EU. Notice! Arc TU = Arc US. Let b be the radius of the ascribed quarter circle. Calculating b. sin(0.5*45) = (0.5*10)/b sin22.5 = 5/b b = 5/sin(22.5) b = 13.0656296488 units. Therefore, ascribed quarter circle perimeter is; ¼*2πb+2b = 0.25*2*π*13.0656296488+2*13.0656296488 = 46.6547023572 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2862917

Blue area is; ½(2.5)²π - (180-2atan(3/4))π(2²)/360 + 0.5*2²sin(180-2atan(3/4)) - (180-2atan(4/3))π(1.5)²/360 + 0.5*1.5²sin(180-2atan(4/3)) + ½(4*3) - 2atan(4/3)π(1.5²)/360 + 0.5(1.5²)sin(2atan(4/3)) - 2atan(3/4)π(2²)/360 + 0.5(2²)sin(2atan(3/4)) = 9.81747704247-3.70918087201+1.92-1.44787749478+1.08+6-2.0864142405+1.08-2.57400443517+1.92 = 12 Area Blue = 12 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2862886

Let the side of the regular pentagon be 1 unit. Calculating Area Red. a² = 1²+1²-2*1*1cos108 a = 1.61803398875 units. b = 108-36-60 b = 12° (c/sin12) = (1.61803398875/sin96) c = 0.33826121272 units. d = 1-c d = 0.66173878728 units. e = 96-60 e = 36° tan36 = f/0.66173878728 f = 0.48078137139 units. Area Red is; ½(df) = 0.5*0.48078137139*0.66173878728 = 0.15907584083 square units. Calculating Area Lilac. sin36 = 0.48078137139/g g = 0.81795412443 units....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2861830

Sir Mike Ambrose is the author of the question. Let the side of the square be 2 units. Therefore; Area square is; 2² = 4 square units. Area S is; Area triangle with two side 0.98921899145 units and 1.02554900641 units, and angle 14.318078177° - Area sector with radius 2 units and angle 28.6361563539° + Area triangle with two side 2 units each, and angle 28.6361563539° = (0.5*0.98921899145*1.02554900641sin14.318078177) - (28.6361563539*2²π÷360) + (0.5*2²sin28.6361563539)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2861175

Let the square side be 2 units. tan30 = a/2 a = ⅓(2√(3)) unit. sin60 = 2/b b = ⅓(4√(3)) units. c = 2-a c = ⅓(6-2√(3)) units. It implies; Area Shaded is; 2(½*(⅓(6-2√(3)))²sin60) + 2*⅓(6-2√(3)) = ⅓(4√(3)) square units. Area of the composite shape is; Area Shaded + 2(½*(⅓(4√(3)))²sin60) = ⅓(4√(3)) + ⅓(8√(3)) = 4√(3) square units. Therefore; Fraction Shaded is; ⅓(4√(3)) ÷ 4√(3) = ⅓

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2861163

Sir Mike Ambrose is the author of the question. Shaded area coordinates are; Coordinate X is; X(½, (9/4)) Coordinate Y is; Y((-3/2), (17/4)) Coordinate Z is; Z((5/2), (33/4)) Therefore shaded area exactly in square units is; Area trapezoid with parallel side (17/4) unity and (33/4) units, and height 4 units - Area trapezoid with parallel side (17/4) units and (9/4) units, and height 2 units - Area under the curve at coordinate (½, 0) and ((5/2), 0) = ½*4((17/4)+(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2859617

Let the side of the regular octagon be a. Calculating a. a² = 8 a = √(8) a = 2√(2) units. 2b² = (2√(2))² b = 2 units. c = 2√(2)+2+2 c = (2√(2)+4) units. Where c is the length of the inscribed rectangle. d = c-2(2√(2)) d = (2√(2)+4)-4√(2) d = (4-2√(2)) units. Where d is the width of the inscribed rectangle. Area Rectangle is; cd = (2√(2)+4)*(4-2√(2)) = 8 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2859583

Sir Mike Ambrose is the author of the question. Let the radius of the semi circle be 1 unit. cos20 = a/2 a = 1.87938524157 units. cos20 = 1/b b = 1.06417777248 units. c = a-b c = 0.81520746909 units. Where c is AB, the side of the square. Area Square is; (0.81520746909)² = 0.66456321766 square units. Calculating AF. sin20 = d/2 d = 0.68404028665 units. 0.68404028665 = e 1.87938524157 = 0.81520746909 cross multiply. 1.87938524157e = 0.81520746909*0.6840402...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2858517

Sir Mike Ambrose is the author of the question. Let the side of the inscribed square be 2 units. Therefore; Area S is; Area triangle with two side 1.03528 units and 0.53589838486 units, and angle 45° = 0.5*1.03528*0.53589838486sin45 = 0.1961531464 space units. Area R is; Area triangle with two side (2√(2)-2) units and 0.65961372961 units, and angle 72.367664609° = 0.5*(2√(2)-2)*0.65961372961sin72.367664609 = 0.26038499708 space units. Area T is; Area sector with...

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