Mathematics Questions and Solution

https://onlineedumath.com/update/mathematics-questions-and-solution/2868178

sin30 = (4√(3))/a a = 8√(3) units. a is AC. AB = ⅓(a) AB = ⅓(8√(3)) units. cos30 = b/⅓(8√(3)) b = ⅓(8√(3))*½√(3) b = 4 units. sin30 = c/⅓(8√(3)) c = ⅓(4√(3)) units. d = 4+b d = 8 units. tan60 = e/8 e = 8√(3) units. cos60 = 8/f f = 16 units. Let r be the radius of the inscribed circle. 16r+8r+8√(3)r = 8*8√(3) (24+8√(3))r = 64√(3) r = (4√(3)-4) units. r = 2.9282 units. g = r+c g = (4√(3)-4)+⅓(4√(3)) g = ⅓(12√(3)-12+4√(3)) g = ⅓(16√(3)-12) units. It i...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2866770

Calculating x, side length of the inscribed regular hexagon. a = 2x units. a is twice the side length of the inscribed regular hexagon. b = ⅙*180(6-2) b = 120° b is the single interior angle of the regular hexagon. c = ½(6) c = 3 units. It implies; d² = 6²+3²-2*3*6cos120 d = 7.9372539332 units. d = 3√(7) units. Notice. d = 2x 2x = 3√(7) x = ½*3√(7) x = 1.5√(7) units. x = 3.9686269666 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2866283

Sir Mike Ambrose is the author of the question. Area blue exactly is square units decimal form is; ½(Area triangle with height (2√(2)+2√(2+√(3))) units and base 12.9282032303sin75 units - Area triangle with height 2 units and base (2sin150) units - Area trapezoid with two parallel side (2√(2)+2√(2+√(3))) units and (2√(2+√(3))) units, and height √(2) units) =½((0.5*(2√(2)+2√(2+√(3)))*12.9282032303sin75) - (0.5*2*2sin150) - (0.5*√(2)((2√(2)+2√(2+√(3)))+(2√(2+√(3))))) = ½(41.7846096909...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2866273

Radius, r of one of the five congruent, inscribed circles is; 8r²+√(3)r-1=0 r = (√(35)-√(3))/16 m. Therefore; Total area of the five congruent, inscribed yellow circles is; 5πr² = 5π( (√(35)-√(3))/16)² = 5π(35-2√(105)+3)/256 = 5π(38-2√(105))/256 = 5π(19-√(105))/128 m² = 1.07416074883 m²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2865779

a² = 4²+3² a = √(25) a = 5 units. a is MN, the side length of the inscribed square. b² = 2(5²) b = 5√(2) units. b is MK, the diagonal of the inscribed square. tanc = 4/3 c = atan(4/3)° c is angle ALK. d = atan(3/4)° d is angle CLM. e = ½(90)+d e = (45+atan(3/4))° e is angle CLN. f = 180-45-e f = 135-(45+atan(3/4)) f = 90-atan(3/4) f = atan(4/3)° f is angle CNL. Calculating x (CL). (x/sin(atan(4/3))) = (5√(2)/sin45) x = 8 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2864642

since tanx = 3/4, it implies; Let AB = 4 units. Let AD = 3 units. tana = 4/3 a = atan(4/3)° a is angle ADB. b = 180-a b = 90+90-atan(4/3) b = (90+atan(3/4))° b is angle BDC. It implies; The required angle, y is; y = angle BCD = angle DCB since BD = CD. Therefore; y = ½(180-b) y = ½(180-(90+atan(3/4)) y = ½(90+90-90-atan(3/4)) y = ½(90-atan(3/4)) y = ½atan(4/3)° y = 26.5650511771°

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....

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