Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2793942

Calculating h. tan60 = h/a a = ⅓(√(3)h) units. tan30 = h/b b = √(3)h units. It implies; a + b = 100 ⅓(√(3)h)+√(3)h = 100 √(3)h+3√(3)h = 300 4√(3)h = 300 √(3)h = 75 h = ⅓(75√(3)) units. h = 25√(3) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2792503

Sir Mike Ambrose is the author of the question. Let the ascribed quarter circle radius be 4 units. Therefore; Area blue is; Area triangle with two side 1 unit and 3 units, and angle 2asin(1/3)° = ½*3sin(2asin(1/3)) = 0.94280904158 space units. Area red is; Area triangle with two side 1 unit and 3 units, and angle (180-2asin(1/3))° + Area triangle with two side 2 units and √(44/3) units, and angle 100.02498786° = ½*3sin(180-2asin(1/3)) + ½*2*√(44/3)sin100.02498786 = 0.942809...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2791220

Let a be the radius of the circle. 10² = 2a²+2a²cosd 50 = a²+a²cosd --- (1). 10² = (2a)²+e² e = √(100-4a²) units. √(100-4a²)² = 2a²-2a²cosd 100-2a² = 2a²-2a²cosd 2a²cosd = 4a²-100 cosd = (2a²-50)/a² --- (2). Substituting (2) in (1). 50 = a²+a²(2a²-50)/a² 100 = 3a² a² = ⅓(100) a = ⅓(10√(3)) units. a = 5.7735026919 units. Again, a is the radius of the circle and also the width of the blue area. cose = 5/a cose = 5/⅓(10√(3)) cose = 15/(10√(3)) cose = 3/(2√(3)) cos...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2789357

Calculating Silver Area. a² = 6²+(0.5*16)² a = √(100) a = 10 cm. a is the radius of the inscribed sector. tanb = 8/6 b = atan(4/3)° c = 2b c = 2atan(4/3)° c is the angle of the inscribed sector. Area silver is; Area square with side length 16 cm - Area rectangle with length 8 cm and width 6 cm - Area sector with radius 10 cm and angle 2atan(4/3). = (16*16)-(8*6)-(2atan(4/3)π*10*10/360) = 256-48-(5atan(4/3)π/9) = 208-(5atan(4/3)π/9) = ⅑(1872-(5atan(4/3)π)) cm² = 115...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2787686

Notice! AB = BC Let AB = BC be a. b²+(0.5(8+10))² = a² b = √(a²-81) units. c = ½(10+8)-8 c = 1 unit. √(a²-81) - 1 9 - √(a²-81) Cross Multiply. a²-81 = 9 a² = 90 a = √(90) a = 3√(10) units. Calculating x. x²+(3√(10))² = 10² x² = 100-90 x² = 10 x = √(10) units. x = 3.1622776602 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2782725

Calculating Area Red. tana = 5/10 a = atan(½)° b = 2a b = 2atan(½)° c = b-45 c = 8.1301023542° d² =10²+10²-2*10*10cos8.1301023542 d = 1.4177804018 units. e = ½(180-8.1301023542) e = 85.9349488229° f² = 2*5² f = 5√(2) units. g = 10-f g = 2.9289321881 units. tanh = 10/5 h = atan(2)° j = 2h j = 2atan(2)° k = j-90 k = 36.8698976458° It implies, area red is; Area triangle with height 1.4177804018 units and base 2.9289321881sin85.9349488229 units + Area s...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2776225

½(xy)=24 xy = 48 ------ (1) x²+y²=10² ------- (2) And; x²+y²=(x+y)²-2xy Therefore; (x+y)²-2xy=100 And; xy = 48 (x+y)²-2(48)=100 (x+y)²=196 x+y=14 ----- (3) x = 14-y ------ (4) Substitute (4) in (1) y(14-y)=48 y²-14y+48=0 (y-7)²=-48+49 y = 7±1 Therefore; y = 8 units or 6 units. It implies; For y = 8 units, x = 14-y x = 14-8 x = 6 units. Calculating r, radius of the inscribed circle. ½(6r)+½(8r)+½(10r)=½(8*6) 3r+4r+5r=24 12r = 24...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2762397

Calculating length OB. a² = 6²+2² a = √(40) a = 2√(10) units. a is AC. tanb = 6/2 b = atan(3)° √(40)² = 2√(50)²-2√(50)²cosc 40 = 100-100cosc cosc = 6/10 c = acos(3/5)° c is angle AOC. d = ½(180-c) d = ½(180-acos(3/5)) d = 63.4349488229° d is angle OAC and angle OCA. e = b-d e = atan(3)-63.4349488229 e = 8.1301023542° e is angle OCB. Therefore, the required length, ? (OB) is; Let it be f. f² = √(50)²+2²-2*2√(50)cos8.1301023542 f = 5.0990195136 units. Ag...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2760466

Shaded area is; Area sector with radius 10 units and angle acos(0.7) - Area triangle with height √(51) units and base 7 units - area sector with radius 3 units and angle 81.37307344° - area sector with radius10 units and angle 17.2538531174° + area triangle with two sides length 10 units, and angle 17.2538531174° = (acos(0.7)*1))π÷360) - ½(√(51)*7) - (9π*81.37307344)/360 -(100π*17.2538531174)/360 + ½(100)sin17.2538531174 = 39.7699415092 - 24.9949994999 - 6.39102624298 - 15.0568272777 +...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2755522

Sir Mike Ambrose is the author of the question. Let the side length of the square be 10 units. Area square is; 10² = 100 square units. Therefore; Area R is; Area semi circle with radius 5 units - Area sector with radius 5 units and angle (180-2atan(2))° - Area kite with diagonal lengths √(10²+5²) units and √(200-200cos(2atan(½))) units + Area sector with radius 10 units and angle 2atan(½)° - 2(area quarter circle with radius 5 units - area triangle with height and base 5 unit...

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Mathematics Question and Solution

Mathematics Question and Solution

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Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

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