Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3127317

cosa = (2√(5))/(3√(5)) a = acos(⅔)° a is angle PRT. tana = 25/b b = 25/(tan(acos(⅔))) b = 22.360679775 units. b = 10√(5) units. Therefore, area triangle TRP is; ½*25b = ½*25*10√(5) = 125√(5) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3127294

Let the radius of the ascribed half circle be 1 unit. a² = 2x² a = √(2)x units. b²+1² = a² b = √(2x²-1) units. c = 1-b c = (1-√(2x²-1)) units. d = ½(c) d = ½(1-√(2x²-1)) units. e²+x² = 2² e = √(4-x²) units. It implies; x ~ 2 ½(1-√(2x²-1)) ~ x Cross Multiply. (1-√(2x²-1)) = x² 1-x² = √(2x²-1) (1-x²)² = 2x²-1 1-2x²+x⁴ = 2x²-1 x⁴-4x²+2 = 0 Let x² = p. p²-4p = -2 (p-2)² = -2+(2)² (p-2)² = 2 p = 2±√(2) It implies; p = (2+√(2)) units. Or p = (2-√(2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3126622

Let a be the radius of the circle. b² = 2(2)² b = √(8) b = 2√(2) cm. c = (2√(2)-a) cm. It implies; 2a² = c² 2a² = (2√(2)-a)² 2a² = 8-4√(2)a+a² a²+4√(2)a-8 = 0 (a+2√(2))² = 8+(2√(2))² (a+2√(2))² = 16 a = -2√(2)±√(16) a = -2√(2)±4 Therefore; a = (4-2√(2)) cm. Area inscribed circle is; πa² = π(4-2√(2))² = π(16-16√(2)+8) = 8π(3-2√(2)) cm² = 4.31209667562 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3126411

Let the bigger inscribed semi circle radius be 2 units. Let the big inscribed semi circle radius be r. Let the inscribed circle radius be q. Calculating r. (2+r)² = 2²+(4-r)² 4+4r+r² = 4+16-8r+r² 4r = 16-8r 12r=16 3r=4 r=4/3 unit. Calculating q. (2+(4/3))²=(2+q)²+((4/3)+q)² (10/3)²=4+4q+q²+(16/9)+(8q/3)+q² (100/9)-(16/9)-4=2q²+4q+(8q/3) (84/9)-4=2q²+4q+(8q/3) 84-36=18q²+36q+24q 48=18q²+60q 6q²+20q-16=0 3q²+10q-8=0 Therefore; q≠-4...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3126404

Let a be the radius of the circle. 2πa = 14π a = 7 cm. Calculating x. cos49 = b/7 b = 7cos49 b = 4.59241320293 cm. c = 2b c = 9.18482640587 cm. c is CD d = 180-49-37.5 d = 93.5° (9.18482640587/sin93.5) = (e/sin37.5) e = 5.60181656513 cm. Notice. e = AB. f = ½(AB) f = 2.80090828257 cm. Therefore, angle x is; cosx = 2.80090828257/7 x = acos(2.80090828257/7) x = 66.4137096823° Calculating AD. g = 49-37.5 g = 11.5° h = 180-2(11.5) h = 157°...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3125852

Calculating Total Shaded Area, Area Grey, Purple and Blue. Notice. KLMN is a square. Calculating area grey. a = 6-4 a = 2 units. sinb = 2/6 b = asin(⅓) b = 19.4712206345° c²+2² = 6² c = 4√(2) units. Area Grey is; (6*2)-½(4√(2)*2)-(19.4712206345π*6²/360) = 0.22608138033 square units. Calculating area purple. d = (90-b) d = 70.5287793655° e = ½(d) e = 35.2643896828° f = e+b f = 35.2643896828+19.4712206345 f = 54.7356103173° g = ½(180-f) g = 62.6321948414° h = 90-g h = 27.3678051587...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3125739

Sir Mike Ambrose is the author of the question. tan30 = a/3 a = √(3) units. b = a+3 b = (3+√(3)) units. tan30 = c/b ⅓√(3) = c/(3+√(3)) c = ⅓(3√(3)+3) units. c = (1+√(3)) units. It implies; (1+√(3)) ~ 3 d ~ √(3) Cross Multiply. 3d =(√(3)+3) d = ⅓(√(3)+3) units. Let e be the side length of the smallest inscribed square. f = d-e f = (⅓(√(3)+3)-e) units. Calculating e. e ~ 3 (⅓(√(3)+3)-e) ~ √(3) Cross Multiply. √(3)e = √(3)+3-3e (3+√(3))e = (3+√(3))...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3125595

Let a be the blue length. b² = 4²+3² b = √(25) b = 5 units. Let alpha be c. cos(2c) = 4/5 cos(c+c) =4/5 (cosc)²-(sinc)² = 4/5 --- (1). cosc = 3/a --- (2). a² = 3²+d² d = √(a²-9) units. sinc = d/a sinc = √(a²-9)/a --- (3). Substituting (2) and (3) in (1). (3/a)²-(√(a²-9)/a)² = 4/5 (9/a²)-((a²-9)/a²) = 4/5 9-(a²-9) = ⅕(4a²) 18-a² = ⅕(4a²) 90-5a² = 4a² 9a² = 90 a² = 10 a = √(10) units. a = 3.16227766017 units. Again, a is the required blue len...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3125501

Let a be the radius of the circle. b = 7-3 b = 4 cm. c = (7-a) cm. d = (3-a) cm. e = c+d e = 7-a+3-a e = (10-2a) cm. Calculating a. e² = (2a)²+4² (10-2a)² = 4a²+16 100-40a+4a² = 4a²+16 100-40a = 16 40a = 84 10a = 21 a = 2.1 cm. Again, a is the radius of the circle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3125497

Notice. Triangle ACD and Triangle BCD are right-angled triangles. Let AD = 2 units. AB = 6 units. It implies; BD = 6-2 BD = 4 units. a² = 2(0.5*4)² a = 2√(2) units. a is BE = CE. tanb = (2/4) b = atan(½)° b is angle DAF. Therefore; DF = 1 units. Area triangle AFD is; ½(AD)*(DF) = ½*2*1 = 1 square units. Calculating area triangle CFE. It implies; Area trapezoid with parallel lengths 2 unit and 4 units respectively and height 2 units - Area trapezo...

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