Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3116838

Calculating Length x (AD). Let a be the diameter of the green inscribed circle. Therefore; x ~ a (b+x) ~ 2a Cross Multiply. 2ax = ab+ax 2ax-ax = ab ax = ab x = b units. Therefore, length x is; x² = 9*4 x = √(36) x = 6 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3116679

Let OF be r, radius of the ascribed circle. a²+r² = (2+3)² a = √(25-r²) units. b = a+r b = (r+√(25-r²)) units. c = r-a c = (r-√(25-r²)) units. Therefore; 4*(2+3) = (r-√(25-r²))(r+√(25-r²)) 20 = r²-(25-r²) 20 = 2r²-25 2r² = 45 r² = (45/2) r = √(45/2) units. Recall. b = (r+√(25-r²)) units. And r = √(45/2) units. b = √(45/2)+√(25-√(45/2)²) b = (√(45)/√(2))+√(25-(45/2)) b = ½√(90)+√(½(5)) b = ½√(90)+½√(10) b = ½(3√(10))+½√(10) b = ½(4√(10)) b = 2√(10) unit...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3116410

Total area of the ascribed composite plane shape is; (¼*12*12π) + (½*6*6π) = 36π+18π = 54π cm² Calculating the inscribed shaded area. Let re be the radius of the inscribed shaded half circle. a = (12-x) cm. Therefore; (12-x)² = r²+x² 144-24x+x² = r²+x² r² = (144-24x) cm. It implies; 12² = (12-x)²+r² Notice. r² = (144-24x) cm. 144 = 144-24x+x²+(144-24x) 0 = -24x+x²+144-24x x²-48x+144 = 0 (x-24)² = -144+(-24)² (x-24)² = 576-144 x = 24±√(432) Therefore; x...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3116402

a² = 3²+4² a = √(25) a = 5 units. a is the hypotenuse of the ascribed right-angled triangle. tanb = 3/4 b = atan(3/4)° tanc = 4/3 c = atan(4/3)° Let d be the radius of the inscribed circle. Calculating d. 3d+4d+5d = 3*4 12d = 12 d = 1 unit. e = 2d e = 2 units. e is the diameter of the inscribed circle. f = 3-1-x f = (2-x) units. g = (1+x) units. Calculating x. 4 ~ (1+x) 5 ~ (2-x) Cross Multiply. 4(2-x) = 5(x+1) 8-4x = 5x+5 9x = 3 x = ⅓ units....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3116399

Let BH be a. It implies; 25 ~ a a ~ 4 Cross Multiply. a² = 100 a = 10 units. b² = 25²+10² b = √(725) units. b is AB, the side length of the ascribed square. c² = 10²+4² c = √(116) units. d = (10+x) units. Calculating x √(116) ~ (10+x) 10 ~ √(725) Cross Multiply. 290 = 100+10x 10x = 190 x = 19 units. Or. cose = 4/√(116) e = acos(10/√(116) Therefore; cose = √(725)/f Where f = (x+10) units. cos(acos(10/√(116))) = √(725)/f 10/√(116) =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3116021

Notice. The angle formed at the meet point (vertex) of length 4 units and x units is equal the angle formed at the meet point (vertex) 9 units and x units. Therefore, as a result, two similar plane shapes, quadrilaterals are observed. Calculating x. x ~ 9 4 ~ x Cross Multiply. x² = (9*4) x = √(36) x = 6 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3115944

Sir Mike Ambrose is the author of the question. Let the square side be 1 unit. Therefore; Area B is; Area triangle with height 1 unit and base sin(53.13010235416) units - Area triangle with height 1 unit and base (0.71428571429sin(53.13010235416)) units. = ½*1(sin(53.13010235416)) - ½*1*(0.71428571429sin(53.13010235416)) = 0.4 - 0.28571428572 = 0.11428571428 square units. Area A is; Area triangle with height 1 unit and base 0.5 units + Area triangle with height 0....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3115937

Calculating Area Blue. (2/sin120) = (1/sina) a = 25.6589062733° b = 180-120-a b = 60-25.6589062733 b = 34.3410937267° Therefore, shaded blue area is; Area sector with radius 2 units and angle 34.3410937267° - Area triangle with height 2 units and base sin34.3410937267 units. = (34.3410937267π*2*2/360)-(½*2sin34.3410937267) = 1.19873030853-0.56411839885 = 0.63461190968 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3115931

a = ½(4+6) a = 5 units. a is radius. 10² = x²+b² b = √(100-x²) units. c = 10-½(4) c = 8 units. x² = c²+d² x² = 8²+d² d = √(x²-64) units. Calculating x. Therefore; 8 ~ √(x²-64) √(x²-64) ~ 2 Cross Multiply. √(x²-64)² = 16 x²-64 = 16 x² = 80 x = √(80) x = 4√(5) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3115438

Notice. r, radius of the ascribed quarter circle is (25/√(2)) units. r = ½(25√(2)) units. (AD)² = 2(½(25√(2)))² (AD)² = 2*¼*625*2 AD = √(625) AD = 25 units. It implies; AB+BC+CD = AD 10+BC+4 = 25 BC = 25-14 BC = 11 units. a = 25-2(4) a = 17 units. b = ½(a) b = ½(17) units. sinc = ½(17)/(½(25√(2))) c = asin(17√(2)/50)° c = 28.7397952917° d = 2c d = 2asin(17√(2)/50)° e = ½(90-d) e = (45-asin(17√(2)/50))° f = ½(180-e) f = ½(180-45+asin(17√(2)/50)) f = (0.5(135+asin(17√(2)/50)))° g = f...

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