Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1896554

AP = √(20²+10²) AP = √(500) AP = 10√(5) cm. OA (radius of the semi circle) is: ½(AP) = 5√(5) cm. tana = 20/10 a = atan(2)° Where a is angle MAO = angle AMO. b = (180-2atan(2))° Where b is angle OMN = angle ONM. c = 180-2b c = 180-2(180-2atan(2)) c = 4atan(2)-180 c = 73.73979529169° Where c is angle MON. sin(0.5*73.73979529169) = d/(5√(5)) d = 3√(5) cm. Where d is half MN. MN = 2d MN = 6√(5) cm. It implies; Area Blue, Area Triangle DMN is; 0.5*10*6√(5)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1896448

a² = 5²-4² a = 3 units. b² = 3²+1² b = √(10) units. Where b is the square side length. Area inscribed square is; b² = √(10)² = 10 square units. Calculating x. c² = √(10)²-3² c = 1 unit. tand = (3/1) d = atan(3)° e = atan(1/3)° f = 5-1 f = 4 units. Therefore; x² = √(10)²+4²-8√(10)cos(atan⅓)) x² = 26-8√(10)(3/√(10)) x² = 26-24 x² = 2 x = √(2) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1896331

Hypotenuse of the ascribed right-angled triangle is; √(40²+h²) Therefore; 40h = 4h+(40*4)+4√(40²+h²) 40h = 4h+160+4√(1600+h²) 10h = h+40+√(1600+h²) (9h-40)² = 1600+h² 81h²-720h+1600 = 1600+h² 80h² = 720h 8h = 72 h = 9 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1895288

r² = (r-1)²+(r-2)² Where r is the radius of the circle. r² = r²-2r+1+r²-4r+4 r²-6r+5 = 0 r²-r-5r+5 = 0 r(r-1)-5(r-1) = 0 Therefore; r ≠ 1 unit. r = 5 units. r-1 5-1= 4 units. Where 4 units is half the side length of the square. It implies; Square side length is; 4*2 = 8 units. Area Red Square is; 8² = 64 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1895253

a =1 b = 0.5 Cross Multiply. b = 0.5a ----- (1). ½(a)+½(a)+½(1*0.5)+½(1-b) = 1 2a+0.5+1-b = 2 2a-b = 0.5 b = 2a-0.5 ---- (2). Equating (1) and (2). 2a-0.5 = 0.5a 1.5a = 0.5 a = (1/2)/(3/2) a = ⅓ units. b = 0.5a b = 0.5*(⅓) b = ⅙ units. c² = (⅓)²+(⅙)² c = 0.37267799625 units. d² = 0.37267799625²+1²-0.37267799625*2cos(atan2) d = 0.89752746786 units. e = √(0.5²+1²)-0.37267799625 e = 0.7453559925 units. f = 0.7453559925²+0.5²-2*0.5*0.7453559925cos(180-atan2) f = 1.06718737291 units. g =...

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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1894594

Let the two congruent angles be a. Let the base of the ascribed triangle be b. tana = 8/b b = 8/(tana) ----- (1). tan2a = 18/b b = 18/(tan2a) ----- (2). Equating (1) and (2). 8/(tana) = 18/(tan2a) And tan2a = 2tana/(1-tan²a) Therefore; 8/(tana) = 18/(2tana/(1-tan²a)) Let p be tana. 8/p = 18/(2p/(1-p²)) 8/p = 18(1-p²)/2p Cross Multiply. 16p = 18p(1-p²) 16p = 18p-18p³ 16 = 18-18p² 18p² =2 p² = ⅑ p = ⅓ And p = tana tana = ⅓ a = atan(⅓) a = 18.43494882292° Calculating b, base of the t...

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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1892805

Using 1 unit base. tan70 = 1/a a = 1/tan70 a = 0.36397023427 units. tan80 = 1/b b = 1/tan80 b = 0.17632698071 units. tan40 = c/a c = 0.36397023427tan40 c = 0.30540728934 units. Therefore, the required angle ? is; Let it be d. tand = (0.17632698071/0.30540728934) d = atan(0.17632698071/0.30540728934) d = 29.99999999959° d = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1892765

Let the radius of the ascribed quarter circle be a. a² = 2(a-1)² a² = 2(a²-2a+1) a² = 2a²-4a+2 a²-4a+2 = 0 (a-2)² = -2+4 a = 2±√(2) a ≠ 2-√(2) a = 2+√(2) units. a = 3.41421356237 units. Where a is AB = BC. BQ = AB-AQ BQ = 2+√(2)-2 BQ = √(2) units. Area Blue, Area Triangle CBQ is; ½*√(2)*(2+√(2)) = ½(2√(2)+2) = (√(2)+1) square units. = 2.41421356237 square units.

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