Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1900719

Let the square side be 3 cm. Therefore; Observing similar triangle ratios. 3 = 2 1 = a Cross Multiply. a = ⅔ units. b = √(3²+1²) b = √(10) cm. c = √(2²+(⅔)²) c = ⅓(2√(10)) cm. d = √(√(10)²+(⅓(2√(10)))²) d = ⅓√(130) cm. e =½(d) e = ⅙√(130) cm. tanf = (√(10)/(2√(10)/3)) f = 56.30993247402° g = 180-56.30993247402-atan(2/(2/3)) g = 52.1250163489° h = 90-52.1250163489 h = 37.8749836511° j² = (⅙√(130))²+(⅓(2√(10)))²-2(⅙√(130))(½(2√(10)))cos56.30993247402 j = 1.90029237517 cm. (1.900292375...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1900311

Let the side of the regular hexagon be a. Calculating a b = 180-60-45 b = 75° (c/sin75) = (2/sin45) c = (1+√(3)) units. d² = 2a²-2a²cos120 d = √(3a²) d = √(3)a ----- (1). e = c+a, and c = (1+√(3)) units. e = (1+√(3)+a) --- (2). Therefore equating (1) and (2). √(3)a = 1+√(3)+a √(3)a-a = √(3)+1 a(√(3)-1) = √(3)+1 a = (√(3+1)/(√(3)-1) a = (√(3)+2) units. It implies; The required length is; a-2 Where a is the side length of the regular hexagon, (√(3)+2) units. = √(3)+2-2 = √(3) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1900162

Let DE = FB = a. Let EG = GH = HF = b. (3b)² = 12²+(12-2a)² b² = ⅑(288-48a+4a²) ---- (1). c = √(144+a²) Where c is CF. d = √((12-a)²-b²) Where d is GC. √(144+a²)² = (2b)²+√((12-a)²-b²)² 144+a² = 4b²+144-24a+a²-b² 3b² = 24a b² = 8a ---- (2) Equating (1) and (2). 8a = ⅑(288-48a+4a²) 72a = 288-48a+4a² 4a²-120a+288 = 0 a²-30a+72 = 0 (a-15)² = 225-72 a = 15±√(153) a ≠ 15+3√(17) a = 15-3√(17) cm. a = 2.63068312315 cm. Calculating b. b² = 8a, and a = 2.63068312315 cm. b² = 8*2.63068312315 b =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1898773

Let the radius of the ascribed quarter circle be 2 units. Let r be the radius of the inscribed circle. Calculating r. (2-r)² = r²+1² 4-4r+r² = r²+1 3 = 4r r = ¼(3) units. r = 0.75 units. a = 2-r a = 2-0.75 a = 1.25 units b² = 1²+1.25² b = ¼√(41) units. b = 1.60078105936 units. c² = (√(41)/4)²-(0.75)² c = √(2) units. Let alpha be x. Calculating tanx. tanx = (opposite/adjacent) tanx = (r/√(2)) tanx = (¾)/√(2) tanx = ⅛(3√(2)) It implies; x (alpha) = 27.9383527296°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1898721

Let AC = 1 unit. (1/sin160) = (CD/sin5) CD = 0.25482634415 units. (BC/sin30) = (1/sin80) BC = 0.50771330594 units. (BD)² = 0.50771330594²+0.25482634415²-2*0.25482634415*0.50771330594cos55 BD = 0.41748304149 units. Calculating angle CBD . Let it be a. (0.25482634415/sina) = (0.41748304149/sin55) a = 30° Therefore angle x is; Angle ABC - a = 80-30 = 50°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1898682

Let the inscribed square side be 1 unit. a² = 1²+1² a = √(2) units. Where a is a diagonal of the inscribed square. It implies; a = r = √(2) units. Where r is the radius of the ascribed semi circle. Calculating angle x. b = 180-45-60 b = 180-105 b = 75° Therefore; x = ½(180-75) x = ½(105) x = 52.5°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1898641

Let the square side be r. Therefore r is the length of the red rectangle and also the radius of the inscribed quarter circle. Let the base of the red area be a. b = √(16-a²) Therefore; r² = (r-a)²+b² r² = r²-2ar+a²+16-a² 2ar = 16 ar = 8 Therefore red area/area red rectangle is ar. = 8 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1897185

EF = 2r. Where r is the radius of the circle. r = √((21/2)²+6²) = ½(3√(65)) units. Therefore EF = 2r = 2*½(3√(65)) = 3√(65) units. EF = 24.1867732449 units.

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.

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