Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2486193

Let a be the radius of the quarter circle. It implies; a² = a²+4²-2*a*4cosb 8cosb = 16 acosb = 2 cosb = 2/a --- (1). 8² = a²+4²+2*a*4cosb 64 = a²+16+8acosb 48 = a²+8acosb --- (2). Substituting (1) in (2) to get a, radius of the quarter circle. 48 = a²+8a(2/a) 48 = a²+16 a² = 48-16 a² = 32 a = √(32) a = 4√(2) units. Again, a is the radius of the quarter circle. Therefore, area blue (area quarter circle) is; ¼(4√(2))²π = ¼*4*4*2π = 8π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2485266

Let AB = AC = 1 unit. It implies; a² = 2-2cos(24+12) a = 0.6180339887 units. a is BC = AD. b = ½(180-24-12) b = ½(144) b = 72° b is angle ACB = angle ABC. c² = 0.6180339887²+1-2*0.6180339887cos12 c = 0.4158233817 units. c is CD. (0.4158233817/sin12) = (0.6180339887/sind) d = 18° d is angle ACD. f = b-d f = 72-18 f = 54° f is angle BCD . g² = 0.4158233817²+0.6180339887²-2*0.4158233817*0.6180339887cos54 g = 0.5027541398 units. g is BD. Therefore, the requir...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2485183

Notice! The regular polygon has 12 sides (regular dodecagon). a = (180(12-2))/12 a = 1800/12 a = 150° a is the single interior angle of the regular dodecagon. b = ½(a) b = 75° c = 90-b c = 15° sin15 = 2/d d = 7.7274066103 cm. tan15 = e/d e = 7.7274066103tan15 e = 2.0705523608 cm. f = 2e f = 4.1411047216 cm. f is the side length of the regular dodecagon. cos15 = d/g g = 7.7274066103/cos15 g = 8 cm. It implies; Painted Area is; Area triangle with height...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2484823

30² = 18²+a² a² = 900-324 a = √(576) a = 24 units. b = ½(a) b = 12 units. Therefore, the required length, x is; x² = 18²+b² x² = 18²+12² x² = 324+144 x = √(468) x = 6√(13) units. x = 21.6333076528 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2484573

tana = 7/4 a = atan(7/4)° b = 180-a b = (180-atan(7/4))° c² = 7²+4² c = √(49+16) c = √(65) cm. It implies, area blue triangle is; Area triangle with height √(65) cm and base 4sin(180-atan(7/4)) cm. = ½*√(65)*4sin(180-atan(7/4)) = 2√(65)sin(180-atan(7/4)) = 2√(65)*(7/√(65) = 2*7 = 14 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2483773

Let AB be x cm. BC = 3x cm. AC = AB+BC AC = 4x cm. sin60 = a/16 a = 8√(3) cm. a is AC. Therefore; 4x = 8√(3) x = 2√(3) cm. Again, x is AB. BC = 6√(3) cm. CE = CD = 8 cm. tanb = 8/(6√(3)) b = atan(4/(3√(3)))° tanb = c/x 4/(3√(3)) = c/(2√(3)) c = ⅓(8) cm. d = 180-60-atan(3√(3)/4) d = 67.589089469° ((8/3)/sin67.589089469) = (e/sin60) e = 2.4980718188 cm. Area red is; ½*(8/3)*2.4980718188sin(atan(3√(3)/4) = 2.6393155163 cm² Area green is; ½*(8...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2483309

tan30 = 8/a a = 8√(3) cm. b = ½(a) b = 4√(3) cm. Area Blue is; ½*8√(3)*10-½*10*4√(3) = 40√(3)-20√(3) = 20√(3) cm² Area triangle ABC is; ½*10*8√(3)+½*8√(3)*8 = 40√(3)+32√(3) = 72√(3) cm² Therefore; Area Blue ÷ Area ABC is; 20√(3)/72√(3) = 5/18

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2482706

a = (10-x) cm. 10 cm is the radius of the ascribed semi circle. 12² = a²+b² b² = 144-(100-20x+x²) b² = 44+20x-x² b = √(44+20x-x²) cm. 10² = x²+b² 100 = x²+44+20x-x² 56 = 20x 14 = 5x x = (14/5) cm. x = 2.8 cm. Let r be the radius of the inscribed orange circle. c = r-x c = (r-(14/5)) cm. d = (10-r) cm. It implies; d² = r²+c² (10-r)² = r²+(r-(14/5))² 100-20r+r² = r²+r²-(28/5)r+(196/25) 100-20r = r²-(28/5)r+(196/25) 2500-500r = 25r²-140r+196 25r²+360r-2304 = 0...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2482631

Let a be the radius of the inscribed orange circle. (3+a)² = a²+b² b² = 9+6a+a²-a² b = √(9+6a) cm. c = c-3 c = (√(9+6a)-3) cm. It implies; (6-a)² = a²+c² (6-a)² = a²+(√(9+6a)-3)² 36-12a+a² = a²+9+6a-6√(9+6a)+9 36-12a = 18+6a-6√(9+6a) 6√(9+6a) = 18a-18 √(9+6a) = 3a-3 9+6a = 9a²-18a+9 0 = 9a²-24a 9a = 24 3a = 8 a = (8/3) cm. Again, a is the radius of the inscribed orange circle. Area inscribed orange circle is; πa² = π(8/3)² = ⅑(64)π cm² = 22.3402144255 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2482618

a² = 4²+3² a = √(25) a = 5 units. a is the radius of the semi circle. tanb = (3/4) b = atan(3/4)° c = 90-b c = (atan(4/3))° tan(atan(3/4)) = d/5 (3/4) = d/5 d = 3.75 units. d = ¼(15) units. tan(atan(4/3)) = e/5 (4/3) = e/5 e = ⅓(20) units. Area 1 is; Area triangle with height 5 units and base ¼(15) units - Area sector with radius 5 units and angle atan(3/4)° = ½*5*¼(15)-(atan(3/4)π*25/360) = ⅛(75)-(25atan(3/4)π/360) = 25(135-atan(3/4)π)/360 square units. Area 2 is; Area triangle with...

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