Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2435543

Notice! R is the radius of the ascribed semi circle. Calculating R. a² = 2*√(2)² a² = 4 a = 2 units. a is the diagonal of the inscribed red and white square. √(2)² = 2b² b² = 2/2 b = 1 unit. It implies; R² = 2²+1² R² = 5 R = √(5) units. Again, R is the radius of the ascribed semi circle. tanc = 2/(√(5)+1) c = atan(2/(√(5)+1))° c = 31.7174744115° tan(atan(2/(√(5)+1))) = d/√(5) d = 1.3819660112 units. e = d-b e = 1.3819660112-1 e = 0.3819660112 units. Therefore, white area is; Area tria...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2435037

a² = 2*5² a = √(50) a = 5√(2) cm. tanb = 10/5 b = atan(2)° c = b-45 c = (atan(2)-45)° c = 18.4349488229° d = 90-b d = (90-atan(2))° d = 26.5650511771° e = 180-2d e = 180-2*26.5650511771 e = 126.8698976458° f² = 5²+5²-2*5*5cos126.8698976458 f = 8.94427191 cm. Therefore, Blue Area is; Are triangle with height 5√(2) cm and base 8.94427191sin18.4349488229 cm. = 0.5*5√(2)*8.94427191sin18.4349488229 = 10 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2434402

Calculating x. Notice! AD = 100 cm. Therefore; Area triangle ABD + Area triangle ACD = Area triangle ABC 0.5*2x*100sin60+0.5*x*100sin60 = 0.5*2x*xsin120 100x+50x = x² x = 100+50 x = 150 cm. It implies; AB = 2*x AB = 2*150 AB = 300 cm. AC = x AC = 150 cm. Therefore, calculating BC. Let it be y. y² = 300²+150²-2*300*150cos120 y² = 157500 y = √(157500) y = 150√(7) cm. y = 396.8626966597 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2433959

Let a be the radius of the inscribed circle. Calculating a. (1+a)² = a²+b² b² = (1+a)²-a² b² = 1+2a b = √(1+2a) cm. c = b-1 c = (√(1+2a)-1) cm. It implies; (2-a)² = a²+c² (2-a)² = a²+(√(1+2a)-1)² 4-4a+a² = a²+1+2a-2√(1+2a)+1 2-6a = -2√(1+2a) 6a-2 = 2√(1+2a) 3a-1 = √(1+2a) (3a-1)² = 1+2a 9a²-6a+1 = 1+2a 9a² = 8a 9a = 8 a = (8/9) cm. Again, a is the radius of the inscribed circle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2433075

a = 4+4+x a = (8+x) cm. a is the diameter of the bigger inscribed circle and also the side length of the square. b = ½(a) b = ½(8+x) cm. 6+y = b 6+y = ½(8+x) 12+2y = 8+x x = 2y+4 --- (1). c = y+b c = y+½(8+x) c = ½(2y+8+x) cm. c is the diameter of the smaller inscribed circle. d = ½(c) d = ¼(2y+8+x) cm. d is the radius of the smaller inscribed circle. e = d-y e = ¼(2y+8+x)-y e = ¼(8+x-2y) cm. Therefore; (¼(2y+8+x))² = (x/2)²+(¼(-2y+8+x))² (4y²+16y+2xy+16y+64...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2432122

a = (180(12-2))/12 a = 1800/12 a = 300/2 a = 150° a is the single interior angle of the regular dodecagon. b = 360-150-90 b = 360-240 b = 120° (6√(3))² = 2c²-2c²cos120 108 = 3c² c² = 36 c = 6 units. c is the side length of the regular dodecagon, the square and the regular hexagon. d² = 2*6² d² = 72 d = √(72) d = 6√(2) units. d is the diagonal of the square. e = ½(d) e = ½(6√(2)) e = 3√(2) units. f² = (3√(2))²+6²-2*3√(2)*6cos(105) f = 8.1961524227 units. g = 2(90-½(150)) g = 2(90-75) g =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2431654

Calculating x, side length of the square. a² = 2x² a = √(2)x units. a is BD, diagonal of the square. tanb = x/(0.5x) b = atan(2)° c = 2b c = 2atan(2)° c is angle DEF. d² = 2(0.5x)²-2(0.5x)²cos(2atan(2)) d = 0.894427191x units. d is DF. e = ½(180-2atan(2)) e = 26.5650511771° e is angle EDF. f = 90-45-26.5650511771 f = 18.4349488229° f is angle FDB. Therefore; x, the square side is; 0.5*0.894427191x*√(2)xsin18.4349488229 = 20 0.2x² = 20 (x²/5) = 20 x² = 100 x = √(100) x = 10 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2429576

a = 180-60-40 a = 80° (6/sin80) = (b/sin40) b = 3.916221868 units. (6/sin80) = (c/sin60) c = 5.2763114494 units. Calculating r, radius of the inscribed circle. 5.2763114494r+3.916221868r+6r = 3.916221868*6sin60 15.1925333174r = 20.3492857473 r = 1.3394267646 units. Calculating d, radius of the ascribed circle. e = ½(80-2*80) e = ½(20) e = 10° Therefore; cos10 = 3/d d = 3/cos10 d = 3.0462798357 units. tan40 = 1.3394267646/f f = 1.5962666587 units. Area Bl...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2429544

Let the two congruent blue semi circle's radius be a. 1 = πb² b² = 1/π b = √(1/π) units. b is the radius of the yellow circle. tanc = a/2a c = atan(½)° d = 2c d = 2atan(½)° sin(atan(½)).= √(1/π)/e e = 1.261566261 units. f = e+b f = 1.261566261+√(1/π) f = 1.8257558445 units. It implies (a+1.8257558445)² = (2a)²+a² a²+3.651511689a+3.3333844037 = 5a² 4a²-3.651511689a-3.3333844037 = 0 Resolving the above quadratic equation. a = 1.47707 units. Again, a is the radius of the two congruent...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2428467

Calculating r, radius of the inscribed circle. 4r+5r+3r = (4*3) 12r = 12 r = 1 unit. tana = 4/3 a = atan(4/3)° b = 180-a b = (180-atan(4/3))° c² = 1+1-2cos(180-atan(4/3)) c = 1.788854382 units. d = ½(180-(180-atan(4/3))) d = 26.5650511771° e² = 2(1)² e = √(2) units. f = 45+d f = 45+26.5650511771 f = 71.5650511771° Therefore, area of the red inscribed triangle is; 0.5*√(2)*1.788854382sin71.5650511771 = 1.2 square units. = (6/5) square units.

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