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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1961055

The radius of the inscribed circle is 2 units. a = 2atan(⅓)° b = 2atan(⅕)° c = 180-a-b c = (180-2atan(⅓)-2atan(⅕))° sin(2atan(⅓)) = d/16 d = (48/5) units. cos(2atan(⅓)) = e/16 e = (64/5) units. f = (90-2atan(⅓)-2atan(⅕))° tan(90-2atan(⅓)-2atan(⅕)) = g/(48/5) g = (198/35) units. Therefore; Area Blue exactly in square units is; ½*(48/5)(198/35) = 4752/175 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1961051

8 = 4 a = (8-a) Cross multiply. 64-8a = 4a 64 = 12a a = ⅓(16) cm. b = 8-a b = ⅓(8) cm. c = atan(⅔)° d = atan(3/2)° ⅔ = 8/e e = 12 cm ⅔ = f/8 f = ⅓(16) cm. g = 8-f g = ⅓(8) cm. Area Orange exactly in cm² is; Area square units side 8 cm - Area rectangle with length 4 cm and width ⅓(8) cm. = 8² - 4*⅓(8) = 64 - ⅓(32) = ⅓(192-32) = ⅓(160) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1960663

Let AB be 1 unit. a = 180-24-45 a = 111° a is angle AED. (1/sin111) = (b/sin24) b = 0.43567391896 units. b is ED. c = √(2) units. c is BD. d = √(2)-0.43567391896 d = 0.97853964341 units. d is BE. (CE)² = 0.97853964341²+1-2*0.97853964341cos45 CE = 0.75741388863 units. (0.75741388863/sin45) = (0.97853964341/sine) e = 66° e is angle ECB. f = 90-66 f = 24° f is angle DCE. g = 90+24 g = 114° g is angle ECF. Therefore, the required angle x is; 180-114-24 = 42°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1960629

Let the square side be 2 units. tan30 = a/2 a = 2/√(3) units. a is BK. b = 2-a b = (2-(2/√(3))) units. b is CK. tan30 = (2-(2/√(3)))/c c = (2√(3)-2) units. c is CM. d = 2-c d = 2-(2√(3)-2) d = (4-2√(3)) units. d is DM. Therefore, the required angle x is; tanx = ½(4-2√(3)) x = atan(½(4-2√(3))) x = 15°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1960597

Let AB = CD = 2 units. Let the big inscribed square side length be a. Considering similar triangle ratios, calculating a. (a-2) = (2+a) 2 = a Cross Multiply. a²-2a = 4+2a a²-4a-4 = 0 (a-2)² = 4+4 a = 2±2√(2) a ≠ 2-2√(2) units. a = 2+2√(2) units. Again, a is the big inscribed square side length. BC = a-AB BC = 2+2√(2)-2 BC = 2√(2) units. Notice; Area Blue is; 2² = 4 square units. Area red is; ½*2*2√(2) = 2√(2) units. It implies; Area Red ÷ Area Blue is; 2√(2)÷4 = ½√(2)

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1960512

Let the square side be 2 units. a = atan(2) b = (180-atan(2))° (2/sin(180-atan(2))) = (1/sinc) c = 26.56505117708° d = 180-(180-atan(2))-26.56505117708 d = 36.86989764584° e = 90-2d e = 90-2*36.86989764584 e = 16.26020470832° Therefore, calculating b. b² = 8-8cos16.26020470832 b = 0.56568542495 units. Calculating a. a² = 8-8cos36.86989764584 a = 1.26491106407 units. Therefore; a : b is; 1.26491106407 : 0.56568542495 = 1.26491106407÷0.56568542495 = 2.2360679775 = √(5)

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1958696

Calculating green area. a² = 5²+5² a = √(50) a = 5√(2) units. b = ½(a) b = ½(5√(2)) units. b = 3.53553390593 units. 5² = (5-c)²+(½(5√(2)))² 25 = 25-10c+c²+¼(50) c²-10c+¼(50) = 0 4c²-40c+50 = 0 Notice; c is the radius of the green circle. c = 1.46447 units. Therefore; Green area is; π(1.46447)² = 6.73768699619 square units. Calculating blue area. d = 5-1.46447 d = 3.53553 units. e = 45° f = ½(e) f = 22.5° tan22.5 = g/5 g = 2.07106781187 units. Ar...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1958336

a² = 10²+16²-2*10*16cos120 a = 2√(129) units. Where a is length AC. (3√(129)/sin120) = (16/sinb) b = 37.58908946897° c = 90-b c = 52.41091053103° d = 60-b d = 22.41091053103° e = 90-d e = 67.58908946897° f = 90-60 f = 30° cos37.58908946897 = g/10 g = 7.92405815693 units. i = 30+37.58908946897 i = 67.58908946897° tan67.58908946897 = j/7.92405815693 j = 19.2148198619 units. Where j is the radius of the arc. sin67.58908946897 = 19.2148198619/k k = 20.78460969082 units. l² = (2√(129)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1958330

Let the radius of the inscribed half circle be 1 unit. Area rectangle is; 2*1 = 2 square units. Calculating Shaded Area as a single fraction.. It is; ½(1*2) - 1² + ¼(1)²π - ((180-2atan(½))π*1²/360) + (0.5*1²sin(180-2atan(½))) = 1-1+¼(π)-(π(180-2atan(½))/360)+⅖ = (π/4)+⅖-(π(180-2atan(½))/360) = ((5π+8)/20)-(π(180-2atan(½))/360) = (90π+144-180π+2πatan(½))/360 = (144-90π+2πatan(½))/360 square units. Therefore; Area Shaded ÷ Area Rectangle as a single fraction is; ((144-90π+2πatan(½))...

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