Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3201140

Calculating the circumference of the blue inscribed circle. Let x be the base of the inscribed right-angled triangle with height 10 units. x²+10² = a² a = √(x²+100) units. b = ½(a) b = ½√(x²+100) units. Calculating x. 10² = 10²+(½√(x²+100))²-10√(x²+100)cosy 10√(x²+100)cosy = ¼(x²+100) --- (1). cosy = x/√(x²+100) --- (2). Substituting (2) in (1). 10√(x²+100)*x/√(x²+100) = ¼(x²+100) 10x = ¼(x²+100) 40x = x²+100 x²-40x+100 = 0 (x-20)² = -100+(-20)² (x-20)² = 300 x = 20±√(300) x =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3199587

Let b = 1 unit. Therefore; Side length of the ascribed regular triangle c is; c = 4b c = 4(1) c = 4 units. Therefore d, radius of the inscribed circle is; 3(4d) = 4√(4²-2²) 12d = 4√(12) 3d = 2√(3) d = ⅓(2√(3)) units. d = 1.15470053838 units. Calculating a. e²+1² = (⅓(2√(3)))² e² = ⅓(4)-1 e = √(⅓) e = ⅓√(3) units. e = 0.57735026919 units. It implies; (⅓(2√(3)))² = (⅓√(3))²+a²-2*(⅓√(3))acos30 ⅓(4) = ⅓+a²-(a) 4 = 1+3a²-3a 3a²-3a-3 = 0 a²-a-1 = 0 (a-½)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3199574

Calculating Total Shaded Area, Area Grey, Purple and Blue. Notice. KLMN is a square. Calculating area grey. a = 6-4 a = 2 units. sinb = 2/6 b = asin(⅓) b = 19.4712206345° c²+2² = 6² c = 4√(2) units. Area Grey is; (6*2)-½(4√(2)*2)-(19.4712206345π*6²/360) = 0.22608138033 square units. Calculating area purple. d = (90-b) d = 70.5287793655° e = ½(d) e = 35.2643896828° f = e+b f = 35.2643896828+19.4712206345 f = 54.7356103173° g = ½(180-f) g = 62.6321948414° h = 90-g h = 27.3678051587...

Mathematics Question and Solution

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Calculating Alpha, let it be x. a = 180-24-54 a = 180-78 a = 102° Let the side length opposite/facing angle 24° (the shortest length) of the inscribed triangle with angles 24°, 54° and 102° be 1 unit. It implies; (1/sin24) = (b/sin102) b = 2.40486717237 units. (1/sin24) = (c/sin54) c = 1.98904379074 units. Therefore; d² = 1.98904379074²+2.40486717237²-2*1.98904379074*2.40486717237cos(24+24) d = 1.82709091528 units. (1.82709091528/sin(24+24)) = (2.40486717237/sine)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3198062

Calculating Shaded Area Blue. Notice. 8 cm is the side length of the ascribed square. Therefore; ½(8) = 4 cm is the radius of the both inscribed half circles. tana = 4/8 a = atan(½)° b = 2a b = 2atan(½)° tanc = 8/4 c = atan(2)° d = 2c d = 2atan(2)° Area Blue is; Area sector with radius 8 cm and angle 2atan(½)°-Area triangle with height 8 cm and base 8sin(2atan(½)) cm+Area sector with radius 4 cm and angle 2atan(2)°-Area triangle with height 4 cm and base 4sin...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3198060

Calculating Area Red Circle. Let a be the radius of the inscribed red circle. b = (4+a) m. c = (21+a) m. Therefore; Calculating a. a(4+a)+a(21+a)+a(4+21) = (4+a)(21+a) 4a+a²+21a+a²+25a = 84+25a+a² 50a+2a² = 84+25a+a² a²+25a-84 = 0 a²+28a-3a-84 = 0 a(a+28)-3(a+28) = 0 (a-3)(a+28) = 0 It implies; a ≠ -28 a = 3 m. Therefore, area red inscribed circle is; πa² = π(3)² = 9π m²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3197241

Calculating Area Ascribed Rectangle. Let x be the width of the ascribed rectangle. Let y be the length of the ascribed rectangle. a = 1+2+1 a = 4 units. b = (x-2) units. c = (y-2) units. 4² = (x-2)²+(y-2)² 16 = x²-4x+4+y²-4y+4 8 = x²+y²-4x-4y x²+y² = 8+4(x+y) --- (1). d²+1² = 2² d = √(3) units. e = (x-1) units. f = 2d+2e f = (2x+2√(3)-2) units. f = (2x+1.46410161514) units. f is the diagonal of the ascribed rectangle. It implies; f² = x²+y² (2x+1.464...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3196588

Calculating The Required Angle. Let it be x. a = ½(12)+b a = (6+b) cm. b is the radius of the inscribed sector. Calculating b. c² = 6²+12² c² = 36+144 c = √(180) c = 6√(5) cm. It implies; a = c 6+b = 6√(5) b = (6√(5)-6) cm. b = 7.416407865 cm. Again, b is the radius of the inscribed sector. tand = 12/6 d = 63.4349488229° e = 180-d e = 116.565051177° f = ½(d) f = 31.7174744115° cos31.7174744115 = g/12 g = 10.2078097002 cm. 7.416407865² = 12²+12²-2*...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3196265

Calculating Shaded Fraction. Let the radius of the inscribed circle be 1 unit. Therefore, the radius of the inscribed half circle is 2 units. a² = 2²+(2+2)² a² = 20 a = 2√(5) units. a is the radius of the ascribed quarter circle. Shaded Area is; ½*π(2)²+π(1)² = 2π+π = 3π square units. Area ascribed quarter circle is; ¼*π(2√(5))² = ¼(20π) = 5π square units. Therefore, shaded fraction area is; Shaded Area ÷ Area ascribed quarter circle = 3π÷5π = ⅗

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3196253

Calculating Inscribed Blue Rectangle Area. a = 15+5 a = 20 units. Let b be the length of the blue inscribed rectangle. b² = 2(20)²-2(20)²cos(90-x) b² = 800-800(cos90cosx+sin90sinx) b² = 800-800sinx --- (1). b² = 2(15)²-2(15)²cos(90+x) b² = 450-450(cos90cosx-sin90sinx) b² = 450-450(-sinx) b² = 450+450sinx --- (2). Equating (1) and (2). 800-800sinx = 450+450sinx 350 = 1250sinx sinx = 7/25 x = 16.2602047083° It implies; c² = 15²+20²-2*15*20cos16.2602047083...

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