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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2584013

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a² = 4²+1² a = √(17) units. b = a-1 b = (√(17)-1) units. b = 3.1231056256 units. c² = (√(17)-1)²+3² c² = 17-2√(17)+1+9 c = √(27-2√(17)) units. c = 4.3305644838 units. tand = (√(17)-1)/3 d = 46.1517819676° tane = 1/4 e = 14.0362434679° f² = 3.1231056256²+4²-2*3.1231056256*4cos14.0362434679 f = 1.2308244188 units. (1.2308244188/sin14.0362434679) = (4/sing) g = 52.0181217332° h = 90-g h = 37.9818782668° j = 180-h-d j = 95.8663397656° (3/sin95.8663397656) =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2582274

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a = 10+10 a = 20 cm. a is twice the radius of the quarter circle. b = ½(10) b = 5 cm. c² = 5²+10² c √(125) c = 5√(5) cm. Observing similar plane shape (right-angled triangle) side length ratios. 20 - 5√(5) d - 5 Cross Multiply. 5√(5)d = 100 d = 20/√(5) d = 4√(5) cm. e² = 20²-(4√(5))² e² = 400-80 e = √(320) e = 8√(5) cm. f = e-c f = 8√(5)-5√(5) f = 3√(5) cm. Therefore, area blue is; Area triangle with height 4√(5) cm and base 3√(5) cm + Area triangl...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2582235

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Calculating area of the blue a. b² = 2a² b = √(2)a cm. b is the diagonal of the blue square. c = 7-5 c = 2 cm. d = 7+5 d = 12 units. It implies; (√(2)a)² = 2²+12² 2a² = 148 a² =74 cm² a² is the area of the blue square.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2582086

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a = 2+2 a = 4 units. a is the radius of the quarter circle. b² = 4²-2² b² = 16-4 b = √(12) b = 2√(3) units. Therefore; Area triangle green is; ½*2√(3)*2 = 2√(3) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2582073

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Let the radius of the circle be a. b = (3+a) units. c = (8+a) units. d² = (3+a)²-a² d = √(9+6a) units. Observing similar plane shape (right-angled) side length ratios. a - √(9+6a) (8+a) - (3+a) Cross Multiply. a(3+a) = (8+a)√(9+6a) It implies; a = 12 units. Again, a is the radius of the circle. Diameter is; 2a = 2*12 = 24 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2582056

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Diameter of the ascribed circle is; 2R = 2*3 = 6 units. a²+2² = 6² a = √(32) a = 4√(2) units. Observing similar plane shape (right-angled) side length ratios. 3 - 4√(2) b - 2 Cross Multiply. 4√(2)b = 6 b = 3/(2√(2)) b = ¾√(2) units. b = 1.0606601718 units. tanc = 3/1.0606601718 c = 70.5287793652° d = ½(c) d = 35.2643896826° tan35.2643896826 = r/e Where r is the required radius of the two congruent inscribed circles. e = 1.4142135624r units. e = √(2)r unit...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2582022

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Let the base of the ascribed triangle be 1 unit. a = 180-70-70 a = 40° (b/sin40) = (1/sin70) b = 0.6840402867 units. Considering similar plane shape (isosceles triangle) side length ratios. 1 - 0.6840402867 (0.6840402867+c) - 1 Cross Multiply. 1 = 0.6840402867²+0.6840402867c (1-0.6840402867²) = 0.6840402867c c = 0.7778619133 units. d = 70-40 d = 30° e² = 1²+0.7778619133²-2*0.7778619133cos30 e = 0.507713306 units. Therefore, the required angle, x is; (0.5077133...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2581737

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a = 4+5 a = 9 units. b² = 9²+3² b = √(90) b = 3√(10) units. c = ½(b) c = ½(3√(10)) units. c is the radius of the quarter circle. d² = 5²-(½(3√(10)))² d = √(25-(45/2)) d = √(½(5)) d = ½√(10) units. e = c-d e = ½(3√(10))-½√(10) e = √(10) units. tanf = 3/4 f = atan(3/4)° tang = (0.5*3√(10))/0.5√(10) g = atan(3)° h = 180-f-g h = (180-atan(3/4)-atan(3))= It implies, the required length, x is; x² = 4²+√(10)²-2*4√(10)cos(180-atan(3/4)-atan(3)) x = 4.242640687...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2581189

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a = 180-(180-104)-60 a = 104-60 a = 44° b = 2a b = 88° Calculating x, the required angle. x = 180-88-(180-104) x = 92-76 x = 16° x is the required angle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2580457

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Let a be the radius of the ascribed semi circle. tanb = (3/4) b = atan(3/4)° It implies; 3² = 2a²-2a²cos(atan(3/4)) 9 = 2a²-(8a²/5) 9 = ⅕(2a²) 2a² = 9*5 a² = ½(9*5) a = ½(3√(10)) units. a = 4.7434164903 units. Again, a is the radius of the ascribed semi circle. It implies, x, diameter of the ascribed semi circle is; x = 2a x = 2* ½(3√(10)) x = 3√(10) units.
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