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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3125595

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Let a be the blue length. b² = 4²+3² b = √(25) b = 5 units. Let alpha be c. cos(2c) = 4/5 cos(c+c) =4/5 (cosc)²-(sinc)² = 4/5 --- (1). cosc = 3/a --- (2). a² = 3²+d² d = √(a²-9) units. sinc = d/a sinc = √(a²-9)/a --- (3). Substituting (2) and (3) in (1). (3/a)²-(√(a²-9)/a)² = 4/5 (9/a²)-((a²-9)/a²) = 4/5 9-(a²-9) = ⅕(4a²) 18-a² = ⅕(4a²) 90-5a² = 4a² 9a² = 90 a² = 10 a = √(10) units. a = 3.16227766017 units. Again, a is the required blue len...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3125501

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Let a be the radius of the circle. b = 7-3 b = 4 cm. c = (7-a) cm. d = (3-a) cm. e = c+d e = 7-a+3-a e = (10-2a) cm. Calculating a. e² = (2a)²+4²  (10-2a)² = 4a²+16 100-40a+4a² = 4a²+16 100-40a = 16 40a = 84 10a = 21 a = 2.1 cm. Again, a is the radius of the circle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3125497

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Notice. Triangle ACD and Triangle BCD are right-angled triangles. Let AD = 2 units. AB = 6 units. It implies; BD = 6-2 BD = 4 units. a² = 2(0.5*4)² a = 2√(2) units. a is BE = CE. tanb = (2/4) b = atan(½)° b is angle DAF. Therefore; DF = 1 units. Area triangle AFD is; ½(AD)*(DF) = ½*2*1 = 1 square units. Calculating area triangle CFE. It implies; Area trapezoid with parallel lengths 2 unit and 4 units respectively and height 2 units - Area trapezo...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3125099

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Notice. The inscribed square area is 3 square units. Therefore the single side length of the inscribed square is √(3) units. The radius of the ascribed quarter circle will be; √((√(3))²+(√(3))²) = √(6) units. Notice that the green inscribed triangle is a right-angled triangle. The marked angle will be; 2(90-tan–¹(√(6)/√(3))) = 2(90-tan–¹(√(2))) = (180-2tan–(√(2))) The hypotenuse of the green inscribed right-angle triangle is; √((√(3))²+(√(6))²) = √(9) = 3 uni...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3124738

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Let AG be y. Calculating y. a² = 8²+y² a = √(y²+64) cm. a is AD. It implies; Considering area ADG including the inscribed circle with radius 1 cm. ½(1*8)+½(1*y)+½(1*√(y²+64)) = ½(8*y) 8+y+√(y²+64) = 8y √(y²+64) = 7y-8 y²+64 = 49y²-112y+64 112y = 48y² 48y = 112 y = 112/48 y = ⅓(7) cm. b = y+8 b = ⅓(7)+8 b = ⅓(7+24) b = ⅓(31) units. b is AC. Recall. a = √(y²+64) And y = ⅓(7) cm. a = √((7/3)²+64) a = √((49/9)+64) a = √(⅑(49+(9*64))) a = √(⅑(49+576)) a = √(⅑*625) a = ⅓(25) cm. Again, a...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3124467

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Let the side length of the regular pentagon be 1 unit. tan36 = a/0.5 a = 0.363271264 units. b² = 0.363271264²+0.5² b = 0.61803398875 units. tan72 = c/0.5 c = 1.53884176859 units. Area Green is; 0.5sin108+0.5*1.53884176859*0.5-0.5*0.5*0.363271264-0.5*0.61803398875*0.61803398875sin36 = 0.65716389015 square units. Calculating Area Blue. Let d be the radius of the inscribed blue circle. Calculating d. e = 0.5(90+36) e =63° (d/tan63) + (d/tan45) = a d = 0.3632...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3124331

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Let the single side length of the square be x. Therefore; Area rectangle of length x unit and width 1 cm + area triangle of height x cm and width (x-2) cm + 2(area triangle of height (x-1) cm and width 1 cm) + 2(area triangle of height ½√(x²+(x-2)²) cm and width 5 cm) = Area square of single side length x cm. It will be; x+½(x(x-2))+½(2(x-1))+¼(10√(x²+(x-2)²)) = x² 2x+x²-2x+2x-2+5√(2x²-4x+4) = 2x² x²+2x-2+5√(2x²-4x+4) = 2x² 5√(2x²-4x+4) = x²-2x+2 5√(2(x²-2x+2)) = x²-2x...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3123965

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Let a be the side length of the regular hexagon. 3² = 1²+a²-2acosb 2acosb = a²-8 cosb = (a²-8)/(2a) --- (1). c²+(a²-8)² = (2a)² c² = 4a²-(a⁴-16a²+64) c = √(-a⁴+20a²-64) units. sinb = c/(2a) sinb = √(-a⁴+20a²-64)/(2a) --- (2). (2√(3))² = 1²+a²-2acos(240-b) 11 = a²-2a(cos240cosb+sin240sinb) 11 = a²-2a(-½(cosb)-½√(3)sinb) 11 = a²+acosb+a√(3)sinb --- (3). Substituting (1) and (2) in (3). 11 = a²+a((a²-8)/(2a))+a√(3)(√(-a⁴+20a²-64)/(2a)) 11 = a²+½(a²-8)+½√(3)√(-a⁴+20a²...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3123716

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Let the radius of the semi circle be 3 units. Calculating area Red. Area red will be; 2(area triangle of two length 3 units and 1 unit, and angle (cos–¹(⅓)), angle they form) + area sector of radius 3 units and angle (180-2cos–¹(⅓)). = ½(2(1*3sin(cos–¹(⅓)))) + ((180-2cos–¹(⅓))*9π÷360) = 3sin(cos–¹(⅓)) + ((180-2cos–¹(⅓))*π÷40) = 5.88695930984 square units. Area of the semi circle is; = ½(9π) = 4.5π square units. Therefore percent of the shaded area from the whole...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3123684

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a = 180-75-27 a = 78° b = 1+½(√(5)+1)  b = 2.61803398875 units. (2.61803398875/sin78) = (c/sin27) c = 1.21511575349 units. sin75 = d/1.21511575349 d = 1.17371168822 units. cos75 = e/1.21511575349 e = 0.31449509901 units. f = b-e-1 f = 2.61803398875-0.31449509901-1 f = 1.30353888974 units. Therefore, the required angle is; Let it be g. tang = d/f g = atan(1.17371168822/1.30353888974) g = 42°
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