Mathematics Question and Solution

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

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

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

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

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

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

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

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°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3123665

Sir Mike Ambrose is the author of the question. Area W to 3 decimal places is; Area triangle of two length 5 units and 7.5 units, and angle 86.4166783015° - Area triangle of two length 5 units and 4 units, and angle 86.4166783015° - Area triangle of two length 2.75 units and 3.5 units, and angle 34.7719440318° = 5.98826978347 ≈ 5.988 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3123589

Sir Mike Ambrose is the author of the question. Let the length of the inscribed rectangle be 2 units. Let the width of the inscribed rectangle be √(3) units. Therefore the single side length of the inscribed square x will be; x+((x)/√(3))+√(3)x=2 x=2√(3)/(4+√(3)) units. Area R is; x² = (2√(3)/(4+√(3)))² = 12/(19+8√(3)) square units. = 0.36522557677 square units. Area S will be; Area triangle of height √(3) units and base 1 unit + Area triangle of two side 2 unit...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3123251

Let r be the radius of the quarter circle. Let x be the interior angle facing 1 cm. a = (x+45)° a is alternate to y, the angle formed by the meet point of r and 2 cm at the circumference of the circle. It implies; y = (x+45)° sinx = 1/r --- (1). cosy = 2/r cos(x+45) = 2/r cosxcos45-sinxsin45 = 2/r ½√(2)(cosx-sinx) = 2/r --- (2). Notice. At (1). Sinx = 1/r Where; 1 cm = opposite. r cm = hypotenuse. Therefore, let b be the adjacent. It implies; b²+1²...

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