Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3139969

a = x° b = (90-x)° R² = 6²+16²-2*6*16cosx R² = 292-192cosx cosx = (292-R²)/192 --- (1). Notice. At (1). (292-R²) is adjacent. 192 is hypotenuse. Let c be opposite. c² = 192²-(292-R²)² c² = 36864-85264+584R²-R⁴ c = √(-R⁴+584R²-48400) units. It implies; sinx = √(-R⁴+584R²-48400)/192 --- (3). Again. R² = 10²+6²-2*10*6cos(90-x) R² = 136-120(cos90cosx+sin90sinx) R² = 136+120sinx --- (4). Substituting (3) in (4). R² = 136+120(√(-R⁴+584R²-48400)/192...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3138417

Let the side length of the square be x. a = (x-4) units. Calculating x. ½(x²) = ½*x(x-4)-10+30 x² = x(x-4)+40 x² = x²-4x+40 4x = 40 x = 10 units. Again, x is the side length of the square. Area Square is; x² = 10² = 100 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3135177

Let AD = BC = 1 unit. a = 180-100-30 a = 50° a is angle BAC. (1/sin50) = (b/sin30) b = 0.65270364467 units. b is AC. Therefore, the required angle theta is, let it be c. (0.65270364467/sinc) = (1/sin100) sinc = 0.65270364467sin100 sinc = 0.64278760969 c = asin(0.64278760969) c = 40°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3134182

Notice. 10 m is the radius of the circle. a = 10-3 a = 7 m. sinb = (7/10) b = asin(7/10)° Calculating length AB. tanb = 10/(AB) AB = 10/(tanb) AB = 10/tan(asin(7/10)) AB = 10.2020406122 m. Or 7 ~ 3 10 ~ x Cross Multiply. 7x = 30 x = ⅐(30) m. x is length Be. It implies; y = 10+x y = 10+⅐(30) y = ⅐(100) m. y is length Bc Therefore; z² +10² = (⅐(100))² z² = (10000/49)-100 z² = (10000-4900)/49 z = √(5100/49) z = ⅐(10√(51)) m. z = 10.2020406122 m. z is the required length AB.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3129461

Let a be the radius of the white inscribed circle. b = (10-a) units. Calculating a. b² = a²+5² (10-a)² = a²+25 100-20a+a² = a²+25 75 = 20a a = ¼(15) units. a = 3.75 units. c = ½(10) c = 5 units. c is the radius of the bigger circle ascribing the white circle. Therefore, area yellow is; πc²-πa² = π(5²-(¼(15))²) = π(25-(225/16)) = π((175/16) = 10.9375π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3127330

Calculating x. Let a be the height of the inscribed isosceles triangle. b = ½(7)+7 b = 10.5 units. c² = 10.5²+a² c = √(a²+110.25) units. Let d be one of the two equal inscribed angles. cosd = 10.5/√(a²+110.25) --- (1). e² = 3.5²+a² e = √(a²+12.25) units. f = (10-√(a²+12.25)) units. It implies; (10-√(a²+12.25))² = √(a²+110.25)²+14²-2*14*√(a²+110.25)cosd 100-20√(a²+12.25)+a²+12.25 = a²+110.25+196-28√(a²+110.25)cosd 100-20√(a²+12.25) = 294-28√(a²+110.25)cosd 50-10√(a²+12.25) = 147-14√...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3126411

Let the bigger inscribed semi circle radius be 2 units. Let the big inscribed semi circle radius be r. Let the inscribed circle radius be q. Calculating r. (2+r)² = 2²+(4-r)² 4+4r+r² = 4+16-8r+r² 4r = 16-8r 12r=16 3r=4 r=4/3 unit. Calculating q. (2+(4/3))²=(2+q)²+((4/3)+q)² (10/3)²=4+4q+q²+(16/9)+(8q/3)+q² (100/9)-(16/9)-4=2q²+4q+(8q/3) (84/9)-4=2q²+4q+(8q/3) 84-36=18q²+36q+24q 48=18q²+60q 6q²+20q-16=0 3q²+10q-8=0 Therefore; q≠-4...

Mathematics Question and Solution

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

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/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/3122646

Let a be the radius of the inscribed circle. b² = 2(10)² b = 10√(2) cm. b is the diagonal of the square. 10²+a² = c² c = √(100+a²) cm. d = b-c d = (10√(2)-√(100+a²)) cm. Calculating a. 2a² = (10√(2)-√(100+a²))² √(2)a = 10√(2)-√(100+a²) √(100+a²) = 10√(2)-√(2)a 100+a² = (10√(2)-√(2)a)² 100+a² = 200-40a+2a² a²-40a+100 = 0 (a-20)² = -100+(-20)² a =20±√(300) a = 20±10√(3) Therefore; a ≠ (20+10√(3)) cm. a = (20-10√(3)) cm. a = 2.67949192431cm. Again, a is the radius of the inscribed circle. C...

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