Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2537563

Let the small inscribed regular triangle side length be x. It implies; Calculating x. 2² = x²+(3x)²-2*x*3xcos60 4 = 10x²-3x² 4 = 7x² x = ⅐(2√(7)) units. Therefore; 3x is; = 3*⅐(2√(7)) = ⅐(6√(7)) units. It implies, area blue is; ½*x*3xsin60 = ½*⅐(2√(7))*⅐(6√(7))*½√(3) = ⅐(3√(3)) square units. = 0.742307489 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2537524

tan60 = √(3)/a a = 1 unit. b = 8-a b = 7 units. tanc = √(3)/7 c = atan(√(3)/7)° d = 2c d = 2atan(√(3)/7)° e = 180-60-d e = (120-2atan(√(3)/7))° f = ½(e) f = (60-atan(√(3)/7))° Calculating R, radius of the bigger circle. It implies; tan(60-tan(√(3)/7)) = R/g g = R/tan(60-atan(√(3)/7)) units. tan60 = R/h h = R/tan60 units. It implies; g+h = 8 (R/tan(60-atan(√(3)/7)))+(R/tan60) = 8 1.5396007178R = 8 R = 8/1.5396007178 R = 5.1961524227 units. Again, R...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2537314

a² = 2²+b² a = √(4+b²) a is AB. b is the height of the yellow inscribed triangle. 0.5*b*c = 6 bc = 12 c = 12/b units. c is the base of the inscribed yellow triangle. d² = b²+(12/b)² d² = (b⁴+12)/b² d = √(b⁴+12)/b units. 2 - √(4+b²) b - e Cross Multiply. e = ½√(4+b²)b units. e is BC. f = ½(e) f = ¼√(4+b²)b units. ¼√(4+b²)b - b g - 2 Cross Multiply. g = ½√(4+b²) units. It implies; Calculating b. 12+(2*b)+(½√(4+b²)*½√(4+b²)b) = ½√(4+b²)b*√(4+b²) 12+2b+¼(4+b...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2537093

Let the side length of the inscribed regular pentagon be 1 unit. a = ½(120) a = 60° b = 180-108 b = 72° c = 180-60-72 c = 48° It implies; (1/sin60) = (d/sin48) d = 0.8581097301 units. e = 1+d e = 1.8581097301 units. e is the side length of the ascribed regular hexagon. f² = 0.8581097301²+1.8581097301²-2*0.8581097301*1.8581097301cos120 f = 2.4048671724 units. Calculating the required angle, Beta. Let it be h. (2.4048671724/sin120) = (1.8581097301/sinh) h = 42...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2536723

Let CE be 1 unit. Observing Sum of Interior Angles of a Triangle. a = 180-64-10 a = 106° a is angle CED. Observing Sine Rule. (1/sin10) = (b/sin106) b = 5.5356854811 units. b is DE. Observing Sum of Interior Angles of a Triangle. c = 180-50+38 c = 92° c is angle BEC. Observing Sine Rule. (1/sin38) = (d/sin50) d = 1.2442624296 units. d is BE. Observing Sum of Interior Angles of a Triangle. e = 180-86-24 e = 70° e is angle AEB. Observing Sine Rule. (1.244262429...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2536518

Let the two equal lengths be a a²+x² = 10² a = √(100-x²) units. b = ½(8+10) b = 9 units. c²+(9-8)² = x² c = √(x²-1) units. Observing similar plane shape (right-angled) side length ratios. 10 - √(100-x²) x - √(x²-1) Cross Multiply. 10√(x²-1) = x√(100-x²) 100(x²-1) = x²(100-x²) 100x²-100 = 100x²-x⁴ x⁴ = 100 x² = 10 x = √(10) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2536469

Radius of the two congruent inscribed white circles a, is; a = ½*4√(3) a = 2√(3) cm. b = 4√(3) cm. b is the radius of the ascribed circle. Area yellow is; 2(2(area sector with radius 4√(3) cm and angle 120° - area triangle with height 4√(3) cm and base 4√(3)sin120 cm) - Area circle with radius 2√(3) cm.) = 2(2((120π*(4√(3))²/360)-(0.5*(4√(3))²sin120))-π(2√(3))²) = 2(2(16π-12√(3))-12π) = 2(32π-24√(3)-12π) = 2(20π-24√(3)) = 8(5π-6√(3)) cm² = 42.5252673802 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2536457

Let a be the radius of the big inscribed circle. Let b be the radius of the bigger inscribed circle. Therefore; b = 2a cm. c = 2a+a c = 3a cm. c is the radius of the ascribed circle. It implies, calculating a. 120+π(2a)²+πa² = π(3a)² 120+4a²π+a²π = 9a²π 120 = 4a²π a²π = 30 a² = (30*7)/22 a² = 105/11 a = √(105/11) cm. a = 3.0895719033 cm. b = 2a And a = 3.0895719033 cm. b = 6.1791438065 cm. c = 3a And a = 3.0895719033 cm. c = 9.2687157098 cm. Therefore;...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2536446

Equation of the curve is; y = ⅔(8x-x²) Calculating Area E. (Area under the curve at x = (3/2) and x = 4) - Area trapezium with parallel sides (13/2) units and (10⅔) units, and height (5/3) units. =(835/36) - ½*(5/2)*((13/2)+(32/3)) = (835/36) - (515/24) = (125/72) square units. Calculating Area F. ½(Area under the curve at x = 0 and x = 8) - Area triangle with height (32/3) units and base 4 units - Area triangle with height 6 units and base 8 units - (Area under the curve...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2535303

Calculating angle x. Notice! x (angle BCD) will sum up the interior reflex angle BCD of the plane shape ABCD to get 360° Plane shape ABCD is a quadrilateral, therefore the sum of its interior angles is 360° It implies; 30+70+20+a = 360 Where a is the interior reflex angle BCD of the plane shape ABCD. a = 360-120 a = 240° Therefore, x, angle BCD is; a+x = 360 240+x = 360 x = 360-240 x = 120°

You Searched for "post"

2242 result found.

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

Mathematics Question and Solution

OnlineEdumath

Socials

Useful Tags

Quick Links

Reach Us

Safety and Abuse Reporting

Thanks for being awesome!

Are you sure you want to report abuse against this website?

Search

You Searched for "post"

2242 result found.

Subscribe to our mailing list for latest updates.

Safety and Abuse Reporting

Thanks for being awesome!

Are you sure you want to report abuse against this website?