Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585818

a = 0.5*5*5sin40 a = 8.0348451211 square units. (b/sin40) = (5/sin70) b = 3.4202014333 units. c = 0.5*3.4202014333*3.4202014333sin40 c = 3.7595933295 square units. Shaded Area is: a-c = 8.0348451211-3.7595933295 4.2752517916 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585632

sin60 = 0.5d/a a = ⅓√(3)d units. a is the radius of the half circle. Therefore, total blue area in terms of d is; 2(60π*⅓√(3)d*⅓√(3)d/360)+(0.5*⅓√(3)d*⅓√(3)dsin120)-(0.5*⅓√(3)d*⅓√(3)dsin60) = 2(πd²/18) = ⅑πd² square units.l

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585623

a = ½(d) units. a is the radius of the green half circle. b = d+a b = ½(3d) units. c² = (½(3d))²-(½(d))² c² = ¼(8d²) c = √(2(d)²) c = √(2)d units. c is the diameter of the small blue semi circle. d = ½(c) d = ½√(2)d units. d = 0.7071067812d units. d is the radius of the small blue semi circle. tane = ½(d)/√(2)d e = atan((½)/√(2)) e = atan(¼√(2)) f² = (3d)²+(√(2)d)²-2*3d*√(2)dcos(0.25√(2)) f = 1.5858373728d units. f is the diameter of the big semi circle. g = ½(f...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585604

Notice! Triangle ABM is equilateral. Calculating a:b. Let AB be 1 unit. c = 90-60 c = 30° cos30 = (0.5*1)/d d = ½÷½√(3) d = 1/√(3) d = ⅓√(3) units. d is the radius of the semi circle. e = 2d e = ⅔√(3) units. e is d, diameter of the semi circle. Therefore; a : b is; 1:⅔√(3) = 3:2√(3) = 3/(2√(3)) = ½√(3)

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585239

a = ⅓(90) a = 30° sin30 = b/4 b = 2 cm. sin60 = c/4 c = 2√(3) cm. d = √(4²-2²)-√(4²-(2√(3))²) d = √(12)-√(4) d = (2√(3)-2) cm. d = 1.4641016151 units. Area Blue is; Area trapezoid with parallel sides 2 cm and 2√(3) cm respectively and height (2√(3)-2) cm + Area sector with radius 4 cm and angle 30° - Area triangle with height 4 cm and base 4sin30 cm. = (½*(2+2√(3))*(2√(3)-2))+(30π*4*4/360)+(½*4*4sin30) = (½*8)+⅓(4π)-4 = 4+⅓(4π)-4 = ⅓(4π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585177

Let a be 1 unit. b = 1+(2*1)+1 b = 4 units. c = ½(b) c = 2 units. 2d² = 4² d² = 8 d = 2√(2) units. e² = 1²+(2√(2))²-2*2√(2)cos45 e = √(9-4) e = √(5) units. (√(5)/sin45) = (1/sinf) f = 18.4349488229° g = 90-2f g = 53.1301023542° h = ½(g) h = 26.5650511771° J² = 2k² j = √(2)k units. k is the radius of the circle. l = (2-√(2)k) units. k² = 1²+(2-√(2)k)² k² = 1+4-4√(2)k+2k² k²-4√(2)k+5 = 0 Calculating k via completing the square approach. (k-2√(2)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585127

Let the side length of the inscribed square be a. Calculating the area of the inscribed red square. 2b² = a² b = ½√(2)a unit. c² = 2a² c = √(2)a unit. c is the diagonal of the inscribed red square. It implies; 10² = (½√(2)a)²+(√(2)a)² 100 = (a²/2)+2a² 100 = ½(5a²) 5a² = 200 a² = 40 a = 2√(10) units. Therefore, area red inscribed square is; a² = 40 square units. Again, a = 2√(10) units is the side length of the inscribed red square.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585091

Calculating area of the inscribed blue half circle. a = ½(6+4) a = 5 units. a is the radius of the ascribed semi circle. b = a-4 b = 1 unit. c²+1² = d² d = √(c²+1) units. c is the radius of the inscribed blue half circle. 5² = c²+√(c²+1)² 25 = c²+c²+1 2c² = 24 c² = 12 c = 2√(3) units. Again, c is the radius of the inscribed blue half circle. Therefore; Area inscribed blue half circle is; ½(c²)π = ½*2√(3)*2√(3)π = 6π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2584218

Notice! Triangle ABC is equilateral. Let AB be 1 unit. a = 180-42-12 a = 126° a is angle BDC. (1/sin126) = (b/sin42) b = 0.8270909153 units. b is CD. c² = 1+0.8270909153²-2*0.8270909153cos48 c = 0.759747123 units. c is AD. Therefore, the required angle, angle CAD is; Let it be d. (0.759747123/sin48) = (0.8270909153/sind) d = 54° Again, d is the required angle, angle CAD.

Mathematics Question and Solution

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

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) =...

You Searched for "delta state office ekrejeta house 2 zenith bank abraka delta state nigeria"

2626 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

Facebook Page

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 "delta state office ekrejeta house 2 zenith bank abraka delta state nigeria"

2626 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?