Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2676449

Sir Mike Ambrose is the author of the question. Area red exactly as a single fraction is; Area triangle red with two lengths ⅓(24-8√(3)) cm and ⅓(18-8√(3)) cm, and angle 60° = ½(⅓(24-8√(3)))(⅓(18-8√(3)))sin60 = ((24-8√(3))(18-8√(3))√(3))/36 = (624√(3)-1008)/36 = (104√(3)-168)/6 = ⅓(52√(3)-84) cm² = 2.02221399786 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2675508

Please, move the above question left/right one time to review the solution. Thank you. Area Square OBCD is; ½(3+2√(2)) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2675320

Let a be the radius of the inscribed circle. b = a+2a b = 3a units. c² = 2(3a)² c = 3√(2)a units. d = 180-45 d = 135° Calculating a. 5² = (3√(2)a)²+a²-2*3√(2)a²cos135 25 = 19a²+6a² 25 = 25a² a = 1 unit. Again, a is the radius of the inscribed circle. Calculating length x. x² = 2a² x² = 2 x = √(2) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2675296

Calculating L. Let a be the width of the rectangle. b = (a-x) units. a(a-x) = 4 --- (1). Observing similar plane shape (right-angled) side length ratios. (a-x) - ½(L) L - a Cross Multiply. a(a-x) = ½(L²) --- (2). Therefore, equating (1) and (2) to get L. ½(L²) = 4 L² = 8 L = √(8) L = 2√(2) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2674965

½(a*a) = 10 a² = 20 a = 2√(5) units. a is the equal adjacent lengths of the inscribed purple isosceles right-angled triangle. Let c be the side length of the square. tand = c/(0.5c) d = atan(2)° e = d-45 e = (atan(2)-45)° cos(atan(2)-45) = 0.5c/(2√(5)) c = 8.4852813742 units. c = 6√(2) units. c is the side length of the ascribed square Are ascribed square (ABCD) is; c² = (6√(2))² = 6√(2)*6√(2) = 72 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2674711

Notice. 3x = 15 It implies; x = 5 m. x is the width of the inscribed green rectangle. The length is 15 m. Therefore, area inscribed green rectangle is 5 *15 = 75 m²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2674127

Let the square side be a. a² = πr² And r = 1 unit, π = 22/7 a² = (22/7) a = √(22/7) units. b = ½(a) b = ½√(22/7) units. c² = 2a² c² = 2√(22/7)² c = √(44/7) c = ⅐(2√(77)) units. c is the diagonal of the square. cosd = ½√(22/7)/1 d = acos(½√(22/7))° d = 27.5750477105° e = 45-d e = 17.4249522895° f = 3e f = 34.849904579° f is the required angle, phi. g² = 2-2cos34.849904579 g = 0.5989126703 units. 2h² = 0.5989126703² h = 0.4234952105 units. Shaded Area...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2673610

Calculating MN, diameter of the yellow semi circle. Let the radius of the yellow semi circle be a. b = ½(3+3+2) b = 4 units. b is the radius of the bigger semi circle. c = b-3 c = 1 unit. Therefore. a² = 1.5²+d² d = √(a²-2.25) units. e = 3+1.5-b e = 4.5-4 e = 0.5 units. f² = 0.5²+√(a²-2.25)² f² = 0.25+a²-2.25 f = √(a²-2) units. It implies; 4² = √(a²-2)²+a² 16 = a²-2+a² 2a² = 18 a² = 9 a = 6 units. Again, a is the radius of the yellow semi circle. The...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2673604

sin30 = 1/a ½ = 1/a a = 2 units. cos30 = ½(a)/b ½√(3) = 1/b √(3)b = 2 b = ⅓(2√(3)) b = ⅓(2√(3)) units. b = 1.1547005384 units. b is the radius of the ascribed semi circle. Calculating c, radius of the inscribed green semi circle. d = 2b-c d = (⅓(4√(3))-c) units. sin30 = c/d ½ = c/(⅓(4√(3))-c) 2c = ⅓(4√(3))-c 9c = 4√(3) c = ⅑(4√(3)) units. Again, c is the radius of the inscribed green semi circle. Area inscribed green semi circle is; ½(c)² = ½*⅑(4√(3))*⅑(4√(3))...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2673119

Let CN = a b = (9+a) m. b is the side length of the square. c = ½(b) c = ½(9+a) m. c = CM = DM. Calculating a. (9+a) - ½(9+a) ½(9+a) - a Cross Multiply. a(9+a) = ¼(9+a)² a = ¼(9+a) 4a = 9+a a = 3 m. Recall. b = (9+a) And a = 3 m. b = 9+3 b = 12 m. Again, b is the side length of the square. c = ½(9+a) And a = 3 m. c = ½(9+3) c = 6 m. Again, c = CM = DM. Shaded Area is; (12*12)-½(12*6)-½(12*9) = 144-36-54 = 144-90 = 54 m²

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

2139 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 "delta state office ekrejeta house 2 zenith bank abraka delta state nigeria"

2139 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?