Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1989974

Let the biggest regular pentagon side length be 2 units. sin72 = a/2 a = 1.90211303259 units. tan72 = (1.90211303259/b) b = 0.61803398875 unit. c = 2b c = 1.2360679775 units. Where c is the side length of the regular red and blue pentagon. Calculating area of the regular red and blue pentagon. (5*1.2360679775²)/(4tan36) = 2.6286555606 square units. d² = 8-8cos108 d = 3.2360679775 units. e = ½(3.2360679775-c) e = 1 unit. f² = 8-8cos36 f = 1.2360679775 units. g² = 2-2cos36 g = 0.618033...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1989748

Angle ABC is; atan(3/4)° Therefore; sin(atan(3/4)) = 3/(BC) BC = 5 units. Angle ACB is; atan(4/3)° Angle EAC is; (180-2atan(4/3))° Calculating CE. (3/sin(atan(4/3))) = (CE/sin(180-2atan(4/3)) CE = 3.6 units. It implies x (BE) is; 5-3.6 = 1.4 units. Or (3/sin(atan(3/4))) = (x/sin(2tan(4/3)-90)) x = 1.4 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1988763

Let the radius of the 3 congruent circles be r. Therefore, the general formula for the area of green section is; ½(πr²)-3(60πr²)/360+2(½*r²*sin60) = ½(πr²)-½(πr²)+r²sin60 = r²sin60 = ½(√(3)r²) square units. The green area does not rely on pie because it can be structured to two congruent equilateral triangles (plane shapes) with single side length r (radius) units each which is equal to the radius of each of the three congruent circles.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1988595

Let BD be 1 unit. Therefore; (1/sin100) = (d/sin70) d = 0.95418889414 units. (1/sin80) = (AB/sin70) AB = 0.95418889414 units. AB is c. It implies; d = c. Calculating a. (a/sin20) = (0.95418889414/sin80) a = 0.33138632523 units. Calculating b. (1/sin100) = (e/sin10) e = 0.17632698071 units. (0.17632698071/sin30) = (f/sin110) f = 0.33138632523 units. It implies; (0.33138632523/sin20) = (b/sin140) b = 0.6228025689 units. Confirming the proof; a+b = c Notice; c = 0.95418889414 units...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1988480

Let r be 1 unit. Calculating R. (R+1)² = x²+(R-1)² Calculating x. x² = R²+2R+1-R²+2R-1 x = √(4R) x = 2√(R) units. It implies; (2R)² = (2√(R))²+2² 4R² = 4R+4 R²-R-1= 0 Solving the quadratic equation. (R-½)² = 1+¼ (R-½) = ±√(5/4) R = ½±½√(5) Therefore; R ≠ ½(1-√(5)) units. R = ½(1+√(5)) units. R = 1.61803398875 units. Therefore; R/r is; ½(1+√(5))/1 = ½(1+√(5))

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1988403

Let the two congruent lengths be 1 unit. (1/sin50) = (a/sin80) a = 1.28557521937 units. (1.28557521937/sin110) = (b/sin60) b = 0.46791111376 units. (1/sin70) = (c/sin80) c = 1.04801052092 units. sin60 = d/2 d = √(3) units. e = d-c e = √(3)-1.04801052092 e = 0.68404028665 units. f² = 0.68404028665²+0.46791111376²-2*0.68404028665*0.46791111376cos70 f = 0.68404028665 units. Notice; e = f Therefore the required angle x, is; 180-2*70 = 180-140 = 40°

Parent's Feedback/Testimonial

https://onlineedumath.com/update/parents-feedbacktestimonial/1987166

Thank you for your feedback/testimonial, Sir Amadin. We, the Management and Educators of Online Edumath are humble and grateful. We will continue working hard to improve our standard because success, wins and working hard is never enough. Sir Amadin is Oghosa's Dad, Oghosa is a proud, smart and clever Online Edumath learner, he will be in year 8 next session. Thank you again, Sir Amadin. Sir Amadin is smart and clever.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1987104

The side length of the congruent Hexagons is 2 unit. Area hexagon is; (6*2²)/(4tan30) = 6√(3)) square units. a² = 8-8cos30 a = √(8-4√(3)) unit. Area ABC is; ½(√(8-4√(3)))²sin60 = (2√(3)-3) square units. (d) Area Hexagon ÷ Area ABC exactly is; (6√(3)) ÷ (2√(3)-3) = 6(2+√(3)) (c) Calculating Area Green to 2 decimal places. b = a + √(2²+2²) b = √(8-4√(3))+2√(2) b = (√(2)+√(6)) units. b = 3.86370330516 units. c² = 8-8cos120 c = 2√(3) units. Area Green is; 2((0.5x2*2sin120) + (0.5*2√(3...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1987096

25 = 49+64-14*8cosa 112cosa = 113-25 a = acos(88/112) a = 38.21321070174° a = Angle ACB. 5b+7b+8b = 56sin38.21321070174 b = (56sin38.21321070174)/20 b = √(3) cm b = 1.73205080757 cm. b = radius of the inscribed circle. tan(0.5*38.21321070174) = √(3)/c c = 5 cm. c = CD. BD = 8-5 BD = 3 cm. Therefore; (BN)² = (2√(3))²+3² BN = √(12+9) BN = √(21) cm. AE = 7-5 AE = 2 cm. Therefore; (AM)² = (2√(3))²+2² AM =√(12+4) AM = 4 cm. It implies; AM*BN is; 4*√(21) = 4√(21) cm²

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