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°

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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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1986629

Let the two equal lengths be 1 unit. (a) Calculating Theta exactly. Let theta be x. Calculating x. x = ½(90-54) x = ½(36) x = 18° sin18 = a/1 a = 0.30901699437 unit. b = 90-18 b = 72° sin72 = c/1 c = 0.9510565163 unit. tan36 = 0.9510565163/d d = 1.30901699437 units. Area Blue is; 0.5(1.30901699437*0.9510565163) = 0.62247457122 square units. e = 270-54-96 e = 120° f = 360-120-72 f = 168° g² = 2-2cos168 g = 1.98904379074 units. h = ½(180-168) h = 6° (1.98904379074/sin54) = (i/sin2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1985915

Radius of the semi circle is 5 units. Calculating r, radius of the inscribed circle. (5-r)² = 3²+r² 25-10r+r² = 9+r² 10r = 16 r = ⅕(8) units. a = 3-r a = ⅕(7) units. b² = 5²-a² b = ⅕(24) units. c = atan((8/5)/8) c = atan(⅕)° d = 2c d = 2atan(⅕)° tan(2tan(⅕)) = e/(5+(7/5)) e = ⅓(8) units. f = atan(12/5)° g = b-f g = (32/15) units. h² = 50-50cos(180-4atan(⅕)) h = (120/13) units. sin(2atan(⅕)) = (8/3)/i i = (104/15) units. j = h-i j = (448/195) units. k = atan(24/7)° l = (180-atan(2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1985499

AE = CF = (12√(2)-12) cm. AE = CF = 4.97056274848 cm. EF = 12√(2)-2(4.97056274848) EF = 7.02943725152 cm. sin60 = a/7.02943725152 a = 6.08767123412 cm. a = height of triangle EFG. It implies; Length BG is; 6√(2)-6.08767123412 = 2.39761014012 cm.

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