Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3138420

Let the length of the rectangle be a. Let the width of the rectangle be (6+b) units. ½(2a-4)(6+b)-70 = ½(6a)+½(a-4)b 12a+2ab-24-4b-140 = 6a+ab-4b 6a+ab = 164 Therefore; a(6+b) = 164 Notice. a is the length of the rectangle. (6+b) is the width of the rectangle. Therefore; Length*Width = a(6+b) = 164 It implies; Area Rectangle = 164 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3138417

Let the side length of the square be x. a = (x-4) units. Calculating x. ½(x²) = ½*x(x-4)-10+30 x² = x(x-4)+40 x² = x²-4x+40 4x = 40 x = 10 units. Again, x is the side length of the square. Area Square is; x² = 10² = 100 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3138413

Calculating R, radius of the circle. 12a = 3*4 a = 1 unit. b = ½(3+4) b = 3.5 units. c = b-3 c = 0.5 units. d = 180-60 d = 120° e² = R²-3.5² e =√(R²-12.25) units. f² = 1²+0.5²-2*1*0.5cos120 f = ½√(7) units. f = 1.32287565553 units. (1.32287565553/sin120) = (1/sing) g = 40.8933946491° h = 90+g h = 130.8933946491° R is; R² = √(R²-12.25)²+1.32287565553²-2*1.32287565553*√(R²-12.25)cos130.8933946491 10.5 = 1.73205080756√(R²-12.25) (10.5)² = (1.732050807...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3137623

Calculating R, radius of the quarter circle. a = 3+4 a = 7 units. b² = 2(4)² b = 4√(2) units. c = R-b c = (R-4√(2)) units. d = 180-45 d = 135° R is; R² = 7²+(R-4√(2))²-2*7(R-4√(2))cos135 R² = 49+R²-8√(2)R+32+7√(2)(R-4√(2)) 8√(2)R = 81+7√(2)R-56 √(2)R = 25 R = ½(25√(2)) units. R = 17.6776695297 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3137622

Let BF be a. b² = 18²+a² b = √(324+a²) units. b is AB. c = (a+14) units. It implies; calculating a. (a+14) ~ √(324+a²) √(324+a²) ~ 18 Cross Multiply. 18(a+14) = 324+a² 18a+252 = 324+a² a²-18a+72 = 0 a²-12a-6a+72 = 0 a(a-12)-6(a-12) = 0 (a-12)(a-6) = 0 It implies; a ≠ 12 units. a = 6 units. Again, a is BF. Recall. b = √(324+a²) And a = 6 units. b = √(324+36) b = √(360) b = 6√(10) units Again, b is AB. Recall Again. c = (a+14) And a = 6 units. c = 6+14 c = 20 units Again, c is BH. I...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3137614

Let the side length of the inscribed regular triangle be a. 2b² = a² b = √(a²/2) b = ½√(2)a units. c = 2b c = √(2)a units. c is the side length of the ascribed square. d² = b²+c² d² = (√(2)a)²+(½√(2)a)² d² = 2a²+¼(2a²) d² = ¼(10a²) d = ½√(10)a units. tane = √(2a)/(½√(2)a) e = atan(2)° f = 180-45-60-atan(2) f = 11.5650511771° Calculating a. 0.5*a*0.5√(10)asin11.5650511771 = 24 0.15849364905a² = 24 a² = 151.425625846 a = 12.3055120107 units. Again, a is the is the side length of the inscr...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3137609

Let AB be 1 unit. Let BC be a. Therefore; CD = (1-a) units. It implies; tan60 = b/(0.5a) 2b = √(3)a b = ½√(3)a units. b is the height of the bigger inscribed equilateral triangle. c = 1-b c = ½(2-√(3)a) units. d = ½(a) units. tan60 = 2e/(1-a) 2e = √(3)-√(3)a e = ½(√(3)-√(3)a) units. e is the height of the big inscribed equilateral triangle. f = b-e f = ½√(3)a-½(√(3)-√(3)a) f = ½(2√(3)a-√(3)) units. g = ½(a)+½(1-a) g = ½(1) g = ½ units. It implies;...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3136722

Let angle AOB be a. Let b = AD = OB = OD. b is the radius of the quarter circle. Calculating b. 2² = 2b²-2b²cosa 2b²cosa = 2b²-4 b²cosa = b²-2 cosa = (b²-2)/b² --- (1). 4² = b²+(2b)²-4b²cosa 16 = 5b²-4b²cosa 4b²cosa = 5b²-16 cosa = (5b²-16)/(4b²) --- (2). Equating (1) and (2). (b²-2)/b² = (5b²-16)/(4b²) 4(b²-2) = 5b²-16 4b²-8 = 5b²-16 b² = 8 b = √(8) b = 2√(2) units. Again, b is the radius of the quarter circle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3136720

Let r = 1 unit. a = (R+1) units. b = (R-1) units. c²+b² = a² c²+(R-1)² = (R+1)² c² = R²+2R+1-(R²-2R+1) c = √(4R) c = 2√(R) units. d = R+2-R d = 2 units. e = R+R e = 2R units. Calculating R. (2R)² = 2²+(2√(R))² 4R² = 4+4R R²-R-1 = 0 (R-½)² = (-½)²+1 (R-½)² = (5/4) R = ½±½√(5) It implies; R ≠ ½(1-√(5)) units. R = ½(1+√(5)) units. Therefore; R÷r is; ½(1+√(5))÷1 = ½(1+√(5))

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3135861

Notice. The ascribed quadrilateral is regular (square). Let a be it's side length. ½*ab = 3 ab = 6 b = (6/a) units. b is the base of the red area. ½*ac = 4 ac = 8 c = (8/a) units. c is the base of the green area d = a-(6/a) d = (a²-6)/a units. d is the height of the yellow area. e = a-(8/a) e = (a²-8)/a units. e is the base of the yellow area. Therefore; Calculating a. ½*d*e = 5 ½*(a²-6)/a*(a²-8)/a = 5 ((a²-6)(a²-8))/a² = 10 (a⁴-14a²+48) = 10a² a⁴-24a²...

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