Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3229436

Calculate Area Green. Green Area is; Area square with length 4 units - Area quarter circle with radius 2 units - Area square with length 2 units - 3(Area square with length 1 unit) - Area square with length 3 units + Area quarter circle with radius 3 units - Area square with length 2 units + Area quarter circle with radius 2 units - Area quarter circle with radius 1 unit. = 16 - ¼(4π) - 4 - 3 - 9 + ¼(9π) - 4 + ¼(4π) - ¼(π) = 16 - 20 + ¼(9π - π) = ¼(8π) - 4 = 2π - 4 = 2(π-...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3229073

Calculating angle x. Let the radius of the inscribed half circle be 1 unit. Therefore, 2 units is the radius of the ascribed quarter circle. a = (2-y) units. b² = 1²+y² b = √(1+y²) units. b is BD. c = 2(OB) Where OB = 2 units, the radius of the ascribed quarter circle, and also the diameter of the inscribed half circle. c = 2*2 c = 4 units. c is double the radius of the ascribed quarter circle. Calculate y. It implies; 4 ~ √(1+y²) √(1+y²) ~ y Cross Multiply....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3228563

Calculating a. tan60 = a/b √(3) = a/b b = a/√(3) b = ⅓(√(3)a) units. tan60 = 2a/c √(3) = 2a/c c = 2a/√(3) c = ⅓(2√(3)a) units. It implies; b+a+2a+c = 1 ⅓(√(3)a)+a+2a+⅓(2√(3)a) = 1 ⅓(3√(3)a)+3a = 1 √(3)a+3a = 1 (3+√(3))a = 1 a = 1/(3+√(3)) a = (3-√(3))/6 a = ⅙(3-√(3)) units. a = 0.21132486541 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3228559

Calculating x, the required angle. a = ⅛*180(8-2) a = ⅛*180*6 a = ½*45*6 a = 135° a is the single interior angle of the regular octagon. b = ⅕*180(5-2) b = ⅕*180*3 b = 36*3 b = 108° b is the single interior angle of the regular pentagon. c = ½(a) c = ½(135) c = 67.5° d = 360-2b-c d = 360-2(108)-67.5 d = 360-216-67.5 d = 76.5° Therefore x, the required angle is; x = 180-d x = 180-76.5 x = 103.5°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3228162

Calculating the red shaded area. Notice. 10 cm is the radius of each of the four equal inscribed circles. tana = 10/(4*10) a = atan(¼)° a = 14.0362434679° b = 3*10 b = 30 cm. (10/sin14.0362434679) = (30/sinc) c = 133.313856658° d = 180-a-c d = 180-14.0362434679-133.313856658 d = 32.6498998741° e = 90-d e = 90-32.6498998741 e = 57.3501001259° f = 180-2a f = 180-2*14.0362434679 f = 151.927513064° Therefore, the required red shaded area is; Area rectangle...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3227864

Calculating Angle x. Let 1 units be BC = CD = AD. a² = 2-2cos80 a = 1.28557521937 units. a is BD. b = 80-½(180-80) b = 80-50 b = 30° b is angle ABD. Therefore, the required angle x is; (1.28557521937/sinx) = (1/sin30) x = asin(0.64278760969) x = 40° Where x is angle BAD.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3227716

Calculating angle x. Let length blue be 1 unit. a = 180-20-10 a = 150° (1/sin10) = (b/sin150) b = 2.87938524157 units. c = 180-50-20-30-10 c = 180-110 c = 70° (2.87938524157/sin70) = (d/sin(30+10)) d = 1.96961550602 units. e² = 1+1.96961550602-2*1.96961550602cos50 e = 1.53208888623 units. Therefore, the required angle x is; (1.53208888623/sin50) = (1/sinx) x = asin(0.5) x = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3227706

Calculating Parallelogram ABCD Perimeter. 4² = x²+y²-2xycos60 16 = x²+y²-xy 16+xy = x²+y² --- (1). 6² = x²+y²-2xycos120 36 = x²+y²+xy 36-xy = x²+y² --- (2). Equating (1) and (2). 16+xy = 36-xy 2xy = 20 xy = 10 x = 10/y --- (3). Calculating y. Substituting (3) in (1). 16+(10/y)y = (10/y)²+y² 26 = (100/y²)+y² 26y² = 100+y⁴ y⁴-26y²+100 = 0 (y²-13)² = -100+(-13)² (y²-13)² = 169-100 y² = 13±√(69) It implies; y ≠ √(13-√(69)) units. y = √(13+√(69)) units. y...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3227155

Sir Mike Ambrose is the author of the question. Calculating Shaded Area ÷ Purple Area to 2 decimal places. Let AB, side length of the regular octagon be 5 units. Radius, r of the inscribed circle is; 5*(2/5) r = 2 units. a = ⅛(180*6) a = 135° 2b² = 25 b = ½(5√(2)) units. c = 180-0.5(135)-45 c = 67.5° sin22.5 = ½(5√(2))/d d = 9.23879532511 units. Where d is the side length of the regular triangle (purple area). Area purple is; 0.5*9.23879532511*9.2387953...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3226924

Calculating Total Inscribed Shaded Area. Let x be the radius of the five congruent inscribed shaded circle. a = (14.9-2x) cm. b = 2x cm. Calculating x. (2x)² = x²+(14.9-2x)² 4x² = x²+222.01-59.6x+4x² x²-59.6x+222.01 = 0 Therefore; x ≠ 55.6076 x = 3.99244 x ≈ 4 cm. c = 6x c = 24 cm. c is the length of the ascribed rectangle. Total Shaded Area is; 3(area equilateral triangle with side 8 cm)+3(area circle with radius 4 cm)+½(area circle with radius 4cm) = 3...

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