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...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3226920

Calculating Area Blue Inscribed Circle. Let a be the radius of the blue inscribed circle. b = (a+5) cm. c = 12-a-5 c = (7-a) cm. d = (16-a-5) d = (11-a) cm. It implies; (a+5)² = (7-a)²+(11-a)² a²+10a+25 = 49-14a+a²+121-22a+a² a²+10a+25 = 170-36a+2a² a²-46a+145 = 0 (a-23)² = -145+(-23)² (a-23)² = 384 a = 23±√(384) a = 23±8√(6) It implies; a ≠ 23+8√(6) units. a = 23-8√(6) units. a = 3.40408205773 units. Area Blue Inscribed Circle is; πa² = π(23-8√(6))²...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3226397

Sir Mike Ambrose is the author of the question. Calculating Exactly Angle Alpha - Angle Beta. Let Alpha be x. Let Beta be y. (6+6√(2))² = 8²+10²-2*8*10cosa 36+72√(2)+72 = 164-160cosa 160cosa = 164-36-72-72√(2) 160cosa = 56-72√(2) 20cosa = 7-9√(2) cosa = (7-9√(2))/20 a = acos((7-9√(2))/20)° a = 106.642318459° 8² = 10²+(6+6√(2))²-2*10(6+6√(2))cosb 64 = 100+36+72√(2)+72-(120+120√(2))cosb (120+120√(2))cosb = 144+72√(2) cosb = (144+72√(2))/(120+120√(2)) cosb = ⅕(3√(2)) b...

Mathematics Question and Solution.

https://onlineedumath.com/update/mathematics-question-and-solution/3226027

Calculating Area Yellow. a = 22+½(24) a = 34 units. Considering a careful, smart analysis of the quarter circle and the inscribed irregular yellow pentagon area. Therefore, yellow area is; 2(area triangle with base ½(24) units and height 34 units) + Area rectangle with length 22 units and base ½(24) units. 2(½*½(24)*34)+(½(24)*22) = (12*34)+(12*22) = 12(34+22) = 12(56) = 672 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3226014

Calculating Area Red Inscribed Square. Let x be the side length of the red square. a = (5-x) cm. b = 5-½(x) b = ½(10-x) cm. Therefore; Calculating x. 5² = (5-x)²+(½(10-x))² 25 = 25-10x+x²+¼(100-20x+x²) 10x-x² = ¼(100-20x+x²) 40x-4x² = 100-20x+x² 5x²-60x+100 = 0 x²-12x+20 = 0 x²-10x-2x+20 = 0 x(x-10)-2(x-10) = 0 (x-10)(x-2) = 0 It implies; x ≠ 10 cm. x = 2 cm. Therefore, area red square is; x² = 2² = 4 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3225253

Sir Mike Ambrose is the author of the question. Area red in its exact simplest decimal square units form is; Area triangle with height and base 1.24899959968 units and 1.20185042515 units + Area triangle with side ⅔ units and (26/15) units, and angle (75+atan(√(3)/7))° = ½(1.24899959968*1.20185042515) + ½(⅔*(26/15))sin(75+atan(√(3)/7)) = 0.75055534995+0.57767089074 = 1.32822624069 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3224810

Calculating Blue Shaded Area. tana = 10/6 a = atan(5/3)° b = 135-a b = (45+atan(3/5))° (6/sin(45+atan(3/5))) = (c/sin(atan(5/3))) c = 5.3033008589 units. Therefore, required blue area is; ½*6*5.3033008589sin45 = 11.25 square units.

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Mathematics Question and Solution

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