Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3106934

Sir Mike Ambrose is the author of the question. Let the square side length be 2 units. Calculating Area C and D. Notice. Area C = Area D. C = ½(2²-¼(2*2π)) C = ½(4-π) square units. Therefore; D = ½(4-π) square units. Calculating Area F. F = ¼(1²*π)-½(1*1) F = ¼(π)-½ F = ¼(π-2) square units. Calculating Area E. E = ½(¼*2*2π)-F E = ½(π)-¼(π-2) E = ½(π)-¼(π)+½ E = ¼(π)+½ E = ¼(π+2) square units. Calculating Area A. A = ½*½(2*2)-F A = 1-¼(π-2) A = ½(3)-...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3106087

Area Triangle is; 16tan(sin–¹(1/3)) = 5.65685424949 square units. Or sina = ⅓ a = asin(⅓)° tana = b/4 b = 4tan(asin(⅓)) units. c = 2b c = 8tan(asin(⅓)) units. Therefore, area triangle is; ½*4*c = 2c = 16tan(asin(⅓)) = 5.65685424949 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3105346

Calculating angle a. Let the length of the ascribed rectangle be 2 units. Therefore it's width is 1 unit. b² = 2²+1² b = √(5) units. b is the diagonal of the ascribed rectangle. tanc = 1/2 c = atan(½)° tand = 1/0.5 d = atan(2)° e = 2d e = 2atan(2)° f = ½(180-2atan(2)°) f = (90-atan(2))° g² = 2(0.5)²-2(0.5)²cos(2atan(2)) g = 0.894427191 units. h = 90-f-c h = 90-(90-atan(2))-atan(½) h = (atan(2)-atan(½))° h = 36.8698976458° j² = 0.894427191²+√(5)²-2*0.894...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3104876

Calculating x. Notice. The composite plane shape is drawn to scale. 2(60+x)+x = 180 120+2x+x = 180 3x = 60 x = 20° Or (60-x)+2(60+10) = 180 60-x+140 = 18 200-x = 180 x = 200-180 x = 20°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3104249

It will be; Let the single side length of the regular heptagon be 5 units. Therefore; (1/x) + (1/y) will be; (1÷(√(50-50cos129))) + (1÷(5+10cos51)) = 0.19934168501≈ 0.2 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3103921

Sir Mike Ambrose is the author of the question. Let the side length of the introduced equilateral triangle inscribing the circle be 2 units. Calculating r, radius of the circle. cos30 = 1/r r = 1/cos30 r = 2/√(3) r = ⅓(2√(3)) units. Area circle is; πr² = π(⅓(2√(3)))² = ⅓(4π) square units. = 4.18879020479 square units. tan30 = a/1 a = ⅓√(3) units. b = r-a b = ⅓(2√(3))-⅓√(3) b = ⅓√(3) units. tan60 = ⅓√(3)/c c = ⅓ units. cos60 = (⅓)/d d = ⅔ units. or si...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3103006

Sir Mike Ambrose is the author of the question. Let the ascribed circle's radius be 1 unit. Area ascribed circle is; π(1)² = π square units. Calculating shaded area. Notice. The side length of the inscribed regular heptagon is equal the radius of the circle, 1 unit. Therefore, shaded area is; 6(area equilateral triangle with side length 1 unit)-Area sector with radius 1 unit and angle 120°-2(area sector with radius 0.5 units and angle 120°)-2(area equilateral triangle with...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3102550

Sir Mike Ambrose is the author of the question. Let AB be 2 units. Area ABCD is; 2² = 4 square units. Let the side length of the ascribed square be x. Calculating x. a = (x-2) units. b = (x-1) units. It implies; x² = a²+b² x² = (x-2)²+(x-1)² x² = x²-4x+4+x²-2x+1 x²-6x+5 = 0 Therefore; x = 5 units. Again, x is the side length of the ascribed square. c = x-2 c = 3 units. d = x-1 d = 4 units. tane = 3/4 e = atan(3/4)° f = 90-2e f = 90-2atan(3/4) f...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3102466

Let the two equal smaller inscribed circle radius be r. Let the radius of the smallest inscribed circle be q. The radius is the biggest inscribed circle is 0.5 units. Therefore; x+2r=0.5 r=½(0.5-x) ----- (1) (0.5+r)²=(0.5+x)²+(x+r)² 0.25+r+r²=0.25+x+x²+x²+2xr+r² r=x+2x²+2xr r(1-2x)=x+2x² r=(x+2x²)/(1-2x) ------- (2) Equating (1) and (2). (x+2x²)/(1-2x)=½(0.5-x) 4x²+2x=(0.5-x)(1-2x) 4x²+2x=0.5-x-x+2x² 2x²+4x-0.5=0 x²+2x-0.25=0 x²+x=0.25+1² (x...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3100770

Length EF will be; √(49+50-70√(2)cos(tan-¹(¾)+45)) EF = √(85) units.

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