Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2521779

Sir Beg is the author of the question. Let the square side length be 1 unit. 1² = 2a² a = √(1/2) a = ½√(2) units. a is AD = CD. Let the radius of the inscribed circle be b. Therefore; c² = 2b² c = √(2)b cm. d² = 2(1)² d = √(2) units. d is the diagonal of the square. Calculating b, radius of the inscribed circle. a+b+c = d ½√(2)+b+√(2)b = √(2) (1+√(2))b = ½√(2) b = ½√(2)/(1+√(2)) b = (½√(2)-1)/(-1) b = ½(2-√(2)) units. b = 0.2928932188 units. Again, b is the...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2521701

Sir Beg is the author of the question. Let AB be a cm, the side length of the ascribed equilateral triangle ABC. b² = 9 b = 3 cm. b is the side length of the inscribed square and the base of the inscribed triangle. c²+(½(a))² = a² c² = a²-¼(a²) c² = ¼(3a²) c = ½√(3)a cm. c is the height of the ascribed equilateral triangle ABC. d = c-3 d = ½√(3)a-3 d = ½(√(3)a-6) cm. d is the height of the inscribed triangle. It implies; Calculating a, the side length of the ascribed reg...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2520350

Let the single side length of the square be x. Therefore; ½(x*(3x/4))+½(4√(x²+(3x/4)²)+½(8x)+½(¼(x)(x-4))=x² (3x²)/8+2√(x²+(9x²/16))+4x+(x²/8)-(x/2)=x² 3x²+16√(x²+(9x²/16))+32x+x²-4x=8x² 16√(x²+(9x²/16))=4x²-28x 256(x²+(9x²/16))=(4x²-28x)² 256x²+144x²=16x⁴-228x³+784x² 16x⁴-228x³+784x²-400x²=0 16x⁴-228x³+384x²=0 x²-14x+24=0 x²-12x-2x+24=0 x(x-12)-2(x-12)=0 (x-2)(x-12)=0 Therefore; x≠2 x=12 units. Therefore, the side length of the square, ABCD is; 12 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2520309

Exact enclosed area R will be; Area sector with radius 2 unit and angle (90-2tan–¹(½)) - Area sector with radius 1 unit and angle (180-4tan–¹(½)) + Area isosceles triangle with two equal lengths 1 unit and angle (180-4tan–¹(½)). =((90-2tan–¹(½))x4π÷360) - ((180-4tan–¹(½))xπ÷360) + 0.5sin(180-4tan–¹(½)) = 1.1235011088 square unit.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2520300

tan30 = 6/a a = 6√(3) units. a is AB. cos30 = 6/b b = 4√(3) units. b is CR. c = a-b c = 2√(3) units. c is DR. d = 180-30 d = 150° d is angle DRQ. It implies, shaded area is; Area triangle with height 2√(3) units and base 6sin150 units. = ½*6*2√(3)sin150 = 6√(3)*½ = 3√(3) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2520273

a² = 6²+6²-2*6*6cos120 a = 6√(3) units. tanb = 3/(6√(3)) b = atan(2/(2√(3))) b = 16.102113752° c = 90-16.102113752 c = 73.897886248° d = 30+16.102113752 d = 46.102113752° It implies; cos46.102113752 = e/6 e = 4.1602514717 units. Therefore, area blue is; 0.5*4.1602514717*6sin73.897886248 = 11.9911209755 square units. ≈ 12 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2520241

Let the radius of the inscribed circle be a. tanb = 4/3 b = atan(4/3)° c = ½(b) c = ½(atan(4/3))° d = (3-a) units. It implies; tan(½(atan(4/3))) = a/(3-a) ½ = a/(3-a) 2a = 3-a a = 1 unit. Again, a is the radius of the inscribed circle. e = 4-1 e = 3 units. f = √(4²+3²)-(3-1) f = 5-2 f = 3 units. tang = 3/4 g = atan(3/4)° h = ½(180-g) h = ½(180-atan(3/4)) h = 71.5650511771° tanj = 3/3 j = atan(1) j = 45° It implies, the required angle, x is; x =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2520234

Let the inscribed square side length be a. 2² = 1²+b² b² = 4-1 b = √(3) units. c = 2b+a c = (a+2√(3)) units. c is BC. d = (a+1) units. sin60 = (a+1)/e ½√(3) = (a+1)/e √(3)e = 2a+2 e = ⅓(2√(3)a+2√(3)) units. e is CQ. f = a+e f = (a+⅓(2√(3)a+2√(3))) units. f is AC. Therefore, calculating a, side length of the square. Equating c and f. (a+2√(3)) = (a+⅓(2√(3)a+2√(3))) 2√(3) = ⅓(2√(3)a+2√(3)) 6√(3) = 2√(3)a+2√(3) 2√(3)a = 6√(3)-2√(3) 2√(3)a = 4√(3) a = 2 uni...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2520208

Let x be 1 unit. a - 1 1 - √(2) Cross Multiply. √(2)a = 1 a = ½√(2) units. It implies; Area yellow is; ½*1*½√(2) = ¼√(2) square units. Area Blue is; ½(√(2)+½√(2))*1 = ½(½(3√(2))) = ¼(3√(2)) square units. Therefore; Blue Area ÷ Yellow Area is; ¼(3√(2))÷¼√(2) = 3

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2515413

Let the side length of the two congruent inscribed regular hexagon be 1 unit each. Calculating the radius, a, of one of the two congruent inscribed yellow circles. tan60 = a/(0.5) a = 0.8660254038 units. a = ½√(3) units. Therefore, area inscribed yellow circles is; 2πr² = 2*π*(½√(3))² = 2*π*¼*3 = ½(3π) square units. Calculating the radius of the ascribed semi circle. b² = 2-2cos120 b = √(3) units. c = 2b c = 2√(3) units. d² = c²+0.5² Where d is the radius of th...

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