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...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2515207

a = 180-60 a = 120° b² = 4²+6²-2*4*6cos120 b = 2√(19) units. b = 8.7177978871 units. b is the side length of the inscribed regular triangle ABC (equilateral triangle). (8.7177978871/sin120) = (4/sinc) c = 23.4132244464° d = 60+c d = 83.4132244464° Therefore, length BD is; (BD)² = 6²+8.7177978871²-2*6*8.7177978871cos83.4132244464 BD = 10 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2515184

a² = 3²+4² a = √(25) a = 5 units. 3b+4b+5b = 3*4 12b = 12 b = 1 unit. b is the radius of the inscribed circle. tanc = 3/4 c = atan(¾)° d = ½(180-atan(¾)) d = 71.5650511771° tane = 4/3 e = atan(4/3)° f = ½(180-atan(4/3)) f = 63.4349488229° sin71.5650511771 = 1/g g = 1.0540925534 units. sin63.4349488229 = 1/h h = 1.1180339888 units. j = 90-71.5650511771 j = 18.4349488229° k = 90-63.4349488229 k = 26.5650511771° l = 180-26.5650511771-18.4349488229 l =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2514912

Let the bigger square side length be 1 unit. a² = 2(1)² a = √(2) units. a is the diagonal of a bigger square. Let b be the side length of the big square. c = (1+b) units. It implies; Calculating b. √(2)² = (1+b)²+b² 2 = 1+2b+b²+b² 2b²+2b-1 = 0 Resolving the quadratic equation via completing the square approach to get b, side length of the big square. 2b²+2b-1 = 0 b²+b = ½ (b+½)² = ½+(½)² (b+½)² = ¼(3) b = -½±½√(3) It implies; b = ½√(3)-½ b = ½(√(3)-1) units...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2513313

Let the side length of the regular pentagon be 1 unit. a² = 1+0.5²-2*0.5cos108 a = 1.2486060205 units. (1.2486060205/sin108) = (0.5/sinb) b = 22.3861775592° c = 180-108-b c = 49.6138224408° d = c-b d = 49.6138224408-22.3861775592 d = 27.2276448816° e = 180-2c e = 180-2*49.6138224408 e = 80.7723551184° f = 180-27.2276448816-e f = 72° (1.2486060205/sin72) = (g/sin27.2276448816) g = 0.6006698572 units. h² = 0.6006698572²+0.5²-2*0.6006698572*0.5cos49.6138224408...

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