Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2740015

Area of the isosceles triangle is; The radius of the semi circle is 10 units. Therefore the single side length of the square is; 2 * 10 = 20 units. Therefore red area is; Area triangle with two side length 20 units each, and angle (90-2atan½)° = ½(20²)sin(90-2atan½) = ½(240) = 120 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2740003

Let the side length of the congruent square be 1 unit. Therefore area of the shape is; 8 square unit. Shaded area is; Area triangle with two length √(2) units and √(5)/4 units, and angle (135-atan(0.5))° = ½(√(2)*(√(5)/4))sin(135-atan(0.5)) = 3/8 square units. Therefore shaded fraction is; Area shaded or Area blue ÷ Area of the complete shape. = (3/8) ÷ 8 = 3/64

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2739995

Yellow area is; 12(area equilateral triangle with radius 2 units) = 12(½(4)sin60) = 24sin60 = 12√(3) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2738459

Let k = 1 unit. a = 5+1 a= 6 units. a is the side length of the square. b = ½(a) b = 3 units. c = b-1 c = 2 units. tand = 6/2 d = atan(3)° e = (180-atan(3))° It implies; (3/sin(180-atan(3))) = (1/sinf) f = 18.4349488229° g = 180-(180-atan(3))-f g = 53.1301023542° h = 90-g h = 36.8698976458° j² = 3²+6²-2*3*6cos36.8698976458 j = 4.0249223595 units. (6/sink) = (4.0249223595/sin36.8698976458) o = 63.4349488229° l = 180-63.4349488229 l = 116.565051177...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2738452

Let the radius of the ascribed circle be 1 unit. Therefore; Area blue is; 12(area sector with radius 1 unit and angle 60° - area equilateral triangle with single side length 1 unit). = 12((60*π/360)-½(sin60)) = 12((π/6)-¼(√(3))) = (2π-3√(3)) square units. Area yellow is; Area circle with radius 1 unit - Area blue. = π - (2π-3√(3)) = (3√(3)-π) square units. Therefore; Area blue : Area yellow is; (2π-3√(3)) : (3√(3)-π) = (2π-3√(3))/(3√(3)-π)

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2738443

Purple area is; Area triangle with height 4 units and base 8 units - (area square with side 4 units - area quarter circle with radius 4 units) - area sector with radius 4 units and angle (180-2atan(½))° + area triangle with two side 4 units each, and angle (180-2atan(½))°. = ½(4*8) - (4²-¼(4²π)) - (180-2atan(½))*4²π÷360 + ½(4²)sin(180-2atan(½)) = 16 -16 + 4π - 17.7143794847 + 6.4 = 4π - 11.314379485 = 1.25199112936 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2738432

Sir Mike Ambrose is the author of the question. Let the single side length of the congruent regular hexagons be 1 unit. Therefore; Area red is; Area trapezium with two parallel side length ½(3√(3)) units and 2√(3) units, and height ½ units + Area triangle with height 2√(3) units and base 2 units. ½*½(½(3√(3))+2√(3)) + ½(2√(3)*2) = (7√(3))/8 + 2√(3) = (23√(3)/8) square units. Area orange is; Area trapezium with two parallel side length 1 unit and 2 units, and height ½√(3)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2738027

8² = (3+a)²+(5+a)² Where a is the radius of the inscribed circle. 64 = 9+6a+a²+25+10a+a² 64 = 34+16a+2a² a²+8a-15 = 0 (a+4)² = 15+16 a = -4±√(31) It implies; a = √(31)-4 a = 1.5677643628 units. Area triangle is; 0.5(3+1.5677643628)(5+1.5677643628) = 0.5*4.5677643628+6.5677643628 = 15 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2737593

Let the side length of the ascribed regular hexagon be 2 units. Calculating Area B. ½*1*1sin60 = ¼√(3) square units. a = ⅙*180(6-2) a = 120° a is the single interior angle of the ascribed and inscribed regular hexagon. b² = 2-2cos120 b = √(3) units. c² = 2²+2²-2*2*2cos120 c² = 12 c = 2√(3) units. d = c-b-1 d = 2√(3)-√(3)-1 d = (√(3)-1) units. sin30 = e/(√(3)-1) e = ½(√(3)-1) units. Calculating Area A. ½(1+2)*½(√(3)-1) = ¼(3√(3)-3) square units. Therefor...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2737135

Notice. ABCD is a square. Let AB equal 1 unit. sin70 = 1/a a = 1.0641777725 units. a is AE. b = 90-70 b = 20° b is angle DAE. c = 90-45-20 c = 25° c is angle BAF. cos25 = 1/d d = 1.103377919 units. d is AF. e² = 1.103377919²+1.0641777725²-2*1.103377919*1.0641777725cos45 e = 0.8302778924 units. e is EF. Therefore, the required angle is; Let it be f. (0.8302778924/sin45) = (1.0641777725/sinf) f = 65° Again, f is the required angle.

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