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Mathematics Question and Solution

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

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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

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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.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2736984

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Let the radius of the quarter circle be a. b = (a-4) units. c = (a-2) units. It implies; a² = (a-4)²+(a-2)² a² = a²-8a+16+a²-4a+4 a²-12a+20 = 0 a²-2a-10a+20 = 0 a(a-2)-10(a-2) = 0 (a-2)(a-10) = 0 It implies; a = 10 units. Again, a is the radius of the quarter circle. Therefore, area quarter circle is; ¼*10*10π = 25π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2736972

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Sir Mike Ambrose is the author of the question. Area red is; Area triangle with height (18/7) units and base (24/7) units. = ½((18/7)*(24/7)) = (216/49) square unit. Area orange is; Area triangle with two side length (24/7) units each, and angle 73.739795292° = ½(24/7)²sin73.739795292 = 5.6424489796 square units. Therefore; Area red ÷ Area orange exactly in decimal form is; (216/49) ÷ 5.6424489796 = 0.78125 = 78125/10000 = 125/160 = 25/32

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2736960

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Sir Mike Ambrose is the author of the question. Area purple is; Area trapezium with two parallel side length (58/15) cm and (46/5) cm, and height 4 cm. = ½((58/15)+(46/5))*4 = (392/15) cm² Area brown is; Area triangle with height 8 cm and base 12 cm - Area triangle with height (14/5) cm and base (56/15) cm. = ½(8*12) - ½((14/5)*(56/15)) = 48 - (392/75) = (3208/75) cm² It implies; Area purple ÷ Area brown exactly is; (392/15) ÷ (3208/75) = 245/401

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2735876

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Calculating area shaded blue. a² = (2+√(3))²+(3+2√(3))² a² = 4+4√(3)+3+9+12√(3)+12 a² = 28+16√(3) a = √(28+16√(3)) units. a = 7.4641016151 units. a is the hypotenuse of the ascribed right-angled triangle. Let b be the radius of the inscribed half circle. c = (2+√(3))-b c = (3.7320508076-b) units. d = √(28+16√(3))-(3+2√(3)) d = 7.4641016151-6.4641016151 d = 1 unit. Calculating b. (3.7320508076-b)² = 1²+b² 13.9282032303-7.4641016151b = 1 12.9282032303 = 7.4641016151b...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2735755

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Calculating red angle. Let the equal two red angles be x each. An irregular hexagon is drawn/derived. It implies; 90+90+109+(360-15)+2x = 180(6-2) 289+345+2x = 720 634+2x = 720 2x = 720-634 2x = 86 x = 43° Where x is the required angle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2735481

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Blue area is; 8(area quarter circle with radius 4 cm - area triangle with height and base 4 cm each) -   8(area quarter circle with radius 2 cm - area triangle with height and base 2 cm each) = 8(4π-8) - 8(π-2) = 32π-64-8π+16 = 24π-48 = 24(π-2) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2735475

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a = ½(12) a = 6 cm. a is the radius of the inscribed half circle. c = (6+b) cm. b is the radius of the inscribed circle. d = (12-b) cm. Calculating b. (6+b)²+6² = (12-b)² 36+12b+b²+36 = 144-24b+b² 72+12b = 144-24b 36b = 72 b = 2 cm. Again, b is the radius of the inscribed circle. e²+6² = 12² e² = 144-36 e = √(108) e = 6√(3) cm. Painted Area is; 2(area sector with radius 12 cm and angle 60°) - Area triangle with height 6√(3) cm and base 12 cm + Area half circle...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2735470

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tan60 = a/2 a = 2√(3) units. a is the height of the inscribed red regular triangle. b = 2+a b = (2+2√(3)) units. Area green is; (½*4*(2+2√(3)))-(½*2√(3)*4) = 4+4√(2)-4√(3) = 4 square units.
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