Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2741533

Sir Mike Ambrose is the author of the question. Let the width of the rectangle be x and the length be y. Let the unknown inscribed length of the rectangle be z. Calculating z. √(74)²+√(58)²=√(34)²+z² Therefore z is; z = √(98) units. Calculating x (width). x² = 156-2√(98)*√(58)cos68.2 x² = 100.003444042 x = 10.0001722006 units. Calculating y (length). y = √(34)cos30.9651378262 + √(58)cos23.199577149 y = 11.999876013 units. Therefore area rectangle is; xy...

Mathematics Question and Solution

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

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/2731933

Let the single side length of the blue area/square be x. Therefore; 5² = x² + (x+½(5-x))² 5² = x² + (½(5+x))² 25 = x² + ¼(25+10x+x²) 100 = 4x²+25+10x+x² 5x²+10x-75=0 x²+2x-15=0 (x+1)²=15+1 x = -1±√(16) x = -1±4 Therefore; x ≠ -5 x = 3 cm Therefore blue area/area square is; x² = 3² = 9 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2721532

Calculating shaded red area. Let a be the diameter of the inscribed big half circle. Therefore, 2a is the diameter of the bigger half circle or a is the radius of the bigger half circle. a² = 6²+b² b = √(a²-36) cm. 6 - c a - √(a²-36) Cross Multiply. c = (6√(a²-36))/a cm. Again. a - √(a²-36) √(a²-36) - d Cross Multiply. d = (a²-36)/a cm. e = a+d e = (2a²-36)/a cm. Calculating a. 10² = ((2a²-36)/a)²+((6√(a²-36))/a)² 100 = ((4a⁴-144a²+1296)/a²)+((36a²-1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2721240

Angle ABC = ½(angle BAC) since angle ACD and angle BCD are equal and length BC = length AD + length AC It implies; a +½(100)+100 = 180 a = 180-150 a = 30° a is angle ACB. b = ½(a) b = 15° b is angle BCD = angle ACD Therefore, the required angle, alpha is; Let it be c. It implies; 100+15+c = 180 c = 180-115 c = 65° Again, c is the required angle alpha.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2718323

Calculating area of the bigger square. Let the bigger square side length be a. b² = 4 b = 2 cm. b is the side length of the big square. c = (a+2) cm. d = (a-2) cm. Calculating a², area of the bigger square. (a-2)²+(a+2)² = 10² a²-4a+4+a²+4a+4 = 100 2a²+8 = 100 2a² = 92 a² = ½(92) a² = 46 cm² Again, a² is the area of the bigger square.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2715326

Let a be the radius of the circle. a² = 6²+8²+2*6*8cosb a² = 100+96cosb --- (1). a² = 16²+8²-2*16*8cosb a² = 320-256cosb --- (2). Equating (1) and (2). 100+96cosb = 320-256cosb cosb = 220/352 b = 51.3178125465° It implies, using (2). a² = 320-256cos51.3178125465 a = 12.6491106407 units. Again, a is the radius of the circle. Therefore, area of the circle is; πa² = π(12.6491106407)² = 160π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2712637

Notice. MN = CQ = 6 units. MQ = 6+2 MQ = 8 units. a² = 8²+6² a = √(100) a = 10 units. a is CM. 3 - 6 r - 10 Cross Multiply. 6r = 30 r = 5 units. r is the radius of the inscribed circle. 5² = 3²+b² b = √(25-9) b = √(16) b = 4 units. c = r+b c = 9 units. c is BQ. d = c+6 d = 15 units. d is BC. tane = 6/8 e = atan(3/4)° e is angle CMQ = angle BAC. tan(atan(3/4)) = 15/f ¾ = 15/f 3f = 4*15 f = 20 units. f is AB. g² = 20²+15² g = √(400+225)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2710974

Calculating area of the inscribed red square. Let a be the side length of the inscribed red square. 2b² = a² b = √(a²/2) b = ½√(2)a units. 10² = a²+(½√(2)a)²-2a*½√(2)acos(45+90) 100 = ½(3a²)+a² 100 = ½(5a²) 5a² = 200 a² = 40 square units. Where a² is the area of the inscribed red square.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2693501

Let the side length of the inscribed regular hexagon which is equal the side length of the ascribed regular nonagon be 1 unit. a = ⅙*180(6-2) a = 120° a is the single interior angle of the inscribed regular hexagon. b = ⅑*180(9-2) b = 140° b is the single interior angle of the ascribed regular nonagon. c = 140-120 c = 20° d = 180-20-60 d = 100° (1/sin100) = (e/sin20) e = 0.3472963553 units. (1/sin100) = (f/sin60) f = 0.8793852416 units. g = 1-f g = 0.1206147584 u...

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Mathematics Question and Solution

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Mathematics Question and Solution

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