Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2626884

Let the length and width of the rectangle be y and z respectively. Therefore; 7+7+x+y+z+7+7+x+y+z=66 Where 7 is the single side length of the square. It implies; 14+14+2x+2y+2z=66 Therefore; 2(x+y+z)=38 x+y+z=19 ------ (1) And; x+z=7 Therefore; x = 7-z ------- (2) z+z+2y=30 2z+2y=30 z+y = 15 y = 15 - z ------- (3) Substituting (2) and (3) in (1) (7-z)+(15-z)+z= 19 22-2z+z=19 z = 22-19 z = 3 Notice; At (3). y = 15 -z, and z = 3 Therefore; y =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2626874

Let YC, radius of the small circle be r. Notice; XY = CD = AD = AB = BC = single side length of square ABCD. Calculating r. πr² - ½(πr²) - 90π(XY)²/360 + ½(XY)² = 1 XY will be; √(r²+r²) = √(2r²) XY = √(2)r Therefore; πr² - ½(πr²) - π(√(2)r)²/4 + ½(√(2)r)² = 1 2πr² - πr² - πr² + 2r² = 2 2r² = 2 r² = 1 r (radius of the small circle) is; r = 1 unit. Therefore; XY = CD = AD = AB = BC = √(2)r And r = 1 XY = CD = AD = AB = BC = √(2) unit. Therefore area squar...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2626857

Shaded Area is; (13*8)-½(11*8)-2(½(13*4))+½(2+13)*8 = 104-44-52+60 = 164-96 = 68 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2626841

Calculating r, radius of the inscribed semi circle. 49r²-168r+144 = 0 Therefore; r = 12/7 units. It implies; Calculating x. x = 5-2(12/7)-(23/20) x = 59/140 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2625826

Let a be the side length of the inscribed square. Calculating a², area of the inscribed square. tan60 = a/b b = ⅓(√(3)a) cm. tan30 = a/c c = √(3)a cm. Calculating a. It implies; b+a+c = 13 ⅓(√(3)a)+a+√(3)a = 13 (4√(3)+3)a = 39 a = (39(4√(3)-3))/39 a = (4√(3)-3) cm. It implies; a² = (4√(3)-3)² = 48-24√(3)+9) = (57-24√(3)) cm² = 3(19-8√(3)) cm² = 15.4307806183 cm² Again, a² is the area of the inscribed square.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2625487

Let a be the radius of the inscribed circle. b = (2a+4) units. b is the side length of the square. c = ½(b) c = (a+2) units. d = c-a d = 2 units. e = c-3 e = (a+2)-3 e = (a-1) units. Calculating a. It implies; a² = 2²+(a-1)² a² = 4+a²-2a+1 2a = 5 a = 2.5 units. Again, a is the radius of the inscribed circle. Recall. b = (2a+4) units. And a = 2.5 units. b = 2(2.5)+4 b = 9 units. Again, b is the side length of the ascribed square. Area Yellow is; b²...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2625427

Let a be the side length of the square. a² = 98 a = √(98) a = 7√(2) cm. tanb = (0.5*7√(2))/(7√(2)) b = atan(1/2)° c = 90-b c = atan(2)° d = 45-b d = (45-atan(1/2))° e = 180-(45-atan(1/2))-atan(2) e = 98.1301023542° f = 135-e f = 36.8698976458° (7√(2)/sin98.1301023542) = (g/sin45) g = 7.0710678119 cm. Therefore, area shaded is; 0.5*7.0710678119*7√(2)sin36.8698976458 = 21 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2625393

Let a be the side length of the square. tanb = a/0.5a b = atan(2)° c² = a²+(0.5a)² c = 1.1180339887a m. Observing similar plane shape (right-angled) side length ratios. 1.1180339887a - a 0.5a - d Cross Multiply. d = (0.5a²)/1.1180339887a d = 0.4472135955a m. e = c-d e = 1.1180339887a-0.4472135955a e = 0.6708203932a m. Calculating a. Recall. Area triangle ABC = 9 m² 0.5*0.6708203932a*0.5asin(atan(2)) = 9 a² = 60 a = √(60) a = 2√(15) m. Notice! AB =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2625079

Let a be the radius of the ascribed circle. a² = (0.5*6)²+b² b = √(a²-9) units. c = a-b c = (a-√(a²-9)) units. d = ½(c) d = ½(a-√(a²-9)) units. d is the radius of the small inscribed circle. e = a+b e = (a+√(a²-9)) units. f = ½(e) f = ½(a+√(a²-9)) units. f is the radius of the big inscribed circle. It implies; f = 2d ½((a+√(a²-9))) = 2*½(a-√(a²-9)) ((a+√(a²-9)) = 2(a-√(a²-9)) a+√(a²-9) = 2a-2√(a²-9) a = 3√(a²-9) a² = 3²(a²-9) 81 = 9a²-a² 8a² = 81 a = √(81/8...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2625009

Sir Mike Ambrose is the author of the question. Area Green is; Area triangle two lengths 16/3 units and ⅓√(181) units, and angle 138.0127875°, the angle they form + Area triangle two lengths 8 units and ⅓√(85) units, and angle 102.528807708°, the angle they form. = ½(16/3)(√(181)/3)sin(138.0127875) + ½(8)(√(85)/3)sin(102.528807708) = 8 + 12 = 20 square units.

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