Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2718556

Calculating x. a²+4² = 5² a² = 25-16 a = 3 units. Therefore; 3*(x+5) = 4*(2x-5) 3x+15 = 8x-20 5x = 35 x = 7 units. It implies; b = 3+x+5 And x = 7 units. b = 3+7+5 b = 15 units. c = 4+2x-5 And x = 7 units. c = 4+2*7-5 c = 18-5 c = 13 units. d = ½(c) d = 6.5 units. e = ½(b) e = 7.5 units. f = e-3 f = 4.5 units. Calculating g, radius of the circle. g² = d²+f² g² = 6.5²+4.5² g = ½(5√(10)) square units. = 7.9056941504 units. Or g² = 7.5²+(6...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2718337

Calculating the required angle, x. a = ⅐*180(7-2) a = ⅐(900) a is the single interior angle of the regular heptagon. It implies, the required angle, x is; Notice. Length IC is parallel to length EK, meaning 'angle ICE is alternative to angle CEK'. Therefore; x = ½(180-a) x = ½(180-⅐(900)) x = ½*⅐(1260-900) x = ½*⅐(360) x = ⅐(180)° x = 25.7142857143°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2718331

Calculating shaded area. Let the equal lengths of the ascribed trapezoid be a. Calculating a. a² = 6²+(18-a)² 0 = 36+324-36a 36a = 360 a = 10 units. Therefore, shaded area is; ½(10+18)*6-½(6*18) = 84-54 = 30 square units.

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/2718315

Calculating area of the square. sina = 28/35 a = asin(28/35)° b = a-45 b = (asin(28/35)-45)° c²+28² = 35² c² = 35²-28² c = 21 units. cos(asin(28/35)-45) = 21/d d = 21.2132034356 units. d is the diagonal of the square. Let e be the side length of the square. Therefore, square area, e² is; 2e² = 21.2132034356² e² = ½(450) e² = 225 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2717461

Let the side length of the big inscribed regular triangle be a. It implies, the side length of the biggest inscribed regular triangle is 3a. Calculating a. 2² = a²+(2a)²-2*a*2acos120 4 = 5a²+2a² 7a² = 4 a = ⅐(2√(7)) units. 3a is; = 3*⅐(2√(7)) = ⅐(6√(7)) units. Therefore, area inscribed blue triangle is; ½*⅐(2√(7))*⅐(6√(7))sin60 = ⅐*6*½√(3) = ⅐(3√(3)) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2717022

Let a be the radius of the inscribed half circle. Let b be the radius of the inscribed circle. c = (12+a) units. d = (12-a) units. Calculating a. (12+a)² = 12²+(12-a)² 48a = 144 3a = 9 a = 3 units. Again, a is the radius of the inscribed half circle. e = (3+b) units. f = (3-b) units. g² = (3+b)²-(3-b)² g² = 12b g = √(12b) units. h = (12-√(12b)) units. Calculating b. (12+b)² = (12-√(12b))²+(12-b)² 48b = 144-24√(12b)+12b 4b = 12-2√(12b)+b 2√(12b) = 12-3...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2716941

a² = 10²+5² a = √(125) a = 5√(5) units. a is the side length of the square. b² = 2a² b² = 2(5√(5))² b = √(250) b = 5√(10) units. b is the diagonal of the square. c = 90+atan(10/5) c = (90+atan(2))° d = 90-atan(5/10) d = (90-atan(0.5))° d = atan(2)° cos(atan(2)) = e/10 e = 4.472135955 units. f = 5√(5)-e f = 6.7082039325 units. 2 = g/4.472135955 g = 8.94427191 units. h = 90-atan(8.94427191/6.7082039325) h = atan(4/3)° j = atan(3/4)° k = 180-45-j k = 98....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2716933

Calculating x. a =⅕*180(5-2) a = ⅕*180*3 a = 108° a is the single interior angle of the regular pentagon. b = ½(180-108) b = ½(72) b = 36° c = ½(180-2(180-108)) c = ½(180-144) c = ½(36) c = 18° d = 180-18-108 d = 108-126 d = 54° Therefore, the required angle, x is; x = 45+b-d x = 45+36-54 x = 81-54 x = 27°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2716120

Let a be the radius of the ascribed half circle. tanb = a/(0.5a) b = atan(2)° b is angle OCA = angle DCE. c = atan(½)° c is angle OAC. d² = a²+(½(a))² d = √(¼(5a²)) d = ½√(5)a² units. d = 1.1180339887a units. d is AC. cos(atan(½)) = e/(2a) e = 1.788854382a units. e is AE. f = e-d f = 0.6708203933a units. f is CE. Calculating a. 0.5*0.6708203933a*0.5asin(atan(2)) = 3 0.3a² = 6 3a² = 60 a² = 20 a = √(20) a = 2√(5) units. Again, a is the radius of the ascrib...

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