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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2545752

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Notice! The inscribed blue area is a square. Let the inscribed blue area side length be a. Calculating a. tan63 = a/b b = a/tan63 --- (1). tan43 = a/c c = a/tan43 --- (2). It implies; (2)+a+(1) = 10 (a/tan43)+a+(a/tan63) = 10 1.07236871a+a+0.5095254495a = 10 2.5818941595a = 10 a = 3.8731254584 cm. Again, a is the side length of the inscribed blue square (square). Therefore, area blue is; a²  = 3.8731254584² = 15 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2545733

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a = (10-b) cm. tan30 = (10-b)/10 10 = 10√(3)-√(3)b b = (10√(3)-10)/√(3) b = ⅓(30-10√(3)) cm. tan30 = ⅓(30-10√(3))/c 3c = 30√(3)-30 c = (10√(3)-10) cm. d = 10-c d = 10-(10√(3)-10) d = (20-10√(3)) cm. Therefore; Area blue is; ½*d*10 = ½*(20-10√(3))*10 = 50(2-√(3)) cm² = 13.3974596216 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2545391

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sin45 = 12/a a = 12√(2) cm. a is AC. Notice ! AB = 2BC Let BC = b Therefore; AB = 2b cm. AC = AB+BC AC = 2b+b AC = 3b cm. And AC = 12√(2) cm. It implies; 12√(2) = 3b b = 4√(2) cm. Again, b is BC. AB = 2b  And b is 4√(2) cm. AB = 2*4√(2) AB = 8√(2) cm. c = 180-30-45 c = 105° (12/sin105) = (d/sin30) d = (6/sin105) cm. d = 6.2116570825 cm. e = 8√(2)-d e = (8√(2)sin105-6)/sin105 cm. e = 5.1020514165 cm  (((8√(2)sin105-6)/sin105) /sin30) = (f/sin1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2543787

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Calculating r, radius of the circle. a² = 4 a= 2 cm. a is the side length of the small square. b = (r-2) cm. It implies; r² = 4²+(r-2)² r² = 16+r²-4r+4 4r = 20 r = 5 cm. Again, r is the radius of the circle. Calculating x, side length of the big square. c = (x-5) cm. It implies; 5² = (x-5)²+√(21)² 25-21 = x²-10x+25 x²-10x+21 = 0 Resolving the above quadratic equation via factorization approach to get x. x²-7x-3x+21 = 0 x(x-7)-3(x-7) = 0 (x-3)(x-7) = 0...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2543745

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c² = 7²+24² c = 25 units. a² = (7+25)²+24² a = 40 units. b = ½(a) b = ½√(40) units. b = 20 units. d²+20² = 25² d = √(625-400) d = √(225) d = 15 units. It implies, R radius of the circle is; R(15+20+25) = (15*20) 60R = 300 R = 5 units. Calculating x. x² = 20²+5² x = √(425) x = 5√(17) units. x = 20.6155281281 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2542212

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a*15 = 540 a = 108/3 a = 36 cm. a is AB. b = a-R b = (36-R) cm. b is OB. It implies; Calculating R. R² = 15²+(36-R)² R² = 225+1296-72R+R² 72R = 1521 R = 21.125 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2541591

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Calculating the area of the blue inscribed rectangle and the area of the ascribed regular pentagon. πr² = 36π r = 6 cm. r is the radius of the inscribed circle. sin(0.5*108) = 6/a a = 7.416407865 cm. b = r+a b = 13.416407865 units. tan36 = 13.416407865/c c = 18.4661012231 cm. d = 2c d = 36.9322024461 cm. d is the width of the inscribed blue rectangle. Therefore; Area Blue Inscribed Rectangle is; 36.9322024461*50 = 1846.6101223052 cm² ≈ 1850 cm² to 3 significan...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2541439

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a = (1-r) units. b²+r² = (1-r)² b = √(1-2r) units. c = (0.5+r) It implies; (0.5+r)² = r²+√(1-2r)² ¼+r+r² = r²+1-2r 3r = 1-¼ 3r = ¾ r = ¼ units. Or (½+r) = (1-r) Where; AB = BC = 1 unit. AE = BE = ½ units. Therefore; 2r = 1-½ 2r = ½ r = ¼ units. Or r will be; r = ½(½(AB)) r = ¼(AB) And AB = 1 unit. r = ¼(1) r = ¼ unit.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2540565

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Let a be the radius of the inscribed sector. b = (2+a) units. Therefore; (2+a)² = a²+6² 4+4a+a² = a²+36 4a = 32 a = 8 units. Again, a is the radius of the inscribed sector. b = (2+a) And a = 8 units. b = 2+8 b = 10 units. c = 6+b And b = 10 units. c = 6+10 c = 16 units. Therefore, Painted Area is; ½(10+16)*8 = 4*26 = 104 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2540051

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Let AD = CD = a. Calculating a. b² = 2a² b = √(2)a units. b is AC. c²+3² = a² c = √(a²-9) units. c is DE. a² = d²+7² d = √(a²-49) units. e = c-d e = (√(a²-9)-√(a²-49)) units. e is CB. It implies; b² = e²+10² (√(2)a)² = (√(a²-9)-√(a²-49))²+100 2a² = a²-9-2√((a²-9)(a²-49))+a²-49+100 2√((a²-9)(a²-49)) = 42 √((a²-9)(a²-49)) = 21 (a²-9)(a²-49) = 441 a⁴-49a²-9a²+441 = 441 a⁴-58a² = 0 a² = 58 a = √(58) units. Therefore, Orange Area is; ½(a²) = ½*√(58)² =...
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