Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1850096

Please, move a step right to view the analyzed figure. Notice; ABC triangle is Isosceles. a² = 3²-2² a = √(5) units. Where a is AD. tanb = ½(√(5))° b = atan(½(√(5)))° b = 48.18968510422° Where b is angle ACB. c = 0.5(180-48.18968510422) c = 65.90515744789° Where c is angle ABC. d = 90-65.90515744789 d = 24.09484255211° Where d is angle BAD cos65.90515744789 = e/3 e = 1.22474487139 units. Where e = x = BF. tan24.09484255211 = f/1.22474487139 f = 0.54772255751 u...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1849189

Let the ascribed quarter circle be 2 units. 2a² = 2² a² = 2 a = √(2) units. b = (2-√(2)) units. c² = (√(2))²+(2-√(2))² c² = 2+4-4√(2)+2 c² = 8-4√(2) c = √(8-4√(2)) units. d² = √(8-4√(2))²-√(2)² d² = 8-4√(2)-2 d² = 6-4√(2) d = √(6-4√(2)) units. tane = √(2)/(√(6-4√(2))) e = atan(√(2)/(√(6-4√(2)))) e = 67.5° f = 180-45-67.5 f = 67.5° tan67.5 = 2/g g = 0.82842712475 units. h = 2-0.82842712475 h = 1.17157287525 units. Therefore; Area Triangle S is; 0.5*2*0.82842712475 = 0.82842712475 sq...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1849033

Let the square side be 2 units. a = √(2²+1²) a = √(5) units. b = atan(2)° c = atan(½)° d² = 8-8cos150 d = 3.86370330516 units. e = 0.5d e = 1.93185165258 units. f = 90-15-atan(½) f = 48.43494882292° g² = 5+1.93185165258²-2√(5)*1.93185165258cos48.43494882292 g² = 3 g = √(3) units. (√(3)/sin48.43494882292) = (1.93185165258/sinf) f = 56.56505117716° Therefore the required angle alpha is; 180-56.56505117716-atan(2) = 180-120 = 60°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1848404

Calculating AQ using similar plane shapes (triangle) rule. 12 = 6 6 = a cross multiply. 12a = 36 a = 3 cm. Where a is AQ. PQ = √(6²+3²) PQ =√(45) PQ = 3√(5) cm. BQ = 12-3 BQ = 9 cm. Calculating BR using similar plane shapes (triangle) rule. 6 = 9 3 = b cross multiply. 6b = 27 b = ½(9) cm. b = 4.5 cm. Where b is BR. QR = √(9²+4.5²) QR = ½(9√(5)) cm. Calculating angle PQS. tanc = PS/PQ c = atan((4.5√(5))÷(3√(5))) c = atan(3/2)° Where c is angle PQS. Angle AQP = atan(2)° Angle AQD =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1847527

a = atan(2)° b = 180-2a b = (180-2atan(2))° Where b is angle FBP. Notice; CQ = 5 cm. Calculating FP using sine rule. Let FP be c. (c/sin(180-2atan(2))) = (5/sin(atan(2))) c = 4.472135955 cm. Therefore; PQ is; CF-CQ-FP = √(10²+5²)-5-4.472135955 = (3√(5)-5) cm. = 1.7082039325 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1845316

Let the AB be 4 cm. Therefore, radius of the inscribed semi circle is; radius = 2 cm. Let the radius of the inscribed circle be r. Calculating r. (2+r)² = (4-r)²+(2-r)² 4+4r+r² = 16-8r+r²+4-4r+r² 4+4r+r² = 20-12r+2r² r²-16r+16r = 0 (r-8)²+16-(-8)² = 0 (r-8)² = 64-16 (r-8)² = 48 r = 8±4√(3) It implies; r ≠ (8+4√(3)) cm. r = (8-4√(3)) cm. Therefore; MQ = 2+r MQ = 2+(8-4√(3)) MQ = (10-4√(3)) cm. a = 2-r a = 2-(8-4√(3)) a = (4√(3)-6) cm b = acos(a/MQ) b = acos((4√(3)-6))/(10-4√(3))) b = 72...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1843774

Let the inscribed blue quarter circle radius be r. Calculating r. Considering similar triangles rule. r = 4 8 = r (cross multiply). r² = 32 r = 4√(2) cm. Therefore; Area blue (area quarter circle) in terms of π is; ¼(4√(2))²π = ¼(32π) = 8π cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1843756

Calculating the side length of square ABCD. 2x² = 8² x² = 32 x = 4√(2) units. Where x is AB. Let BG = y Semi circle radius r is; r = ½(4√(2)+y) units. Calculating y. sin45 = ½(4√(2)+y)/7 ½(√(2)) = (4√(2)+y)/14 √(2) = (4√(2)+y)/7 y = 7√(2)-4√(2) y = BG = 3√(2) units. Calculating r. r = ½(4√(2)+y) And y = 3√(2) units. r = ½(4√(2)+3√(2)) r = ½(7√(2)) units. a = x-r a = 4√(2)-½(7√(2)) a = ½√(2) units. Calculating BF. (BF)² = r²-a² (BF)² = (½(7√(2)))²-(½√(2))² (BF...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1842275

a² = 10²+10²-2*10*10cos150 a = 19.31851652578 cm. Where a is AF which is equal AE. (19.31851652578/sin150) = (10/sinb) b = 15° Where b is angle BAF. tan15 = c/10 c = 2.67949192431 cm. It implies; Area Green Region is; Area square with side 10 cm - 2(area triangle with height 2.67949192431 cm and base 10 cm. = 10² - ½*2(10*2.67949192431) = 100-26.7949192431 = 73.2050807569 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1841371

Let the small inscribed square side be a. Let b be the side of the big inscribed square. It implies; b = 10-a Therefore; Calculating a. tan(90-75) = a/(10-a) a(1+tan15) = 10tan15 a = (10tan15)/(1+tan15) a = ⅓(15-5√(3)) cm. b = 10-⅓(15-5√(3)) b = ⅓(15+5√(3)) cm. Blue area is; a²+b² = (⅓(15-5√(3)))²+(⅓(15+5√(3)))² = ⅓(200) cm²

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