Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2481368

Calculating x (radius of the small inscribed circle) 28² + x² = (56-x)² 784 + x² = 3136 - 112x + x² 112x = 2351 x = 21 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2479474

a = 108-90 a = 18° cos18 = b/1 b = 0.9510565163 units. c = 1+0.9510565163 c = 1.9510565163 units. c is the side length of the regular ascribed pentagon. sin18 = d/1 d = 0.3090169944 units. e² = 2*1.9510565163²-2*1.9510565163²cos108 e = 3.1568757573 units. Therefore, green area is; Area regular pentagon with side length 1.9510565163 cm - 5(area square with side length 1 unit) - 5(area triangle with height 0.9510565163 units and base 0.3090169944 units. = 2(0.5*1.9510565163*1.9510565163si...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2479159

a = 120π*2*30÷360 a = ⅓(60)π a = 20π cm. a = 62.8318530718 cm. a is the length of arc of the sector and also, the base circumference of the cone. 20π = 2πb b = 10 cm. b is the radius of the base of the circular cone. 30² = 10²+c² c² = 900-100 c = √(800) c = 20√(2) cm. c is the height of the cone. Volume of the cone is; ⅓*10*10π*20√(2) = ⅓(2000√(2))π cm³ = 2961.9219587722 cm³ Calculating the volume outside the sphere. 2*30d+20d = 20*20√(2) 80d = 400√(2) d = 5√(2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2479061

Let the square side be 2a cm. It implies; a is the radius of the inscribed semi circle. Let b equal the radius of the big inscribed quarter circle. Therefore; (a+b)² = a²+(2a)² a²+2ab+b² = 5a² 2ab+b² = 4a² --- (1). (2a-2)² = 2²+b² 4a²-8a+4 = 4+b² b² = 4a²-8a --- (2) From (2). b = √(4a²-8a) --- (3). Substituting (2) and (3) in (1) to get a. 2ab+b² = 4a² 2a√(4a²-8a)+4a²-8a = 4a² 2a√(4a²-8a) = 8a 4a²(4a²-8a) = 64a² 4a²-8a = 16 a²-2a-4 = 0 Resolving the abo...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2478964

Let the two equal lengths of the isosceles triangle be 1 unit each. a = 180-2(12+36) a = 180-96 a = 84° b² = 2-2cos84 b = 1.3382612127 units. c = 180-12-30 c = 138° (d/sin30) = (1.3382612127/sin138) d = 1 unit. It implies; The required angle x, is; x = ½(180-36) x = ½(144) x = 72°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2478768

Let the side length of the regular pentagon be 1 unit. a² = 2-2cos108 108° is the single interior angle of the regular pentagon. a = 1.6180339887 units. a is BE. b = 108-60 b = 48° c = ½(180-48) c = 66° d = 180-60-c d = 54° e = 180-d-b e = 78° f = 180-e f = 102° g = 180-f-72 g = 6° g is angle BEG = angle BGE. h = 180-2g h = 168° h is angle EBG. j = h-72 j = 96° j is angle CBG. k = a-1 k = 1.6180339887-1 k = 0.6180339887 units.. k is BF l² = 0....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2478557

a² = 3²+1² a = √(10) units. a is AB. It implies; AC = ½√(10) units. tanb = 1/3 b = atan(1/3)° tanc = ½√(10)/√(10) c = atan(½)° d² = (½√(10))²+√(10)² d² = ½(5)+10 d = √(25/2) d = ½(5√(2)) units. Observing similar right-angled triangle side length ratios. ½√(10) - ½(5√(2)) e - √(10) Cross Multiply. 10 = 5√(2)e e = √(2) units. sinf = √(2)/(0.5√(10)) f = 63.4349488229° tanh = 3/1 h = atan(3)° h = 71.5650511771° j = 63.4349488229+71.5650511771 j...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2478126

tan30 = a/12 a = 12tan30 a = 4√(3) cm. a is the side length of the square. b = ½(12)+a b = (6+4√(3)) cm. c = ½(a) c = 2√(3) cm. Observing similar right-angled triangle side length ratios. 2√(3) - d (6+4√(3)) - 4√(3) Cross Multiply. 24 = d(6+4√(3)) d = 12/(3+2√(3)) d = 12(3-2√(3))/(-3) d = 4(2√(3)-3) cm. Shaded Area exactly is; Area square with side 4√(3) cm - Area triangle with height 4√(3) cm and base 4(2√(3)-3) cm. = (4√(3))²-½*4(2√(3)-3)*4√(3) = 48-48+24√(3...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2477392

x+50+180-a = 180 a = x+50 --- (1). x+30+180-b = 180 b = x+30 --- (2). 3x+180-2a+180-2b = 180 3x+180-2(a+b) = 0 2(a+b) = 3x+180 --- (3). Therefore, substituting (1) and (2) in (3) to get x. 2(x+50+x+30) = 3x+180 4x+160 = 3x+180 x = 20° Therefore; 3x is; = 3*20 = 60°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2477288

Notice, triangle ABC is isosceles. Let BC, the base of triangle ABC = 1 unit. a = ½(180-45) a = ½(135) a = 67.5° a is angle ABC = angle ACB. b = a-45 b = 22.5° b is angle DBE. c = a-30 c = 37.5° c is angle BCD. d = 180-45-67.5 d = 67.5° d is angle BEC. (1/sin67.5) = (e/sin45) e = 0.7653668647 units. e is CE. f = 180-67.5-37.5 f = 180-105 f = 75° f is angle BDC. (1/sin75) = (g/sin67.5) g = 0.9564704736 units. g is CD. h² = 0.9564704736²+0.7653668647²-...

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