Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2569002

Let the single side length of the square be (x+10) unit. Calculating x. 10²=(5+x)²+5² 100-25-25=10x+x² Therefore; x²+10x-50=0 Resolving the above quadratic equation via completing the square approach to get x. (x+5)²=50+25 x =-5±√(75) Therefore; x ≠ -5-5√(3) x = (5√(3)-5) cm. Square length is; (x+10) And x = (5√(3)-5) cm. = 10+5√(3)-5 cm. = (5+5√(3)) units. Area of the square is; (5+5√(3))² = 100+50√(3) = 50(2+√(3)) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2567562

Calculating x, length AC = length AB. a = 14+4 a = 18 units. b = ½(18) b = 9 units. c²+9² = x² c = √(x²-81) units. d = 9-4 d = 5 units. It implies; 13² = √(x²-81)²+5² 169 = x²-81+25 x² = 169+56 x² = 225 x = 15 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2567548

Let CD be b units. Let BC be a units. c = 2+4 c = 6 units. Observing similar plane shape (right-angled) side length ratios. a - 6 6 - b Cross Multiply. ab = 36 --- (1). Let alpha be d. tand = 4/b --- (2). tan(2d) = 6/a --- (3). tan(2d) = (2tand)/(1-tan²d) --- (4). It implies, substituting (4) in (3). (2tand)/(1-tan²d) = 6/a And at (2), tand = 4/b (8/b)/(1-(16/b²)) = 6/a 8b/(b²-16) = 6/a (8*36)/(b²-16) = 6 48 = b²-16 b² = 64 b = 8 units. Again, b is...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2567496

tana = (0.5*4)/6 tana = ⅓ a = atan(⅓)° b = 2a b = 2atan(⅓)° c = ½(180-b) c = ½(180-36.8698976458) c = 71.5650511771° d = c+½(180-c) d = 125.7825255886° e = 180-d e = 54.2174744115° f = 180-125.7825255886-2atan(⅓) f = 17.3475767656° (6/sin17.3475767656) = (g/sin(2atan(⅓))) g = 12.0737473608 units. Therefore, the required length, x is; sin54.2174744115 = x/12.0737473608 x = 9.7947331922 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2567460

Let a be the side length of the inscribed blue square. b² = 2a² b = √(2)a units. 2c² = a² c = √(a²/2) c = ½√(2)a unit. It implies; 1² = (√(2)a)²+(½√(2)a)² 1 = 2a²+½(a²) 1 = ½(5a²) 5a² = 2 a² = ⅖ a = √(2/5) units. Again, a is the side length of the inscribed blue square. Area inscribed blue square is; a² = √(2/5)² = ⅖ square units. = 0.4 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2567101

For y = (25/16) unit. x is; (25/16) = (5/4)^(2x) 2x = 2 x = 1 unit. Gradient of the curve at x equal 1 unit is; dy/dx = (25/8)ln(5/4) At y = 0 at the tangent, x is; x = (50ln(5/4)-25)/50ln(5/4) units. Therefore; Area Yellow exactly as a single fraction is; Area triangle with height (√(881)/16)sin(180-atan(25/16)) units and base (-(50ln(5/4)-25)/50ln(5/4)) unit. = ½*(-(50ln(5/4)-25)/50ln(5/4))*(√(881)/16)sin(180-atan(25/16)) = (25-50ln(5/4))/(64ln(5/4)) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2566596

Radius of the inscribed circle, r is; √(5)r + r = 2 r = 2/(√(5)+1) unit. r = ½(√(5)-1) unit Therefore; Area Yellow exactly is square units is; 2(area triangle with height (√(5)-1) unit and base ½(√(5)-1) units). = 2(½*(√(5)-1)*½(√(5)-1) = ½(√(5)-1)² = (3-√(5)) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2565882

Sir Mike Ambrose is the author of the question. tan30 = a/6 a = 2√(3) cm. b = (6-2√(3)) cm. sin30 = 2√(3)/c c = 4√(3) cm. 6d+2√(3)d+ 4√(3)d = 6*2√(3) (6+6√(3))d = 12√(3) d = (72√(3)-216)/(-72) d = (3-√(3)) cm. d = 1.2679491924 cm. d is the radius of the inscribed circle. e = 6-d e = (3+√(3)) cm. Or e = 4√(3)-(√(3)(3-√(3))) e = 4√(3)-3√(3)+3 e = (3+√(3)) cm. e = 4.7320508076 cm. f² = 4.7320508076²+6²-2*6*4.7320508076cos60 f = 5.4772255751 cm. (5.4772255751/si...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2565814

Showing that quadrilateral ABCD is cyclic. Notice! For quadrilateral ABCD to be cyclic; Angle ADC will equal 2 times angle ABC. Angle ADC = 2(Angle ABC) It implies; Since Angle ABC = 60° Angle ADC must equal 2(60°) = 120° The working! a² = 10²+6²-2*10*6cos60 a = 2√(19) units. a = 8.7177978871 units. a is AC. Therefore, calculate angle ADC. Let it be b. (2√(19))² = 6²+4²-2*6*4cosb 76 = 52-48cosb 48cosb = 52-76 cosb = -24/48 b = acos(-1/2) b = 120° b is ang...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2565695

Blue area is; ½(9x2)+½(√(68)x√(17)) = 9+½(34) = 9+17 = 26 cm²

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