Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2471810

Shaded Area is; Area circle with diameter 30 cm - 7(area circle with diameter 10 cm) = π(½(30))² - 7π(½(10))² = π(15)² - 7π(5)² = 225π - 175π = 50π cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2471583

Sir Mike Ambrose is the author of the question. Let CD = 2 units. Therefore; Calculating AB. AB = 1 + (2sin18*tan72 + 2sin36)tan54 AB = 1 + 3.07768353717tan54 AB = 1 + 4.23606797749 AB = 5.23606797749 units. It implies; Length AB ÷ Length CD to 2 decimal places is 5.23606797749 ÷ 2 = 2.61803398874 ≈ 2.62

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2470996

x² = (r+1)² - (r-1)² Therefore x = 2√(r) unit. y² = (r+1)² - (2√(r))² Therefore y = √(r²-2r+1) y = √(r-1)² y = (r-1) unit. z = (r+1) - y z = (r+1) - (r-1) Therefore z = 2 units. a = r+r Therefore a = 2r units. A sincere study and analysis of the composite circles deduced a right-angled triangle with two adjacent sides, 2 units and 2√(r) units respectively, and hypotenuse 2r units. Therefore; Calculating r, the radius of the three congruent circles. (2r)² = 2² + (2√(r))² 4r² = 4 + 4r 4r²...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2470982

a = 2atan(2)° b = ½(180-a) b = ½(180-2atan(2))° tan(½(180-2atan(2))) = c/(0.5x) 0.5 = c/(0.5x) c = 0.25x cm. d = x-0.25x d = 0.75x cm. Calculating x. 0.5*x*0.75x = 24 x² = 24/(0.75*0.5) x² = 64 cm. x = 8 cm. Area red is; (8*8)-24-½(4²)π = 64-24-8π = 40-8π = 8(5-π) cm² = 14.8672587713 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2470304

Let AB = a Notice. AB = CD a² = 2²+1²+2*2*1cosx a² = 5+4cosx --- (1). a² = 3²+4²-2*3*4cosx a² = 25-24cosx --- (2). Calculating x. Equating (1) and (2). 5+4cosx = 25-24cosx 28cosx = 20 cosx = (20/28) x = acos(5/7) x = 44.4153085972° y = 180-x y = 180-44.4153085972 y = 135.5846914028° Let BC = b. AD = 3b. (3b)² = 1²+4²+2*4*1cosc 9b² = 17+8cosc b² = ⅑(17+8cosc) --- (3). b² = 2²+3²-2*3*2cosc b² = 13-12cosc --- (4). Calculating b. Equating (3) and (4). ⅑(17+8cosc) = 13-12cosc 17+8cosc...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2469936

Diameter of the inscribed circle is; ½(AB) = ½(8) = 4 units. a = ½(4) a = 2 units. a is the radius of the inscribed circle. Calculating red length. Notice, the red length is the side of the inscribed regular hexagon. Let the red length be b. c² = 2b²-2b²cos120 120° is the single interior angle of the inscribed regular hexagon. c = √(3)b units. d = ½(b) units. e² = 2(2)² e = √(8) e = 2√(2) units. f² = 2(8)² f = 8√(2) units. f is the diagonal of the ascribed square. Therefore, red lengt...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2469365

Let the side length of the regular hexagon be 1 unit. a = ½(180-108) 108° = single interior angle of the regular pentagon. 180° = sum of angles on a straight line. a = ½(72) a = 36° tan36 = b/0.5 b = 0.363271264 units. Therefore, area blue is; ½(0.5a*b) = ½(0.5*0.363271264) = 0.090817816 square units. Calculating Lilac Area. Let the side length of the regular pentagon be c. d = ½(c) units. e² = 2-2cos120 e = √(3) units. f² = 2c²-2c²cos108 f = 1.6180339887c units. Calculating c. f² =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2468855

Let the side length of the regular pentagon be 1 unit. a² = 1²+1² a = √(2) units. b = ½(a) b = ½√(2) units. c = ½(360-108-108) c = ½(360-216) c = ½(144) c = 72° d = 180-108-45 d = 72-45 d = 27° (0.5√(2)/sin108) = (e/sin45) e = 0.5257311121 units. (0.5√(2)/sin108) = (f/sin27) f = 0.3375401519 units. sin18= g/0.3375401519 g = 0.1043056432 units. Length AB = 2e+2g+1 AB = 2.2600735106 units. h = 1-0.3375401519 h = 0.6624598481 units. j = AB-e j = 2.2600735106-2*0.5257311121 j = 1.2086112...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2468509

Let the side length of the ascribed regular pentagon be 1 unit. Let a be the side length of the inscribed regular hexagon. b = 180-108-(108-120) b = 72-60 b = 12° Where; 108° is the single interior angle of the regular ascribed pentagon. 120° is the single interior angle of the inscribed regular hexagon. 180° is the sum of the interior angles of a triangle. (a/sin108) = (c/sin12) c = 0.2186112889a units. d = a+c d = a+0.2186112889a d = 1.2186112889a units. e² = 2a²-2a²...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2468422

Let the radius of the two equal inscribed semi circles be 1 unit each. 2² = a²+1² a² = 3 a = √(3) units. b = 1+1+a b = (2+√(3)) units. b is the height of the ascribed right-angled triangle. It implies, observing similar right-angled triangle side length ratios. c - 1 (2+√(3)) - √(3) Cross Multiply. √(3)c = 2+√(3) c = (2+√(3))/√(3) c = ⅓(2√(3)+3) c = 2.1547005384 units. c is the base of the ascribed right-angled triangle. Therefore, shaded area (blue area) fraction is...

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