Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2725731

Let a be the radius of the inscribed half circle. b = (a-1) units. It implies, calculating a. a² = (a-1)²+5² a² = a²-2a+1+25 2a = 26 a = 13 units. Again, a is the radius of the inscribed half circle. c = a+5 c = 18 units. c is the width of the ascribed rectangle. Recall. b = a-1 And a = 13 units. b = 13-1 b = 12 units. d = 2a d = 26 units. tane = 5/12 e = atan(5/12)° cos(atan(5/12)) = f/26 f = 24 units. g = f+1 g = 25 units. g is the length of the a...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2724460

Sir Mike Ambrose is the author of the question. The Equation of the curve is; y = -¾(x²) + 6x Therefore shaded area is; Area under the curve at point (4, 0) and ((4+√(6)), 0) - Area rectangle with length (15/2) units and width √(6) units. = ½(21√(6)) - (½(15)*√(6)) = ½(21√(6)) - ½(15√(6)) = ½(6√(6)) = 3√(6) square unit.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2724171

Sir Mike Ambrose is the author of the question. Area R exactly, in square cm decimal form is; Area square with side 2√(11) cm - Area triangle with side 2√(11) cm and 4.97194902762 cm, and angle 15° - Area trapezoid with parallel sides 1.97064750999 cm and 2√(11) cm, and height 2√(11) cm. = (2√(11))² - ½(2√(11)*4.97194902762)sin15 - ½(1.97064750999+2√(11))*2√(11) = 44 - 4.26794919252 - 28.5358983847 = 11.1961524228 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2723682

a² = 20²+15² a = √(625) a = 25 units. 25 - 20 15 - b Cross Multiply. 25b = 15*20 b = 12 units. b is the height of the shaded inscribed region, triangle. Again; 25 - 20 20 - c Cross Multiply. 25c = 20*20 c = 16 units. c is the base of the shaded inscribed region, triangle. Therefore, area shaded inscribed region, triangle is; ½*bc = ½*12*16 = 96 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2723527

Let a be the radius of the larger circumscribed circle. Calculating a. cos30 = 4/r ½√(3) = 4/r r = 8/√(3) r = ⅓(8√(3)) units. Therefore, area circumscribed circle is; π(8/√(3))² = ⅓(64π) square units. Calculating the area of the two inscribed green circles. b² = 3²+8²-2*3*8cos60 b² = 73-24 b = √(49) units. b = 7 units. Let c be the radius of the big inscribed green circle. Calculating c. 3c+7c+8c = 3*8sin60 18c = 12√(3) c = ⅓(2√(3)) units. Area big inscr...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2722645

Sir Beg is the author of the question. Notice. The composite plane shape is not drawn to scale. 2a² = 2² a² = 2 a = √(2) cm. a is OC = AC. 2b² = √(2)² b² = 1 b = 1 cm. b is OF = CF = DE, the length of the inscribed rectangle CDEF. Calculating EF = CD, the width of the inscribed rectangle CDEF. Let it be d. c = (1+d) cm. It implies; 2² = 1+(1+d)² 4 = 1+1+2d+d² d²+2d-2 = 0 (d+1)² = 2+1² d = -1±√(3) Therefore; d = (√(3)-1) cm. d = 0.7320508076 cm. Again, d...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2722594

Let a be the radius of the circle and also the width of the rectangle. b = (a-x) units. c = (a-2x) cm. d = a+c d = (2a-2x) cm. d is the length of the rectangle. 12² = (2a-2x)²+(a-x)² 144 = 4a²-8ax+4x²+a²-2ax+x² 144 = 5a²-10ax+5x² --- (1). 2x - (a-x) (a-x) - (2a-2x) Cross Multiply. (a-x)² = 4x(a-x) a-x = 4x a = 5x --- (2). Substituting (2) in (1). 144 = 5(5x)²-10(5x)x+5x² 144 = 125x²-50x²+5x² 144 = 80x² 36 = 20x² 18 = 10x² x² = 9/5 x = √(9/5) x = ⅕(3√(5...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2722585

a = r+r a = 2r units. a is the diameter of the circle. b = 3+1 b = 4 units. c²+(2r)² = 4² c = √(16-4r²) units. It implies; 2√(3) - 4 r - √(16-4r²) Cross Multiply. 4r = 2√(3(16-4r²)) 2r = √(3(16-4r²)) 4r² = 3(16-4r²) 4r² = 48-12r² 16r² = 48 r = √(3) units. Therefore, area circle is; πr² = π√(3)² = 3π square units. Calculating the required angle, theta. Let it be x. c = √(16-4r²) And r = √(3) units. c = √(16-4√(3)²) c = √(16-12) c = √(4) c = 2 u...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2721819

Sir Mike Ambrose is the author of the question. Area yellow exactly in square cm is; Area triangle with height 10√(13)/9 cm and base 5√(13)/6 cm = ½(10√(13)/9*5√(13)/6) = ½((13*50)/(54)) = (13*25)/54 = 325/54 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2721532

Calculating shaded red area. Let a be the diameter of the inscribed big half circle. Therefore, 2a is the diameter of the bigger half circle or a is the radius of the bigger half circle. a² = 6²+b² b = √(a²-36) cm. 6 - c a - √(a²-36) Cross Multiply. c = (6√(a²-36))/a cm. Again. a - √(a²-36) √(a²-36) - d Cross Multiply. d = (a²-36)/a cm. e = a+d e = (2a²-36)/a cm. Calculating a. 10² = ((2a²-36)/a)²+((6√(a²-36))/a)² 100 = ((4a⁴-144a²+1296)/a²)+((36a²-1...

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