Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3161756

Green Area is; Area square with length 4 units - Area quarter circle with radius 2 units - Area square with length 2 units - 3(Area square with length 1 unit) - Area square with length 3 units + Area quarter circle with radius 3 units - Area square with length 2 units + Area quarter circle with radius 2 units - Area quarter circle with radius 1 unit. = 16 - ¼(4π) - 4 - 3 - 9 + ¼(9π) - 4 + ¼(4π) - ¼(π) = 16 - 20 + ¼(9π - π) = ¼(8π) - 4 = 2π - 4 = 2(π-2) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3161751

Calculating Length x. 7² = x²+(x+5)²-2*(5+x)xcosa 49 = 2x²+10x+25-(10x+2x²)cosa (10x+2x²)cosa = 2x²+10x-24 cosa = (2x²+10x-24)/(10x+2x²) --- (1). 7² = x²+5²+2*x*5cosa 49-25-x² = 10xcosa 24-x² = 10xcosa cosa = (24-x²)/(10x) --- (2). Therefore, length x is; Equating (1) and (2). (2x²+10x-24)/(10x+2x²) = (24-x²)/(10x) 20x³+100x²-240x = 240x+48x²-10x³-2x⁴ 2x⁴+30x³+52x²-480x = 0 x³+15x²+26x-240 = 0 Therefore; x = 3 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3161747

Calculating length c. cosd = 9/12 d = acos(3/4)° e = 2d e = 2acos(¾)° Therefore, length c is; cose = c/a cos(2acos(¾)) = c/12 c = 12cos(2acos(¾)) c = 12*⅛ c = ½(3) units. c = 1.5 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3159873

Let a be the diameter of the circle. x² = 6²+6²+2*6*6cosb x² = 72+72cosb --- (1). x² = 14²+a²-2*14acosb x² = 196+a²-28acosb --- (2). Notice. x is length AC b is angle ADC. cosb = 14/a --- (3). Calculating a. Equating (1) and (2). 72+72cosb = 196+a²-28acosb --- (4). Substituting (3) in (4). 72+72(14/a) = 196+a²-28a(14/a) 72+72(14/a) = 196+a²-28(14) 72a+1008 = 196a+a³-392a a³-268a-1008 = 0 Therefore; a = 18 units. Again, a is the diameter of the circ...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3159871

Let a = AB = AD = CD. b² = 2a² b = √(2)a units. b is BD. Notice. Triangle BCD is a right-angled triangle, where BC is the hypotenuse. c² =(√(2)a)²+a² c = 3a² c = √(3)a units. c is BC. cosd = (0.5*10)/a cosd = 5/a --- (1). d is angle CAD. From (1). sind = √(a²-25)/a --- (2). (√(3)a)² = a²+10²-2a*10cos(90-d) 3a² = a²+100-20a(cos90cosd+sin90sind) 2a² = 100-20a(sind) a² = 50-10asind --- (3). Calculating a. Substituting (2) in (3). a² = 50-10a√(a²-25)/a a² = 50-10√(a²-25) 10²(a²-25) =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3158612

Sir Mike Ambrose is the author of the question. Calculating Exactly Green Area. Let x be the side length of each of the three inscribed congruent squares. a² = 2x² a = √(2)x a is the diagonal of the squares. b = 45-30 b = 15° cos15 = c/(√(2)x) c = ½(√(3)+1)x units. c = 1.36602540378x units. d = c+x d = ½(1+√(3))x+x d = ½(3+√(3))x units. d is the height of the ascribed regular hexagon. e² = 2²+2²-2*2*2cos120 e = √(8+4) e = 2√(3) units. e = d. Calculating x....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3157726

Let a be the radius of the quarter circle. b = (a-6) units. c = (a-8) units. Let x be the side length of the square. x² = (a-8)²+(a-6)² x² = a²-16a+64+a²-12a+36 x² = 2a²-28a+100 x = √(2a²-28a+100) units. d = ½(45) d = 22.5° It implies; √(2a²-28a+100)² = a²+(a-8)²-2*a(a-8)cos22.5 2a²-28a+100 = 2a²-16a+64-2a²cos22.5+16acos22.5 36-12a = 14.7820725202a-1.84775906502a² 1.84775906502a²-26.7820725202a+36 = 0 Therefore; a = 13 units. Again, a is the radius of...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3157724

Calculating the radius of the inscribed circle. a²+6² = 10² a = √(64) a = 8 units. Let x be the radius of the inscribed half circle. Calculating x. b = (8-x) units. c = (6-x) units. It implies; x ~ (8-x) (6-x) ~ x Cross Multiply. x² = 48-14x+x² 14x = 48 x = ⅐(24) units. Recall. b = (8-x) And x = ⅐(24) units. b = 8-⅐(24) b = ⅐(32) units. tand = ⅐(24)÷⅐(32) d = atan(3/4)° Let y be the radius of the inscribed circle. tan(0.5atan(3/4)) = y/(⅐(32)) ⅓ = 7y/32 21y = 32 y = (32/21) uni...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3155706

Let the three equal lengths be x. a = x+x+x+8+8 a = (16+3x) units. a is the diameter of the ascribed half circle. b = ½(a) b = ½(16+3x) units. b is the radius of the ascribed half circle. c² = x²+8² c = √(64+x²) units. It implies; Calculating x. √(64+x²) ~ (16+3x) x ~ √(64+x²) Cross Multiply. 16x+3x² = 64+x² 2x²+16x-64 = 0 x²+8x-32 = 0 (x+4)² = 32+(4²) (x+4)² = 48 x = -4±√(48) x = -4±4√(3) It implies; x ≠ -4-4√(3) units. x =4(√(3)-1) units. x = 2.9282032302...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3155704

Calculating r2, radius of the smaller inscribed circle. r2 will be; Let it be x. Therefore; (4+x)²=(8-x)²+(4-x)² 16+8x+x²=64-16x+x²+16-8x+x² x²-32x+64=0 (x-16)²=-64+256 x = 16±8√(3) x ≠ 16+8√(3) x = 16-8√(3) Therefore; x = r2 = 8(2-√(3)) units. Again, r2 is the radius of the smaller inscribed circle.

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