Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2615199

Sir Mike Ambrose is the author of the question. Area Green total, exactly in cm² is; Area triangle with height 24 cm and base 10 cm + Area triangle with height 24 cm and base (24tan15) cm - 2(area triangle with height 10 cm and base (10tan15) cm) = ½*24*10 + ½*24*24tan15 - 2(½*10*10tan15) = 120 + 288(2-√(3)) - 100(2-√(3)) = 120 + 576 - 288√(3) - 200 + 100√(3) = (496 - 188√(3)) cm² = 4(124-47√(3)) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2613734

Let the square side length be a. It implies; Area square is a² Calculating a² (Area of the Square). √(2)² = b²+c² c = √(2-b²) --- (1). √(68)² = (a+b)²+c² 68 = (a+b)²+√(2-b²)² 68 = (a+b)²+2-b² 66 = a²+2ab --- (2). 8² = b²+(c+a)² --- (3). 64 = b²+c²+2ac+a² 64 = b²+(√(2-b²)²+2a√(2-b²)+a² 64 = b²+2-b²+2a√(2-b²)+a² 62 = 2a√(2-b²)+a² (62-a²)² = 4a²(2-b²) 3844-124a²+a⁴ = 8a²-4a²b² 132a²-3844-a⁴= 4a²b² b² = (132a²-3844-a⁴)/(4a²) b = √(132a²-3844-a⁴)/2a --- (3). Substit...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2611915

Calculating yellow inscribed circle area. a² = 6²+8² a = √(100) a = 10 units. 2b² = 10² b² = 50 b = 5√(2) units. b is the radius of the half circle. (5√(2)/sin45) = (8/sinc) c = 53.1301023542° d = 180-c d = 126.8698976458° e = 180-45-d e = 8.1301023542° (5√(2)/sin45) = (f/sin8.1301023542) f = 1.4142135624 units. f = √(2) units. Let g be the radius of the inscribed yellow circle. h = (5√(2)-g) units. j² = 2g² j = √(2)g units. Calculating g, radius of t...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2597388

The radius, r if the small circle will be; (5+r)²=2(5-r)² 25+10r+r²=50-20r+2r² Therefore r²-30r+25=0 (r-15)²=-25+225 r = 15±√(200) r = 15±10√(2) It implies; r ≠ 15+10√(2) r = 15-10√(2) unit. Therefore area blue will be; ¼(100-25π) - (15-10√(2))²π = 3.05304958693 square unit.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2595419

Let the side length of the inscribed green square be a. b = ½(10-a) units. c = b+a c = ½(10-a+2a) c = ½(10+a) units. Therefore, a² (area green inscribed triangle) is; 10² = a²+(½(10+a))² 100 = a²+¼(100+20a+a²) 400 = 4a²+100+20a+a² 5a²+20a-300 = 0 a²+4a-60 = 0 Resolving the above quadratic equation via factorization approach to get a, side length of the inscribed green square. a²+10a-6a-60 = 0 a(a+10)-6(a+10) = 0 (a+10)(a-6) = 0 It implies; a ≠ -10 a = 6 units....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2591226

Calculating the area of the inscribed blue circle. a = ½(14+6) a = 10 units. a is the radius of the ascribed semi circle. b = a-6 b = 4 units. c²+4² = d² c is the radius of the inscribed half circle. d = √(c²+16) units. 10² = √(c²+16)²+c² 100 = c²+16+c² 84 = 2c² c² = 42 c = √(42) units. Again, c is the radius of the inscribed half circle. e = (c+f) e = (√(42)+f) units. f is the radius of the inscribed blue circle. g = (10-f) units. Notice! d = √(c²+16) And c...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2585127

Let the side length of the inscribed square be a. Calculating the area of the inscribed red square. 2b² = a² b = ½√(2)a unit. c² = 2a² c = √(2)a unit. c is the diagonal of the inscribed red square. It implies; 10² = (½√(2)a)²+(√(2)a)² 100 = (a²/2)+2a² 100 = ½(5a²) 5a² = 200 a² = 40 a = 2√(10) units. Therefore, area red inscribed square is; a² = 40 square units. Again, a = 2√(10) units is the side length of the inscribed red square.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2582274

a = 10+10 a = 20 cm. a is twice the radius of the quarter circle. b = ½(10) b = 5 cm. c² = 5²+10² c √(125) c = 5√(5) cm. Observing similar plane shape (right-angled triangle) side length ratios. 20 - 5√(5) d - 5 Cross Multiply. 5√(5)d = 100 d = 20/√(5) d = 4√(5) cm. e² = 20²-(4√(5))² e² = 400-80 e = √(320) e = 8√(5) cm. f = e-c f = 8√(5)-5√(5) f = 3√(5) cm. Therefore, area blue is; Area triangle with height 4√(5) cm and base 3√(5) cm + Area triangl...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2580404

Calculating Area Red. It is; Area sector with radius 5 m and angle 60° - Area triangle with height 5 units and base 5sin60 m + 2(area triangle with height 2.5 m and base 2.5 m) = (60π*5*5/360)-(0.5*5*5sin60)+2(0.5*2.5*2.5sin60) = ⅙(25π)-¼(25√(3))+⅛(25√(3)) = (100π-150√(3)+75√(3))/24 = (100π-75√(3))/24 m² = 7.6773106163 m²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2571617

Calculating x. 7² = x²+5²+2*5*xcosa 49 = x²+25+10xcosa 24 = x²+10xcosa 10xcosa = 24-x² cosa = (24-x²)/(10x) --- (1). b = (x+5) units. 7² = x²+(x+5)²-2x(x+5)cosa 49 = x²+x²+10x+25-(2x²+10x)cosa 24 = 2x²+10x-(2x²+10x)cosa --- (2). Substituting (1) in (2). 24 = 2x²+10x-(2x²+10x)cosa And cosa = (24-x²)/(10x) 24 = 2x²+10x-(2x²+10x)(24-x²)/(10x) 240= 20x²+100x-(2x+10)(24-x²) 240 = 20x²+100x-(48x-2x³+240-10x²) 240 = 20x²+100x-48x+2x³-240+10x² 480 = 2x²+30x²+52x x³+15x²+26x-240= 0 It implies;...

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Mathematics Question and Solution

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Mathematics Question and Solution

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