Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2772515

R is; R² = 2(R-2)² R² = 2(R²-4R+4) R² = 2R²-8R+8 R²-8R+8 = 0 (R-4)² = -8+16 (R-4)² = 8 R = 4±2√(2) Therefore; R ≠ 4-2√(2) cm. R = 4+2√(2) cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2772375

a² = 11*11+10*10-2*11*10cos50 a = 8.92113926968 cm. (8.92113926968/sin50) = (10/sinb) b = 59.16920318624° c = 180-50-59.16920318624 c = 70.83079681376° d = 0.5a d = 4.46056963484 cm. e² = 4.46056963484²+5.5²-2*4.46056963484*5.5cos59.16920318624 e = 5 cm. (5/sin59.16920318624) = (5.5/sinf) f = 70.83079681376° g² = 4.46056963484²+5²-2*4.46056963484*5cos70.83079681376 g = 5.5 cm. (5.5/sin70.83079681376) = (5/sinh) h = 59.16920318624° i = 180-59.16920318624-70.83...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2771965

Calculating the area of the ascribed right-angled triangle. a = (y+2) units. a is the adjacent base of the right-angled triangle. b = (z+2) units. b is the adjacent height of the right-angled triangle. 2 units is the radius of the inscribed circle. c = (y+z) units. c is the hypotenuse of the right-angled triangle. a²+b² = c² (y+2)²+(z+2)² = (y+z)² --- (1). tanx = 2/z --- (2). tanx = 2/(y+2) --- (3). Equating (2) and (3). 2/z =2/(y+2) z = y+2 --- (4). Substitu...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2771184

Radius of the quarter circle is (2+3) = 5 units. Let a be the side length of the inscribed square. b²+3² = a² b = √(a²-9) units. c = 3+b c = (3+√(a²-9)) units. Calculating a. (3+√(a²-9))²+3² = 5² 9+6√(a²-9)+a²-9+9 = 25 9+6√(a²-9)+a² = 25 6√(a²-9) = 16-a² 36(a²-9) = (16-a²)² 36a²-324 = 256-32a²+a⁴ a⁴-68a²+580 = 0 It implies; a = √(10) units a = 3.1622776602 units. Again, a is the side length of the inscribed square. Recall. b = √(a²-9) units And a = √(10)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2771080

tana = 3/4 a = atan(3/4) b² = 3²+4² b = 5 units. cos(atan(3/4)) = 1/c c = 1.25 units. c = (5/4) units. sin(atan(3/4)) = 1/d d = 1.6666666667 units. d = (5/3) units. e = b-c-d e = 5-(5/4)-(5/3) e = (60-15-20)/12 e = 25/12 units. e is AB. sin(atan(3/4)) = f/(25/12) f = (5/4) units. f is BC. cos(atan(3/4)) = g/(25/12) g = (5/3) units. g is AC. Therefore, area triangle ABC (yellow area) is; ½*fg = ½*(5/4)*(5/3) = (25/24) square units. = 1.0416666667 square...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2771070

Calculating angle x. Let BC be 2 units. a = 180-70-40 a = 70° a is angle ACB. cos70 = 1/b b = 2.9238044002 units. b is AB. c = 180-80-50 c = 50° c is angle ADC. d = ½(b) d = 1.4619022001 units. d is AY = DY. cos50 = e/2.9238044002 e = 1.8793852416 units e is CX = DX. f² = 1²+1.8793852416²-2*1*1.8793852416cos(70+50) f = 2.5320888863 units. f is MX. (2.5320888863/sin120) = (1/sing) g = 20° g is angle CXM. h² = 1.8793852416²+1.4619022001²-2*1.4619022001*...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2771049

Sir Mike Ambrose is the author of the question. Let the side length of the regular hexagon be 1 unit. Therefore; Area green is; Area regular hexagon with side 1 unit + Area regular pentagon with side 1 unit + Area square with side 1 unit + Area square with side ½√(3) units. = (6/4tan(180/6)) + (5/4tan(180/5)) + 1² + (½√(3))² = 2.59807621135 + 1.72047740059 + 1 + (3/4) = 6.06855361194 space units. Area red is; Area triangle with two side 1 unit and ½√(3) units, and angle 1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2769787

Calculating radius of the quarter circle. Let a be the radius of the quarter circle. b = ½(DE) b = 2√(2) units. c²+(2√(2))² = a² c² = a²-8 c = √(a²-8) units. d = c-BE d = (√(a²-8)-5) units. e = c-CD e = (√(a²-8)-3) units. It implies; Calculating a. 2√(2) - (√(a²-8)-5) (√(a²-8)-3) - 2√(2) Cross Multiply. 8 = (√(a²-8)-3)(√(a²-8)-5) 8 = a²-8-8√(a²-8)+15 8√(a²-8) = a²-1 64(a²-8) = a⁴-2a²+1 64a²-512 = a⁴-2a²+1 a⁴-66a²+513 = 0 Resolving the above quadratic...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2769445

Calculating r. a = (2+r) units. b = (1+r) units. c = r+r c = 2r units. It implies; (2r)² = (1+r)²+(2+r)²-2(1+r)(2+r)cos60 4r² = 1+2r+r²+4+4r+r²-(2+3r+r²) 4r² = 2r²+6r+5-2-3r-r² 3r² = 3r+3 r²-r-1 = 0 (r-½)² = 1+(-½)² (r-½)² = (5/4) r = ½±√(5/4) r = ½±½√(5) It implies; r ≠ ½(1-√(5)) r = ½(1+√(5)) units (golden ratio). r = 1.6180339887 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2769438

Let x be 2 units. Therefore; Area square is; 8² = 64 square units. Calculating area rectangle. The length is; √(8²+6²) + 2cos(atan(4/3)) = 10+(6/5) = 56/5 units. The width is; 2sin(atan(4/3)) = 8/5 units. Area rectangle is; (56/5)*(8/5) = 448/25 square units. It implies; Area rectangle : Area square is; (448/25) : 64 = 448 : (25*64) = 56 : (25*8) = 14 : (25*2) = 7 : 25

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