Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3122493

Let the ascribed semi circle radius be R. Let the inscribed circle radius be r. Notice; R=2r It implies; R²=1²+r² (2r)²=1+r² 4r²=1+r² 4r²-r²=1 3r²=1 Therefore; r = (√(3)/3) units. Green area/area of the inscribed circle will be; π(√(3)/3)² = (π/3) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3122294

a ~ 6 6 ~ 8 Cross Multiply. 8a = 6*6 2a = 9 a = ½(9) units. b = a+8 b = ½(9)+8 b = ½(25) units. b is the diameter of the ascribed half circle. c = ½(b) c = ¼(25) units. c is the radius of the ascribed half circle. d = c-a d = ¼(25)-½(9) d = ¼(7) units. Let e be the radius of the inscribed circle. f = e-d f = e-¼(7) f = ¼(4e-7) units. g = c-e g = ¼(25)-e g = ¼(25-4e) units. Calculating e. g² = e²+f² (¼(25-4e))² = e²+(¼(4e-7))² (625-200e+16e²)/...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3122166

Radius of the green circle is 4 units, therefore area of the green circle is; π4² = 16π square units. Calculating radius of the red circle. Let the radius of the red circle be r. (4-r)²+(5-r)²=(4+r)² 16-8r+r²+25-10r+r²=16+8r+r² r²-26r+25=0 r²-25r-r+25=0 r(r-25)-1(r-25)=0 Therefore; r≠25 units. r=1 unit. Therefore area red circle is; π1² = π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3122031

Let a be the radius of the circle. c = (a-1) units. Calculating a. 7² = a²+(a-1)²-2a(a-1)cosb 49 = a²+a²-2a+1-(2a²-2a)cosb 48 = 2a²-2a-(2a²-2a)cosb (2a²-2a)cosb = 2a²-2a-48 --- (1). 11² = a²+(a-1)²-2a(a-1)cosb 121 = a²+a²-2a+1-(2a²-2a)cosb 120 = 2a²-2a+(2a²-2a)cosb (2a²-2a)cosb = 120+2a-2a² --- (2). Equating (1) and (2). 2a²-2a-48 = 120+2a-2a² 4a²-4a-168 = 0 a²-a-42 = 0 a²-7a+6a-42 = 0 a(a-7)+6(a-7) = 0 (a-7)(a+6) = 0 Therefore; a = 7 units. Again, a is th...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3121933

Let 1 unit be the side length of the regular hexagon. a = ⅙*180(6-2) a = 120° a is the single interior angle of the regular hexagon. cos75 = 0.5/b b = 1.93185165258 units. c² = 1²+0.5²-2*1*0.5cos120 c = 1.32287565553 units. d = 120-75 d = 45° (0.5/sine) = (1.32287565553/sin120) e = 19.1066053506° f = 120-e f = 100.893394649° g = 360-120-100.893394649-75 g = 64.106605351° h = 90-g h = 25.893394649° j² = 1.32287565553²+1.93185165258²-2*1.32287565553*1.9318516...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3121618

Let the three equal inscribed lengths be y. a = 5+4 a = 9 units. a is the radius of the ascribed half circle. b² = y²+y² b = √(2)y units. Notice. a = b √(2)y = 9 y = ½(9√(2)) units. c = 2y c = 2*½(9√(2)) c = 9√(2) units. It implies; 9² = 5²+(9√(2))²-2*5*9√(2)cosd 81 = 187-90√(2)cosd 90√(2)cosd = 106 d = acos(106/(90√(2)))° d = 33.6110339926° Therefore, the required length, x is; sind = x/(9√(2)) x = 9√(2)sin33.6110339926 x = 7.04556598153 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3121528

Sir Mike Ambrose is the author of the question. Let the radius of the semi circle be 3 units. Therefore shaded area ÷ area square will be; (½(area quarter circle of radius 3 units - area triangle of height and width 3 units respectively) + area triangle of two sides ((3√(2)/2)-(3√(3)tan22.5)/2) units and ((3√(2)/2)-(3√(3)tan22.5)/2)xsin(tan–¹(3)-45)÷sin(180-tan–¹(3))) units, and the angle they form, angle 45°) ÷ (area square of single side length ½(3√(2)) units). {½((9π/4)-(9/2)) +...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3121510

Let the radius of the ascribed quarter circle be R. Let the radius of the inscribed circle be r. Therefore; 2²+(2r)²=R² 4+4r²=R² Therefore; R²-4r²=4 Multiplying through by ¼(π). It implies; ¼(πR²) - πr² = π Notice. ¼(πR²) = area of the ascribed quarter circle. πr² = area of the inscribed circle. Therefore area of the red shaded region in the given quarter circle is; π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3121350

a = (10+x) units. a is the radius of the half circle. (10+10)² = 16²+b² b = √(400-256) b = √(144) b = 12 units. It implies; 12 ~ c 20 ~ 10 Cross Multiply. 20c = 120 c = 6 units. d = 21-6 d = 15 cm. e = 2d-21 e = 30-21 e = 9 units. Calculating x. x*(10+10+x) = 21*9 x²+20x = 189 x²+20x-189 = 0 x²+27x-7x-189 = 0 x(x+27)-7(x+27) = 0 (x+27)(x-7) = 0 Therefore; x = 7 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3121237

Let the radius of the circle be a. sinb = a/(a+a) b = 30° cos30 = c/2a c = √(3)a units. d = 2-c d = (2-√(3)a) units. sin30 = e/d e =½(2-√(3)a) units. cos30 = f/d ½√(3) = f/(2-√(3)a) f = ½(2√(3)-3a) units. g = a+a+a g = 3a units. h = f+g h = ½(2√(3)-3a)+3a h = ½(2√(3)+3a) units. h is the diameter of the half circle. j = ½h j = ¼(2√(3)+3a) units. j is the radius of the half circle. k = a+j k = a+¼(2√(3)+3a) k = ¼(2√(3)+7a) units. l = j-a l = ¼(2√(3)...

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