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)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3121010

Sir Mike Ambrose is the author of the question. Area S exact, or to 3 d.p. will be; Area sector of radius 2√(7) units and angle (180 - 2sin–¹(4sin120÷(2√(7)))) + area triangle of two lengths 2 units and 4 units respectively and angle they form, 120° - area sector of radius 2√(7) units and angle 60° - area triangle of lengths 2√(7) units and 2 units, and angle they form, (180-(60+sin–¹(4sin120÷(2√(7)))) = (28πx(180 - 2sin–¹(4sin120÷(2√(7))))÷360) + (4sin120) - (28πx60÷360) - (2√(7)x(180...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3120902

Let the radius of the red circle be r. Calculating r. It will be; √(13²-5²)=√((9+r)²-(9-r)²)+√((4+r)²-(4-r)²) 12=√(36r)+√(16r) 12=10√(r) √(r)=1.2 r=1.44 units. Calculating angle PQR considering triangle PQR. It will be; Cos–¹((13²+5.44²-10.44²)/(2x13x5.44)) = 50.6923518839 Therefore length RX will be; √(4²+5.44²-2x4x5.44cos50.6923518839) = 4.24551165713 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3120587

Sir Mike Ambrose is the author of the question. Let the radius of the inscribed semi circle be 1 unit. Therefore; Shaded area ÷ Rectangle area in it's simplest decimal form will be; (0.25π-((180-2tan–¹(0.5))xπ÷360)+(0.5sin(180-2tan–¹(0.5)))) ÷ 2 = 0.0782494456 ÷ 2 = 0.0391247228 Or (1-(¼(4-π))-((180-2tan–¹(0.5))π÷360)+(0.5sin(180-2tan–¹)))÷2 = 0.0782494456 ÷ 2 = 0.0391247228

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3120582

Sir Mike Ambrose is the author of the question. Let the radius of one of the two equal circles be 1 unit. Total shaded area ÷ Rectangle area will be; (½(2x4)-π-(1-¼(4-π)-((180-2tan–¹(0.5))π÷360)+(0.5sin(180-2tan–¹(0.5)))))÷(2x4) = (4-π-(¼(π)-((180-2tan–¹(0.5))π÷360)+(0.5sin(180-2tan–¹(0.5)))))÷8 = (4-π-0.0782494456)÷8 = 0.78015790081÷8 = 0.0975197376

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