Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3203677

Let 1 unit be AB = BC = CD. a² = 2-2cos150 a = 1.93185165258 units. a is BD. b = 90-½(180-150) b = 90-15 b = 75° b is angle ABD. c² = 1+1.93185165258²-2*1.93185165258cos75 c = 1.93185165258 units. Notice. AD = BD Therefore, the required angle, angle ADC is; 15+(180-2(75)) = 15+(180-150) = 15+30 = 45°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3202632

Sir Mike Ambrose is the author of the question. Calculating Area Shaded ÷ Area Purple to 2 decimal places. Let AB be 5 units. Radius, r of the inscribed circle is; r = 5*(2/5) r = 2 units. a = ⅛(180*6) a = 135° 2b² = 25 b = ½(5√(2)) units. c = 180-0.5(135)-45 c = 67.5° sin22.5 = ½(5√(2))/d d = 9.23879532511 units. Where d is the side length of the regular triangle (purple area). Area purple is; 0.5*9.23879532511*9.23879532511sin60 = 36.95994598698 square...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3202538

Calculating Area Blue. a = 5+3 a = 8 units. It implies; b*4 = 3*8 4b = 24 b = 6 units. b is BC. c = 4+b c =10 units. d² = 10²+5² d² = 125 d = √(125) d = 5√(5) unit. d is the diameter of the ascribed circle. e = ½(d) e = ½(5√(5)) units. e = 5.59016994375 units. e is the radius of the ascribed circle. tanf = 3/6 f = atan(½)° tang = 5/10 g = atan(½)° h = f+g h = 2atan(½)° j = 180-2h j = (180-4atan(½))° j = 73.7397952917° Area Blue is; Area sec...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3202296

Calculating area of the circle. Let the radius of the circle be r. Therefore; (2r)²=2(√(20))² 4r²=40 r = √(10) units. Therefore area of the circle is; π√(10)² = 10π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3202289

Calculating R, radius of the inscribed circle. Let a be the radius of the inscribed half circle. Calculating a. b²+9² = 15² b² = 225-81 b = √(144) b = 12 units. c = (12-a) units. d = (9-a) units. It implies; (12-a) ~ a a ~ (9-a) Cross Multiply. a² = 108-21a+a² 21a = 108 7a = 36 a = ⅐(36) units. Or 12a+9a = (12*9) 21a = 108 a = ⅐(36) units. Recall. c = (12-a) And a = ⅐(36) units. c = 12-⅐(36) c = ⅐(84-36) c = ⅐(48) units. tane = 9/12 e = at...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3201721

Calculating the inscribed angle beta, let it be y. Let (x+3) units be the side length of the square. a²+x² = 4² a = √(16-x²) units. b = 1+a b = (1+√(16-x²)) units. b is the side length of the square. Notice. b = (x+3) units as well. Therefore; Calculating x. It implies; x+3 = 1+√(16-x²) (x+2)² = 16-x² x²+4x+4 = 16-x² 2x²+4x-12 = 0 x²+2x-6 = 0 (x+1)² = 6+(1)² (x+1)² = 7 x = -1±√(7) It implies; x ≠ -1-√(7) units. x = (√(7)-1) units. x = 1.64575131106 u...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3201150

Sir Mike Ambrose is the author of the question. Calculating Area Blue exactly in square units. Area Blue is; Area triangle with two side length 10½ units and ((7√(5)sin(atan(3/4)))/(2sin(180-atan(2)-atan(3/4))) units respectively, and angle (180-2atan(2))°, the angle they form. = ½(21/2)((7√(5)sin(atan(3/4)))/(2sin(180-atan(2)-atan(3/4)))sin(180-2atan(2)) = ¼(21)(378/99) = ½(441/11) = (441/22) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3201140

Calculating the circumference of the blue inscribed circle. Let x be the base of the inscribed right-angled triangle with height 10 units. x²+10² = a² a = √(x²+100) units. b = ½(a) b = ½√(x²+100) units. Calculating x. 10² = 10²+(½√(x²+100))²-10√(x²+100)cosy 10√(x²+100)cosy = ¼(x²+100) --- (1). cosy = x/√(x²+100) --- (2). Substituting (2) in (1). 10√(x²+100)*x/√(x²+100) = ¼(x²+100) 10x = ¼(x²+100) 40x = x²+100 x²-40x+100 = 0 (x-20)² = -100+(-20)² (x-20)² = 300 x = 20±√(300) x =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3199587

Let b = 1 unit. Therefore; Side length of the ascribed regular triangle c is; c = 4b c = 4(1) c = 4 units. Therefore d, radius of the inscribed circle is; 3(4d) = 4√(4²-2²) 12d = 4√(12) 3d = 2√(3) d = ⅓(2√(3)) units. d = 1.15470053838 units. Calculating a. e²+1² = (⅓(2√(3)))² e² = ⅓(4)-1 e = √(⅓) e = ⅓√(3) units. e = 0.57735026919 units. It implies; (⅓(2√(3)))² = (⅓√(3))²+a²-2*(⅓√(3))acos30 ⅓(4) = ⅓+a²-(a) 4 = 1+3a²-3a 3a²-3a-3 = 0 a²-a-1 = 0 (a-½)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3199574

Calculating Total Shaded Area, Area Grey, Purple and Blue. Notice. KLMN is a square. Calculating area grey. a = 6-4 a = 2 units. sinb = 2/6 b = asin(⅓) b = 19.4712206345° c²+2² = 6² c = 4√(2) units. Area Grey is; (6*2)-½(4√(2)*2)-(19.4712206345π*6²/360) = 0.22608138033 square units. Calculating area purple. d = (90-b) d = 70.5287793655° e = ½(d) e = 35.2643896828° f = e+b f = 35.2643896828+19.4712206345 f = 54.7356103173° g = ½(180-f) g = 62.6321948414° h = 90-g h = 27.3678051587...

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