Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3213003

Calculating Shaded Blue Area. Notice. ½(10) = 5 cm is the radius of the inscribed circle. a² = 2(10)² a = 10√(2) cm. a is the diagonal of the ascribed square. b = ½(a) b = 5√(2) cm. It implies; 10² = 5²+(5√(2))²-2*5*5√(2)cosc 100 = 25+50-50√(2)cosc 50√(2)cosc = 75-100 cosc = -(25)/(50√(2)) c = acos(-¼*√(2)) c = 110.704811055° d = 2c d = 221.409622109° e = 360-d e = 138.590377891° (5/sinf) = (10/sin110.704811055) f = 27.8855668363° g = 2f g = 55.771133...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3212999

Calculating fraction of yellow shaded area. Let the single side length of the square be 2 units. Area yellow (shaded area) is; ½(4) square units. = 2 square units. Area square is; 2x2 = 4 square units. Therefore fraction of shaded area is; Area yellow ÷ Area square = 2÷4 = ½

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3211958

Kindly move the question right/left one time to review the framework/analysis of the question. Thank you nicely. Calculating length DG. Let x be the side length of the square. Let h be the height of triangle ADG. Let a be the height of triangle CDG. It implies; ½(xh ) = 18 xh = 36 cm² Observing Similar Plane Shape Rule. Consider triangle CDG and height a. h ~ a a ~ (x-h) Cross Multiply. a² = xh-h² And xh = 36 cm² a² = 36-h² a²+h² = 36 Length DG...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3211729

Calculating Shaded Area. Let the radius of the semi circle be r. Calculating r. 2+√(5/2)r ~ ½(3r) 2r ~ √(5/2)r Cross multiply. 3r² = 2√(5/2)r + (5/2)r² (½)r² = 2√(5/2)r (¼)r² = √(5/2)r (1/16)r⁴ = (5/2)r² (r²/16) = (5/2) 2r² = 80 r² = 40 r = √(40) r = 2√(10) units. Therefore radius, q of the inscribed circle is; q = ½(r) q = ½(2√(10)) q = √(10) units. It implies shaded area will be; Area ascribed semi circle with radius...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3211722

Calculating the area of each rectangle. Let the length and width of the rectangle be y and z respectively. Therefore; 7+7+x+y+z+7+7+x+y+z=66 Where 7 is the single side length of the square. It implies; 14+14+2x+2y+2z=66 Therefore; 2(x+y+z)=38 x+y+z=19 ------ (1) And; x+z=7 Therefore; x = 7-z ------- (2) z+z+2y=30 2z+2y=30 z+y = 15 y = 15 - z ------- (3) Substituting (2) and (3) in (1) (7-z)+(15-z)+z= 19 22-2z+z=19 z = 22-19 z = 3...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3211720

Sir Mike Ambrose is the author of the question. Calculating Area Orange ÷ Area Triangle ABC. Let the single side length of the regular hexagon be 2 units. Therefore; Area orange is; 1.25264388762 square units. Area triangle ABC is; (9√(3)/4) square units. It implies; Area Orange ÷ Area Triangle ABC is; 1.25264388762÷(9√(3)/4) = 0.32142857143

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3210665

Sir Mike Ambrose is the author of the question. Let alpha be x. Calculating Cot²x and (Red length ÷ Blue length)² as single fractions. Let the side length of the square be 2 units. tan30 = a/1 no a = tan30 a = ⅓√(3) units. b = 2-a b = 2-⅓√(3) b = ⅓(6-√(3)) units. tanx = 1/b tanx = 1/(⅓(6-√(3))) tanx = 3/(6-√(3))° Calculating Cot²x. cotx = 1/tanx cotx = 1/(3/(6-√(3))) cotx = ⅓(6-√(3)) Therefore; cot²x is; (⅓(6-√(3)))² = ⅑(36-12√(3)+3) = ⅑(39-12√(3)) = ⅓(1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3210542

Calculating Red Area. Let x be length AB equal length BD. a²+3² = x² a = √(x²-9) units. It implies; Calculating x ½(4+x)*3 = (3*x)+½(3√(x²-9)) 12+3x = 6x+3√(x²-9) 12-3x = 3√(x²-9) (4-x)² = x²-9 16-8x+x² = x²-9 25 = 8x x = ⅛(25) cm. x = 3.125 cm. Again, x is length AB equal length BD. cosb = 3/(3.125) b = 16.2602047083° c = 90+b c = 106.2602047083° Therefore, the required red area, area triangle ABD is; ½(x²)sinc = 0.5*3.125²sin106.26020470...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3210536

Calculating Length x. a = 12+8 a = 20 units. cosb = 20/24 b = acos(5/6)° c = 90-b c = (90-acos(5/6))° c = 56.4426902381° Therefore, the required length x is; cos56.4426902381 = 8/x x = 8/cos56.4426902381 x = 14.4725445397 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3209421

Sir Mike Ambrose is the author of the question. Calculating Blue Area + Red Area Exactly. Let x be the side length of the regular octagon. a = ⅛*180(8-2) a =135° a is the single interior angle of the regular octagon. b = 135-½(180-135) b = 135-22.5 b =112.5° c² = 2x²-2x²cos135 c = 1.84775906502x cm. Calculating x. 0.5*x²sin135+0.5*0.5x*1.84775906502xsin112.5 = 16+24√(2) 0.78033008589x² = 49.941125497 x² = 64 x = 8 cm. Again, x is the side length of the regular octa...

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