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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2851947

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Sir Mike Ambrose is the author of the question. Area yellow is; Area triangle with two side 8√(2) cm and 2 cm, and angle 45° - Area triangle with height (20/3) cm and base ⅓(20tan(45-2atan(⅓))) cm = ½*2*8√(2)sin45 - ½*(20/3)*⅓(20tan(45-2atan(⅓))) = 8 - (200tan(45-2atan(⅓)))/9 = (72 - 200tan(45-2atan(⅓)))/9 cm² = (2666664 - 1058200)/333333 cm² = 1608464/333333 cm² = (1608464÷5291)/(333333÷5291) = 304/63 cm² = 4.8253968254 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2851935

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Sir Mike Ambrose is the author of the question. Area red is; Area triangle with two side 4 cm and 8 cm, and angle 110°. = 0.5*4*8sin110 = 15.0350819326 cm² Area blue is; Area triangle with height 4.47952822702 cm and base (4.47952822702tan33.67018696) cm = 0.5*4.47952822702*4.47952822702tan33.67018696 = 6.6836970062 cm² Therefore; Area Blue + Area Red in cm² to 2 d. p. is; 6.6836970062 + 15.0350819326 = 21.7187789388 cm² ≈ 21.72 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2850252

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Sir Mike Ambrose is the author of the question. Point P coordinate is; P(4, -3) Point R coordinate is; R(0, ⅓(37)) Point Q (maximum turning point) coordinate is; Q((-3√(259)/14), (4√(259)/7)) Therefore; Area green exactly in square unit decimal form is; Area triangle with height 14.2909320391 units and base (15.8464857653sin(atan(23/6)-atan((8√(259)+42)÷(3√(259)+56))) units. = 0.5*14.2909320391*15.8464857653sin(atan(23/6)-atan((8√(259)+42)÷(3√(259)+56)) = 32.7...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2850241

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Sir Mike Ambrose is the author of the question. The side of the square is; 6.065425 cm Therefore; Area red in cm² to 2 d. p is; Area triangle with height (2.1tan29.11) cm and base 2.1 cm = 0.5*2.1(2.1tan29.11) = 1.22779137323 ≈ 1.23 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2848600

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ac+1 = a ----- (1) bc+1 = 4c ----- (2) 3ab+3 = 7b ---- (3) From (1) ac+1 = a a = 1/(1-c) ----- (4) Sub. (4) in (3) 3ab+3 = 7b 3(1/(1-c))b+3 = 7b 3b/(1-c) = (7b-3) 3b = (1-c)(7b-3) 3b = 7b-3-7bc+3c 7bc+3 = 4b+3c --- (5) Solving (2) and (5) bc+1 = 4c 7bc+3 = 4b+3c It implies; 7bc+7 = 28c 7bc+3 = 4b+3c 4 = 25c-4b -------- (6) b = (25c-4)/4 ----- (7) Sub. (7) in (2). c(25c-4)/4 + 1 = 4c 25c²-4c = 16c-4 25c²-20c+4 = 0 25c²-10c-10c+4 =0 5c(5c-2)-2(5c-2) = 0 (5...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2848589

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Sir Mike Ambrose is the author of the question. Shaded area is; Area under the curve at points (-1, 0) and (7, 0) - Area triangle with height 4 units and base 8 units. = ⅓(208) - ½(4*8) = ⅓(160) space units. Area S is; Area under the curve at points (-2, 0) and (7, 0) - Area triangle with height 4 units and base 8 units - Area triangle with height 1 unit and base (7/4) units - Shaded area. = 72 - ½(4*8) - ½(1(7/4)) - ⅓(160) = (43/24) space units. Area R is; Area tr...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2846881

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Sir Mike Ambrose is the author of the question. Let the side of the regular hexagon be 1 unit. Therefore; Area blue is; 2(area triangle with height 1 unit and base (⅓√(3)) units). = 2(½*1*(⅓√(3))) = ⅓√(3) space unit. Area orange is; Area triangle with height (3-√(3))/6 units and base (√(3)-1) units. = ½*(3-√(3))/6*(√(3)-1) = (2√(3)-3)/6 space units. It implies; Area Blue ÷ Area Orange Exactly is; ⅓√(3) ÷ (2√(3)-3)/6 = 2√(3) ÷ (2√(3)-3) = (2√(3)*(2√(3)+3))...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2846865

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8 = 4 a = (8-a) Cross multiply. 64-8a = 4a 64 = 12a a = ⅓(16) cm. b = 8-a b = ⅓(8) cm. c = atan(⅔)° d = atan(3/2)° ⅔ = 8/e e = 12 cm ⅔ = f/8 f = ⅓(16) cm. g = 8-f g = ⅓(8) cm. Area Orange exactly in cm² is; Area square units side 8 cm - Area rectangle with length 4 cm and width ⅓(8) cm. = 8² - 4*⅓(8) = 64 - ⅓(32) = ⅓(192-32) = ⅓(160) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2845491

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Sir Mike Ambrose is the author of the question. Radius of the inscribed circle, r is; 2r² = 4 r² = 2 r = √(2) units. Therefore; Area R in square unit to 3 d. p. is; Area sector with radius √(2) units and angle atan(0.5√(2))° - Area triangle with height 0.8165 units and base √(2)sin(atan(0.5√(2))) units - Area sector with radius 2 units and angle 25.5286945392° + Area triangle with height 2 units and base (2sin25.5286945392) units. = (atan(0.5√(2))π*√(2)²÷360) - (0.5*0.816...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2845481

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The radius of the inscribed circle is 2 units. a = 2atan(⅓)° b = 2atan(⅕)° c = 180-a-b c = (180-2atan(⅓)-2atan(⅕))° sin(2atan(⅓)) = d/16 d = (48/5) units. cos(2atan(⅓)) = e/16 e = (64/5) units. f = (90-2atan(⅓)-2atan(⅕))° tan(90-2atan(⅓)-2atan(⅕)) = g/(48/5) g = (198/35) units. Therefore; Area Blue exactly in square units is; ½*(48/5)(198/35) = 4752/175 square units.
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