Mathematics Question and Solution

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

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

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

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

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

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.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2843870

Sir Mike Ambrose is the author of the question. Let the side length of the three congruent, inscribed square be x. Calculating x. x + 2 = 2√(3) x = (2√(3)-2) units. Therefore; Area green exactly in square units decimal form is; Area triangle with height (2√(3)-2) units and base (4-2√(3)) units + Area triangle with height (2√(3)-2) units and base ⅓(6-2√(3)) units + Area triangle with height 1.55909609016 units and base ⅓(4√(3)-6)sin(50.103909361) units. = (0.5*(2√(3)-2)(4-2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2843843

Let a be the side of the square. cos30 = b/8 b = 4√(3) cm. sin30 = c/8 c = 4 cm. a = b+12 a = (12+4√(3)) cm. Area Square is; a² = (12+4√(3))² a² = 358.27687752661 cm² d² = 12²+4²-2*4*12cos30 d = 8.76707255797 cm. (8.76707255797/sin30) = (12/sine) e = 180-43.18678543201 e = 136.81321456799° f = 360-e-180 f = 43.18678543201° g² = 8²+8.76707255797²+2*8*8.76707255797cos43.18678543201 g = 6.21165708246 cm. (6.21165708246/sin43.18678543201) = (8.76707255797/s...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2842333

Let the diameter of the inscribed white semicircle be x. Therefore the diameter of the ascribed semicircle is 2x. Calculating x. (3x/2)² = x²+36 (9x²/4)-x² = 36 (5x²/4) = 36 √(5)x = 12 x = ⅕(12√(5)) units. Where x is the radius of the ascribed semicircle. Radius of the inscribed white semicircle is; ½(x) = ⅕(6√(5)) units. Area Blue is; π½(⅕(12√(5)))² - π½(⅕(6√(5)))² = ½(π(⅕(108))) = ⅕(54π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2842327

Sir Mike Ambrose is the author of the question. Area purple is; Area trapezoid with two parallel sides 4 cm and (50/9) cm, and height (7/2) cm - Area triangle with height 4 cm and base 3 cm. = ½(4+(50/9))*(7/2) - ½(4*3) = (602/36) - 6 = 193/18 cm² Area orange is; Area trapezoid with two parallel sides 6 cm and (40/9) cm, and height (7/2) cm - Area triangle with height (40/9) cm and base (40/13) cm. = ½(6+(40/9))*(7/2) - ½((40/9)*(40/13)) = (329/18) - (800/117) = 4277/234...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2840749

Let the height of the yellow area trapezoid be a. Calculating a. tan72 = a/b b = (a/tan72) cm. c = 2b c = (2a/tan72) cm. d = 8+c d = (8+(2a/tan72)) cm. It implies; ½*a(8+(8+(2a/tan72))) = 56.23 a(16+0.64983939247a) = 112.46 0.64983939247a²+16a- 112.46 = 0 Therefore; a = 5.70627 cm. d = (8+(2a/tan72)) d = 11.70815903005 cm. tan18 = e/5.70627 e = 1.85407951502 cm. f = 2e f = 3.70815903005 cm. (g/sin36) = (8/sin108) g = 4.94427191 cm. sin72 = h/8...

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