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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2243242

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Area Blue is; Area trapezium with two parallel side 4 cm and 1 cm, and height 4 cm - Area semi circle with radius 2 cm. = ½*4(4+1) - ½(4π) = 10 - 2π = 2(5-π) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2242676

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a = 180-12-2-46-8 a = 180-68 a = 112° a is angle ABC. Let AC be 1 unit. It implies; (1/sin112) = (b/sin14) b = 0.257447904 units. b is AB. c = 180-8-2 c = 170° c is angle ADC. It implies; (1/sin170) = (d/sin2) d = 0.2009781915 units. d is AD. e² = 0.2009781915²+0.257447904²-2*0.2009781915*0.257447904cos46 e = 0.1865112432 units. e is BD. (0.1865112432/sin46) = (0.2009781915/sinf) f = 50.8173805282° f is angle ABD. It implies; x (angle CBD) = a-f x = 112-50.8173805282 x = 61.1826194718°...

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/2241077

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Online Edumath Educators and Learners are Super Smart and Amazingly, Very Clever. Communicate us to mentor/teach your child/children Mathematics online at affordable tuition, helping them become a lover of Mathematics, and also, a Mathematician. Call/WhatsApp: +2349076614992 or +2348134903121 Email: [email protected] [email protected] [email protected] Facebook Page: Online Edumath Truly, we are super smart and amazingly, very clever!

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2241064

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tana = 12/8 a = atan(3/2)° b = 180-a b = (180-atan(3/2))° (10/sin(180-atan(3/2))) = (4/sinc) c = 19.4400348282° d = 180-b-c d = 180-19.4400348282-(180-atan(3/2)) d = 36.8698976458° e = 90-d e = 90-36.8698976458 e = 53.1301023542° It implies, Area S (shaded area green) is; Area trapezoid with height 12 units, and parallel sides 4 units and 12 units respectively - Area triangle with height 10 units and base 4sin36.8698976458 units - Area sector with radius 10 units and angle 53.1301023542°...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2239203

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

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2237207

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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)) unit). = 2(½*1*(⅓√(3))) = ⅓√(3) space unit. Area orange is; Area triangle with height (3-√(3))/6 unit and base (√(3)-1) unit. = ½*(3-√(3))/6*(√(3)-1) = (2√(3)-3)/6 space unit. It implies; Area Blue ÷ Area Orange Exactly is; ⅓√(3) ÷ (2√(3)-3)/6 = 2√(3) ÷ (2√(3)-3) = (2√(3)*(2√(3)+3)) ÷ ((2√(3)-3)*(2√(3)+3)) = (12+6√(3)) ÷ (12-9) = 6(2+√(3)) ÷ 3 =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2235317

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Area blue is: Area triangle with height 1 cm and base (3sin(109.471220634)) cm. = 0.5*1*3sin(109.471220634) = 1.41421356238 cm² = √(2) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2235311

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Let the radius of the inscribed white semi circle be r. Therefore; r²+(1+r)²=(2+r)² r²+1+2r+r²=4+4r+r² r²-2r-3 = 0 (r-1)² = 3+(-1)² (r-1)² = 4 r = 1±2 r ≠ -1 r = 3 unit. It implies; Area blue is; Area semi circle with radius 5 unit - Area semi circle with radius 3 unit. = ½(π*5²) - ½(π*3²) = ½(25π) - ½(9π) = ½(25π-9π) = ½(16π) = 8π square unit.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2235307

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Radius of the inscribed circle, r is; 2r² = 4 r² = 2 r = √(2) unit. Therefore; Area R in square unit to 3 d. p. is; Area sector with radius √(2) unit and angle atan(0.5√(2))° - Area triangle with height 0.8165 unit and base √(2)sin(atan(0.5√(2))) unit - Area sector with radius 2 unit and angle 25.5286945392° + Area triangle with height 2 unit and base (2sin25.5286945392) unit. = (atan(0.5√(2))π*√(2)²÷360) - (0.5*0.8165*√(2)sin(atan(0.5√(2)))) - (25.5286945392π*2²÷360) + (0.5*2*2sin25.5286...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2234887

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tan60 = a/10 a = 10√(3) cm. a is the height of the equilateral triangle. sin(atan(2)) = b/(10√(3)) b = 15.4919333848 cm. b is a side length of the square. It implies; Area Square is; b² = (15.4919333848)² = 240 cm²
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