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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2377561

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Let the radius of the small yellow inscribed circle be a. Let the radius of the big green inscribed circle be b. Notice! PQ = a+b Calculating radius of the ascribed semi circle. c² = 12²+16² c = √(144+256) c = √(400) c = 20 units. c is the diameter of the ascribed semi circle. d = ½(c) d = ½(20) d = 10 units. d is the radius of the ascribed semi circle. tane = 16/12 e = atan(4/3)° e is angle BAC. sin(atan(4/3)) = f/12 4/5 =f/12 f =48/5 f = 9.6 units. cos(atan(4/3)) = g/12 3/5 = g/12 g =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2377260

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Let the side of the three inscribed congruent regular hexagon be 1 unit. It implies, the side length of the square is; a = 6 units. b² = 2-2cos120 b = √(3) units. c = 1+(½) c = ½(3) units. c = 1.5 units. d = ½+2 d = ½(5) units. d = 2.5 units. e = a-b e = (6-√(3)) units. f = 6-c f = 6-½(3) f = ½(12-3) f = ½(9) units. f = 4.5 units. It implies, the required angle x, is; atan((2.5)/(6-√(3)))+x+atan((6-√(3))/4.5) = 90 x = 90-atan((2.5)/(6-√(3)))-atan((6-√(3))/4.5) x = 16.1558984641°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2376862

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Calculating Area triangle ABC. Let a be the radius of the inscribed circle. b² = 2a² b = √(2)a units. b is EG. c² = a²+(2+a)² c² = a²+4+4a+a² c = √(2a²+4a+4) units. c is CE. Therefore; c² = b²+2²-2*2*bcos135 √(2a²+4a+4)² = (√(2)a)²+4-4*√(2)acos135 2a²+4a+4 = 2a²+4+4 4a = 4 a = 1 unit. a is the radius of the inscribed circle. AC = CG+AG AC = 2+1 AC = 3 units. Notice! CG = CH = 2 units. And a (circle's radius) = DH = DG = AG = DE = AE = 1 unit. tanb = (...

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/2376544

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Online Edumath Educators and Learners are Super Smart and Amazingly, Very Clever. Communicate us to mentor/teach/educate your child/children Mathematics online at affordable tuition, helping them become a lover of Mathematics, and also, a Mathematician. Call/WhatsApp: +2349076614992 or +2348152827445 Email: [email protected] [email protected] [email protected] Facebook Page: Online Edumath We are super smart and amazingly, very clever!

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2376473

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Notice! The area of the regular pentagon and the square are the same. Let it be 4 square units. Therefore the side length of the square a, is; a = √(4) a = 2 units. Calculating b, the side length of the regular pentagon. Notice, b = x c = ⅕(180(5-2)) c = 108° c is the single interior angle of the regular pentagon. d = 108-90 d = 18° cos18 = e/b e = 0.9510565163b units. sin18 = f/d f = 0.3090169944b units. g = 2f+b g = 2(0.3090169944b)+b g = 1.6180339887b units. It implies; ½(b²)sin1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2376368

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Notice. The polygon is regular hexagon. The three circles are congruent. The inscribed triangle is equilateral. Let the side length of the regular hexagon be a. Let the a the congruent circles' radius be b. It implies; a = b+2 --- (1). (2b)² = (b+1)²+a²-2a(b+1)cos60 4b² = b²+2b+1+a²-ab-a 3b² = 2b+1+a²-ab-a --- (2). Substituting (1) in (2) to get b, radius (r) of the three congruent circles. 3b² = 2b+1+(b+2)²-b(b+2)-(b+2) 3b² = 2b+1+b²+4b+4-b²-2b-b-2 3b² = 1+4b+4-b-2 3b² = 3b+3 b²-b-1 = 0...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2376145

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Radius of the semi circle r is; r = ½(10) r = 5 cm. a = 1+5 a = 6 cm. b = 5-1 b = 4 cm. a² = b²+c² c = √(6²-4²) c = √(20) c = 2√(5) cm. cosd = 4/6 d = acos(2/3) d = 48.1896851042° e = 90-d e = 90-48.1896851042 e = 41.8103148958° f = 180-d f = 180-48.1896851042 f = 131.8103148958° It implies; g² = 6²+5²-2*6*5cos131.8103148958 g = 10.0498756211 cm. (10.0498756211/sin131.8103148958) = (5/sinh) h = 21.7667200119° cosi = 1/10.0498756211 i = acos(1/10.04...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2376099

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Shaded Area is; Area quarter circle with radius 80 cm - Area triangle with height 80 cm and base 80 cm - Area quarter circle with radius 60 cm + Area triangle with height 60 cm and base 60 cm + Area quarter circle with radius 60 cm - Area triangle with height 60 cm and base 60 cm - Area sector with radius 40 cm + Area triangle with height 40 cm and base 40 cm. = ¼(80*80π)-½(80*80)-¼(60*60π)+½(60*60)+¼(60*60π)-½(60*60)-¼(40*40π)+½(40*40) = 1600π-3200-900π+1800+900π-1800-400π+800 = 1600π-3200...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2375899

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a = 3+1 a = 4 units. b² = (2√(3))²-r² b = √(12-r²) units. Observing similar plane shape side lengths. 2√(3) - 4 √(12-r²) - 2r Cross Multiply. 4√(3)r = 4√(12-r²) √(3)r = √(12-r²) 3r² = 12-r² 4r² = 12 r = √(3) units. r is the radius of the circle. It implies the required angle theta, let it be x, is; cosx = (2r)/4 And r = √(3) units. cosx = (2√(3)/4 x = acos(√(3)/2) x = 30° Or sinx = r/(2√(3)) And r = √(3) units. sinx = √(3)/(2√(3)) x = asin(√(3)/(2√(3))) x = asin(½) x = 30° Notice, x i...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2373947

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Calculating the interior angles. It the interior angles be a, b and c identifying them clockwise. Notice. The plane shape consist of two composite regular nonagon. It implies; d = ⅑(180(9-2)) d = 140° d is a single interior angle of the composite regular nonagon. Calculating a. a = 180-40 a = 40° Let the side length of the composite regular nonagon be 1 unit. e² = 1²+1²-2*1*1cos100 e = 1.5320888862 units. f² = 1.5320888862²+1²-2*1.5320888862*1cos100 f = 1.969615506 units. Calculating...
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