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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2422413

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Let a be the diameter of the smaller inscribed circle. 2a is the radius of the ascribed semi circle. (2a)² = (0.5*3)²+b² b² = 4a²-2.25 b = √(4a²-2.25) units. (2a)² = (0.5*5)²+c² c² = 4a²-6.25 c = √(4a²-6.25) units. c = a a = √(4a²-6.25) a² = 4a²-6.25 6.25 = 3a² a² = 6.25/3 a = √(25/12) units. a = 5/2√(3) a = ⅙(5√(3)) units. a = 1.443375673 units. Again, a is the diameter of the smaller inscribed circle. c = ½(a) c = ½*⅙(5√(3)) c = (5√(3))/12 units. c = 0.7216878365 units. c is the radius...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2421984

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a² = 2√(6)² a² = 12 a = √(12) a = 2√(3) units. b = ½(a) b = ½(2√(3)) b = √(3) units. 2c² = √(3)² 2c² = 3 c = √(3/2) units. c =½√(6) units. d = 2√(6)-½√(6) d = ½(4√(6)-√(6)) d = ½(3√(6)) units. d is the side length of the inscribed square. Observing similar triangles side length ratios. ½√(6) - e 2√(6) - ½(3√(6)) Cross Multiply. ¼(18) = 2√(6)e 9 = 4√(6)e e = ⅜√(6) units. tanf = 2√(6)÷½√(6) f = atan(4)° g = f-45 g = (atan(4)-45)° h = 180-45-(atan(4)-45) h = (90+atan(¼))° tan(atan(4)-45)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2421763

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a = 90+20 a = 110° sin20 = b/1 b = sin20 units. b = 0.3420201433 units. cos20 = c/1 c = cos20 units. c = 0.9396926208 units. Calculating d, the square side length observing similar plane shape side lengths (right-angled triangle) ratios. 1 - 0.3420201433 d - 0.9396926208 Cross Multiply. 0.9396926208 = 0.3420201433d d = 2.7474774197 units. Therefore y is; d-1 y = 2.7474774197-1 y = 1.7474774197 units. Calculating x. x² = 1²+1²-2*1*1cos110 x = √(2.6840402867) x = 1.6383040886 units. It...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2421438

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Let a be 1 unit. c² = 1²+1²-2*1*1cos120 c² = 2-2cos120 c = √(3) units. c = 1.7320508076 units. It implies; 1+b = c 1+b = √(3) b = (√(3)-1) units. b = (1.7320508076-1) units. b = 0.7320508076 units. Notice. a = 1 unit is the side length of the regular hexagon. b = (√(3)-1) units is the side length of the three congruent inscribed squares. Therefore; b/a is; (√(3)-1)/1 = (√(3)-1) = 0.7320508076 ≈ 0.73 to 2 decimal places.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2421406

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a² = 30²+10²-2*30*10cos120 a = 36.0555127546 cm. (36.0555127546sin120) = (30/sinb) b = 46.1021137521° cosb = 10/d Where d is PQ. cos46.1021137521 = 10/d d = 10/cos46.1021137521 d = 14.4222051019 cm.

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/2419797

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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/2419738

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Notice, AB = 4 units is the height of the inscribed regular pentagon. Calculating Area Blue. a = ⅕(180(5-2)) a = 108° a is the single interior angle of the inscribed regular pentagon. b = ½(a) b = ½(108) b = 54° c = ½(180-108) c = ½(72) c = 36° d = b-c d = 54-36 d = 18° Therefore; tan18 = e/4 e = 1.2996787849 units. f = 2e f = 2*1.2996787849 f = 2.5993575699 units. f is the side length of the inscribed regular pentagon. g = (4-r) units. g is the radius of the inscribed circle. r is the...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2419488

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Calculating x, the diameter of the semi circle. Notice, we have a cyclic quadrilateral in the complete circle. Therefore; x² =11²+y² y² = x²-121 --- (1). cosa = (11/x) --- (2). y² = 7²+2²+2*2*7cosa --- (3). Therefore, substituting (2) and (1) in (3) to get x, the diameter of the semi circle. x²-121 = 7²+2²+2*2*7(11/x) x²-121 = 53+28(11/x) x³-121x = 53x+308 x³-174x-308 = 0 Resolving the above polynomial equation to get x, the diameter of the half circle. x³-14x²+14x²-174x-308 = 0 It imp...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2418395

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Notice. Area Blue is an inscribed half circle. Let r be the radius of area blue. r² = (r-1)²+(r-2)² r² = r²-2r+1+r²-4r+4 0 = r²-6r+5 Resolving the above quadratic equation via completing the square approach to get r, radius of the inscribed half circle. (r-3)² = -5+(-3)² (r-3)² = 4 r-3 = ±√(4) r = 3±2 Therefore; r ≠ 3-2 = 1 cm. r = 3+2 r = 5 cm. It implies, area blue inscribed semi circle is, ½(πr²/ = 0.5*π(5)² = ½(25π) cm² = 12.5π cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2418181

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Calculating r, radius of the two congruent circles. r² = 2²+a² a = √(r²-4) units. r² = 1²+b² b = √(r²-1) units. c = 2+1+1 c = 4 units. d = b-a d = (√(r²-1)-√(r²-4)) units. Therefore, r, radius of the two congruent circles is; (2r)² = c²+d² (2r)² = 4²+(√(r²-1)-√(r²-4))² 4r² = 16+r²-1-2√((r²-1)(r²-4))+r²-4 2r²-11 = -2√((r²-1)(r²-4)) 4((r²-1)(r²-4)) = (11-2r²)² 4(r⁴-5r²+4) = 121-44r²+4r⁴ -20r²+16 = 121-44r² 24r² = 105 8r² = 35 r = √(35)/√(8) r = ¼√(70) units. r = 2.0916500663 units.
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