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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2359159

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a² = 6²-1² a = √(35) units. b = atan(√(35)) cos(atan(√(35))) = c/2 c = ⅓ unit. Where c is CE. d = atan(√(35))-60 d = 20.40593177314° Where d is angle FCE. tan20.40593177314 = e/(1/3) e = 0.12400483955 units. Where e is EF.  f = 6-(1/3) f = ⅓(17) units. Where f is AE. Therefore; Length AF is; √((⅓(17))²+(0.12400483955)²) = 5.66802331605 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2358759

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Sir Mike Ambrose is the author of the question. Let the side of the regular pentagon be 1 unit. It implies; Area red is; ½*sin54*cos54 = 0.23776412908 square units. Area orange is; ½*((sin54-cos54sin72)/sin72)((sin54sin72sin108+sin²54)/sin72sin108) = 0.20143700267 square units. Area green is; ½*((sin54)⁴/(sin108(sin72)²)) = 0.24898982848 square units. Therefore; Area Red ÷ (Area Green - Area Orange) is; 0.23776412908 ÷ (0.24898982848 - 0.20143700267) = 0.23776412908 ÷ 0.04755282581 = 5....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2355523

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Radius of the inscribed small semi circles, r is; (3+r)² = 3²+(6-r)² 36 = 18r r = 2 cm Therefore; Area blue is; Area semi circle with radius 6 cm - Area semi circle 2 cm - Area circle with radius 3 cm. = ½*6²π - ½*2²π - 3²π = ½(36π) - ½(4π) - 9π = 18π - 2π - 9π = 18π - 11π = 7π cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2355510

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tan60 = a/5 a = 5√(3) cm. Where a is AD. b = ½(a) b = ½(5√(3)) cm  c = atan((½(5√(3))/5) c = atan(½√(3)) Where c is angle AGB d = atan(5√(3)/5) d = atan(√(3)) Where d is angle CAG. e = 180-c-d e = 180-atan(½√(3))-atan(√(3)) e = 79.10660535087° Where e is angle AFG. Therefore; (5/sin79.10660535087) = (f/sin(atan(√(3)))) f = 4.40958551844 cm. Where f is FG. It implies; Shaded Area is; Area triangle with height 4.40958551844 cm and base 5sin(atan(0.5√(3))) cm. = 0.5*4.40958551844*5sin(a...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2355501

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Let the square side be a. tan60 = a/b b = a/tan60 cm. Where b is BS. Calculating a. 2b+a = 6 2(a/tan60)+a = 6 a((2/tan60)+1) = 6 a = 6/((2/tan60)+1) a = 6(2√(3)-3) cm. Area Shaded STR is; ½(a²) = ½(6(2√(3)-3))² = 54(7-4√(3)) cm² = 3.87702556513 cm²

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/2355119

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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/2355101

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a² = 2*6² a² = 2*36 a = √(2*36) a= 6√(2) units. Let the radius of the small inscribed circle be r. Calculating r. b² =2r² b = √(2)r units. It implies; b+r+6 = a √(2)r+r+6 = 6√(2) (√(2)+1)r = 6√(2)-6 r = (6√(2)-6)/(√(2)+1)  r = (12-6√(2)-6√(2)+6)/1 r = (18-12√(6)) r = 6(3-2√(2) units. r = 1.0294372515 units. It implies; Area green is; Half area square with side 6 units - Area sector with radius 6 units and angle 45° - Half area square with side 1.0294372515 units - Area sector with radi...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2354568

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a = atan(2)° b = 180-2a b = (180-2atan(2))° Where b is angle FBP. Notice; CQ = 5 cm. Calculating FP using sine rule. Let FP be c. (c/sin(180-2atan(2))) = (5/sin(atan(2))) c = 4.472135955 cm. Therefore; PQ is; CF-CQ-FP = √(10²+5²)-5-4.472135955  = (3√(5)-5) cm. = 1.7082039325 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2354558

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a² = 10²+10²-2*10*10cos150 a = 19.31851652578 cm. Where a is AF which is equal AE. (19.31851652578/sin150) = (10/sinb) b = 15° Where b is angle BAF. tan15 = c/10 c = 2.67949192431 cm. It implies; Area Green Region is; Area square with side 10 cm - 2(area triangle with height 2.67949192431 cm and base 10 cm. = 10² - ½*2(10*2.67949192431) = 100-26.7949192431 = 73.2050807569 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2353582

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Let r be the radius of the ascribed semi circle. Therefore; ½(r) is the radius of the inscribed circle. (2r)² = 2a² a² = 2r² a = √(2)r cm. b = ½(a) b = ½(√(2)r) cm. c² = a²+b² c² = (√(2)r)²+(½(√(2)r))² c² = 2r²+½(r²) c = √(½(5r²)) c = √(2.5)r cm. d = 2+c d = (2+√(2.5)r) cm. b² = e²+2² (½(√(2)r)²-2² = e² e² = ½(r)²-4 e = √(0.5r²-4) cm. It implies; Calculating r. (2r)² = d²+e² 4r² = (2+√(2.5)r)²+√(0.5r²-4)² 4r² = 4+4√(2.5)r+2.5r²+0.5r²-4 r² = 4√(2.5)r r = 0 Or r = 4√(2.5) units. r = 6.3...
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