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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2258428

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Notice. The green circles are congruent (their radius is equal). Let it be r. Calculating r. a = (1-r) units. b = (1+r) units. c = (1-r-r-r) = (1-3r) units. It implies; Observing Pythagoras Rule. b² = a²+c² (1+r)² = (1-r)²+(1-3r)² 1+2r+r² = 1-2r+r²+1-6r+9r² 4r = 1-6r+9r² 9r²-10r+1 = 0  Resolving the above quadratic equation observing factorization approach. 9r²-9r-r+1 = 0 9r(r-1)-1(r-1) = 0 (9r-1)(r-1) = 0 Therefore; r ≠ 1 unit. r = ⅑ units. It implies, radius of the green congruent cir...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2257715

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Calculating Red Area. tana = 5/10 a = atan(½)° b = 2a b = 2atan(½)° tanc = 10/5 c = atan(2)° d = 2c d = 2atan(2)° It implies; Area Red is; Area sector with radius 5 units and angle 2atan(2)° - (2(area triangle with height 10 units and base 5 units)-area sector with radius 10 units and angle 2atan(½)°) - 2(Area quarter circle with radius 5 units-area triangle with height and base 5 units respectively) = (2atan(2)π*5*5/360) - (2(0.5*10*5)-(2atan(½)π*10*10/360)) - 2((¼*π*5*5)-½*5*5) = 27.6...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2257504

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a² = 2*6² a² = 72 a = 6√(2) cm. a is diagonal of the square. Let r be the radius of the inscribed circle. It implies; b² = 2r² b = √(2)r cm. Therefore; 4+r+b = a 4+r+√(2)r = 6√(2) Calculating r. (1+√(2))r = 6√(2)-4 r = (6√(2)-4)/(1+√(2)) r = ((6√(2)-4)(1-√(2)))/(-1) r = (6√(2)-12-4+4√(2))/(-1) r = (10√(2)-16)/(-1) r = 2(8-5√(2)) cm. r = 1.8578643763 cm exactly in decimal.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2256984

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Area Orange is; 87/10 square unit. Please, move the above question/figure left, one time to review the solution. Thank you.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2256354

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Radius, r of one of the five congruent, inscribed circles is; 8r²+√(3)r-1=0 r = (√(35)-√(3))/16 m. Therefore; Total area of the five congruent, inscribed yellow circles is; 5πr² = 5π( (√(35)-√(3))/16)² = 5π(35-2√(105)+3)/256 = 5π(38-2√(105))/256 = 5π(19-√(105))/128 m² = 1.07416074883 m²

We are smart and clever!

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Please, browse through the link below. Screenshot slide video of our teaching and learning activities with our super smart and amazingly, very clever learners. It is in memory of Late Sound Sultan (Legendary Advocate Musician). Rest on, Sound Sultan. https://www.facebook.com/100085657676431/posts/pfbid0sABrt5eAC5KgmzRhTc37TRVcb8LUWugQU2zjuwnBQdJc32AudxpytFQ1F7Uti6k3l/?app=fbl

We are smart and clever!

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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 +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/2255662

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Notice; The semi circle has inscribed quadrilateral which is a cyclic quadrilateral when the semi circle is made a complete circle considering it's diameter. Calculating Alpha. Let it be a. It implies; 60+b = 180 b = 120° Therefore; a+b = 180 a = 180-b Where b = 120°, and a is Alpha. It implies; a = 180-120 a (Alpha) = 60°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2255632

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cos30 = r/(1-r) Where r is the radius of each of the three congruent inscribed circles. It implies; 2r = √(3) - √(3)r r = √(3)/(2+√(3)) r = (2√(3)-3)/(1) r = (2√(3)-3) unit. r = 0.46410161514 unit.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2255511

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Area blue is; Area square with side 10 cm - Area triangle with height 5 cm and base 10 cm - Area triangle with height 10 cm and base (4.71404520791sin45) cm - Area triangle with height 5 cm and base (4.71404520791sin45) cm - Area triangle with height 5 cm and base (2.2360679775sin(atan(2))) cm = 10² - ½(10*5) -½(10*4.71404520791sin45) - ½(5*4.71404520791sin45) - ½(5*2.2360679775sin(atan(2))) = 100 - 25 - ⅓(50) - ⅓(25) - 5 = 70 - ⅓(50) - ⅓(25) = ⅓(210-75) = ⅓(135) cm² = 45 cm²
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