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Screenshot of today's teaching and learning activities with some of our super smart and clever learners; Oghosa (year 7), Adesuwa (year 5), Derick (year 6), Ogheneruemu (year 4) and Oghenezino (year 4). Educators: Mr. Daniel, Mr. Mudiaga We are smart and clever.

We are smart and clever!

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Screenshot of some of our numerous teaching and learning activities with our very talented, smart and clever learners. Online Edumath Educators and Learners are Smart and Clever!

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1868239

Let the side length of the square be 2 units. Let the radius of the small inscribed circle be x. Let the radius of the big inscribed circle be y. Calculating y. (2-y)² = y²+y² 2-y = √(2)y 2 = √(2)y+y It implies; y = 2/(√(2)+1) units. y = 0.82842712475 units Where y is EG. Calculating x. (2+x)² = 2(2-x)² 4+4x+x² = 2(4-4x+x²) 4+4x+x² = 8-8x+2x² x²-12x+4 = 0 Solving the quadratic equation using completing the square approach to get x. (x-6)² = -4+(-6)² (x-6)² = 32 x-6 = ±4√(2) x = 6±4√(2)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1868158

Radius of the inscribed circle, r is; r = 2 cm. Therefore; Shaded Region is; area sector with radius 2 cm and angle 120°- area triangle with two equal lengths 2 cm and the angle they form, 120° + Area quarter circle with radius 2 cm - 2(area sector with radius 2 cm and angle 60°) + Area equilateral triangle with sides 2 cm each. = ⅓(*4π)-½(4sin120)+(¼*4π)-2((60π*4)/360)+(½*4sin60) = ⅓(4π)-√(3)+π-⅓(4π)+√(3) = π cm² Or Shaded Region is; Quarter the area of the inscribed...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1868088

Calculating shaded area considering semi circle PQ. Diameter of semi circle PQ will be 10cm, therefore radius is 5cm. Shaded area will be; Area semi circle PQ of radius 5cm and angle 180° - 2(area sector of radius 5cm and 60° - area equilateral triangle of equal side lengths 5cm each and angle 60° each angle) - 2(area sector of radius 5cm and 60° - area equilateral triangle of equal side lengths 5cm each and angle 60° each angle). =(5²π)/2 - (2(5²π*60)/360 - (5²sin60)/2) - (2(5²π*6...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1867952

Let angle CAD = a Let angle BAD = b Let AD = c tana = 3/c ----- (1). tanb = 2/c ----- (2). Adding (1) and (2). Therefore; tana+tanb = 5/c ----- (3). Notice; tan(a+b) = tana+tanb/(1-tanatanb) tana+tanb = tan(a+b)(1-tanatanb) And, a+b = 45° It implies; tana+tanb = tan(45)(1-tanatanb) tan45 = 1 Therefore; tana+tanb = (1-tanatanb) ----- (4). Substituting (1), (2) and (3) in (4) to get c (height of triangle ABC) 5/c = 1-(3/c)(2/c) 5/c = 1-(6/c²) (5/c)+(6/c²) = 1 (5c+6)/c² = 1 c²-5c-6...

We are smart and clever!

https://onlineedumath.com/update/we-are-smart-and-clever/1867356

Screenshot of today's teaching and learning activities with some of our very talented, smart and clever learners; Prince (year 4) and Oghenezino (year 4). Educator: Mr. Oyoma We are smart and clever.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1867311

Let the radius of one the the two congruent inscribed circles be a. Therefore; a = 2 cm. tanb = (2/4) b = atan(½)° c = (180-2atan(½))° c = 126.86989764584° Shaded Region is; Half area rectangle with length 4 cm and width 2 cm - Area square with side 2 cm + Area quarter circle with radius 2 cm - Area sector with angle 126.86989764584° and radius 2 cm + Area triangle with two equal lengths 2 cm and angle they form, 126.86989764584° = ½(4*2)-(2*2)+¼(2²π)-(126.86989764584π*2²)360+(0.5*2*2si...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1867163

AC = √(12²+12²) AC = 12√(2) cm. AE = √(12²+6²) AE = 6√(5) cm. Angle BAE = atan(½)° Angle CAE = 45-atan(1/2) Angle CAE = 18.43494882292° cos18.43494882292 = (AF)/12√(2) AF = 16.099689438 cm. EF = AF-AE EF = 16.099689438-6√(5) EF = 2.683281573 cm. Angle BCG = 90-45-18.43494882292 Angle BCG = 26.56505117708° tan26.56505117708 = (BG)/12 BG = 6 cm. CF = √(6²-2.683281573²) CF = 5.366563146 cm. (BF)² = 16.099689438²+12²-24*16.099689438cos(atan(1/2)) BF = 7.58946638441 cm. (7.58946638441/sin...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1866286

Let AB = a. Let BE = b. It implies; 2a+b = 392 b = 392-2a BC = ½(b-70) BC = ½(392-2a-70) BC = ½(322-2a) Therefore; BC = DE = 161-a c = ½(70)+161-a c = 196-a Where c is half BE. d² = a²-(196-a)² d² = a²-38416+392a-a² d² = 392a-38416 d = √(392a-38416) Where d is the height of the triangle ABE. e² = √(392a-38416)²+35² e² = 392a-38416+1225 e² = 392a-37191 e = √(392a-37191) Where e is AC = AD. It implies; Calculating a (AB = AE) AB+BC = AD+CD a+161-a = √(392a-37191)+70 161-70 = √(392a-3719...

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We are smart and clever!

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Mathematics Question and Solution

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Mathematics Question and Solution

Mathematics Question and Solution

We are smart and clever!

Mathematics Question and Solution

Mathematics Question and Solution

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