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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2262927

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a = 3x+x a = 4x a is the square side. it implies; Area Square is; a² = (4x)² = 16x² square units. Calculating Shaded Area. b = 45° tanc = (4x/3x) c = atan(4/3)° d = 180-b-c d = (180-45-atan(4/3))° Therefore; (4x/sin(180-45-atan(4/3))) = (e/sin(atan(4/3)) e = 3.2324881426x units. It implies; Shaded Area is; 0.5*3.2324881426x*4xsin45 = 4.5714285715x² square units. Shaded Ratio is; 4.5714285715x²÷16x² = 0.2857142857

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2262781

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Let the inscribed purple regular heptagon side be 1 unit. a = ⅐(180*5) a = ⅐(900)° b = 180-⅐(900)) b = ⅐(360)° c = 180-2(⅐(360)) c = ⅐(540)° (1/sin⅐(540)) = (d/sin⅐(360)) d = 0.8019377358 unit. e = 2(0.8019377358)+1 e = 2.60387547161 units. f = 90-½(⅐(540)) f = ⅐(360)° cos⅐(360) = g/2.60387547161 g = 1.62348980186 units. h = 2g h = 3.24697960372 units. Where h is the base of the ascribed quadrilateral. cos(180/7) = i/0.8019377358 i = 0.72252093395 unit. j = 2i j = 1.4450418679 units...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2261681

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Calculating x. (6+12)² = (6+x)²+(12+x)² 18² = 36+12x+x²+144+24x+x² 324 = 180+36x+2x² 2x²+36x+180-324 = 0 2x²+36x-144 = 0 x²+18x-72 = 0 Resolving the quadratic equation. (x+9)² = 72+81 x+9 = ±√(153) x = -9±√(153) It implies; x ≠ -9-√(153) x = (-9+√(153)) units. x = 3.3693168769 units. Calculating area green. a = 6+x a = 6+(-9+√(153)) a = (-3+√(153)) units. b = 12+x b = 12+(-9+√(153)) b = (3+√(153)) units. Green Area (area right-angled triangle) is; ½(ab) =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2261651

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Let the square side be a. b² = 3²+4² b = √(25) b = 5 units. tanc = 3/4 c = atan(¾)° d = 90-c d = 90-atan(¾) d = atan(4/3)° It implies; Calculating a, the square side. tan(d) = a/(5-a) tan(atan(4/3)) = a/(5-a) (4/3) = a/(5-a) 3a = 20-4a 7a = 20 a = (20/7) units. Therefore; a² (square area) is; (20/7)² = (400/49) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2260176

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Let a be the diameter of the semi circle. a² = b²+14² b² = a²-196 b = √(a²-196) It implies; b² = 6²+6²-2*6*6(-cosc) a²-196 = 72+72cosc a²-196-72 = 72cosc 72cosc = a²-268 cosc = (a²-268)/72 --- (1). cosc = 14/a --- (2). Equating (1) and (2). cosc = cosc (a²-268)/72 = 14/a a³-268a = 1008 a³-268a-1008 = 0 Resolving the polynomial equation. a = 18 units. It implies; Radius, r of the semi circle is; ½(a) = ½(18) = 9 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2260119

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Let BD = 1 unit. It implies; CD = 2BD = 2 units. a = 180-60 a = 120° a is angle ADB. b = 180-120-45 b = 15° b is angle BAD (1/sin15) = (c/sin120) c = 3.346065215 units. c is AB. d² = 3.346065215²+3²-2*3.346065215*3cos45 d = 2.4494897428 units. d is AC. (2.4494897428/sin45) = (3/sine) e = 60° e is angle BAC. f = e-b f = 60-15 f = 45° f is angle CAD. Therefore, the required angle, angle ACB is; Let it be g. g = 180-f-60 g = 180-45-60 g = 180-105 g = 75°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2259284

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Let BC be a (square side). tanb = 1/2 b = atan(½)° sinatan(½) = c/a c = 0.4472135955a d = 2c d = 0.894427191a d is CE. ((½)a)² = 2²+e² e² = (a²/4)-4 e = √(¼(a²-16)) e = ½√(a²-16) f = e+(a/2) f = ½√(a²-16)+(a/2) f = ½(√(a²-16)+a) f is CF. It implies; Observing Pythagoras Rule. d² = 2²+f² (0.894427191a)² = 4+(½(√(a²-16)+a))² 0.8a² = 4+¼((a²-16)+2a√(a²-16)+a²) 0.8a² = 4+¼(a²-16)+½(a√(a²-16))+¼(a²) 0.8a²-½(a²) = ½(a√(a²-16)) 1.6a²-a² = a√(a²-16) 0.6a² = a√(a²-16) 0.6a = √(a²-16) 0.36a² = a²-...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2259044

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Calculating PQ. a = 8+2 a = 10 cm. a is the diameter, AC of the semi circle. b = ½(AC) b = ½(10) b = 5 cm. b is the radius of the semi circle. c = b-2 c = 5-2 c = 3 cm. 5² = 3²+d² d² = 25-9 d = √(16) d = 4 cm. d is the radius of the circle. 4² = 5²+5²-2*5*5cose 50cose = 50-16 cose = (34/50) e = acos(17/25)° f = 2e f = 2acos(17/25)° It implies; Length PQ is; (PQ)² = 5²+5²-2*5*5cos(2acos(17/25)) (PQ)² = 50-50cos(2acos(17/25)) (PQ)² = 53.76 PQ = √(53.76) PQ = 7.3321211119 cm. PQ = 7.33 cm...

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/2258577

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Area S, exactly in square unit is; Area triangle with height 2 unit and base (2sin120) unit + Area triangle with height 2 unit and base 2√(2) unit - Area sector with radius 2 unit and angle (30+atan(√(2)))° = (½*2*2sin120) + (½*2*2√(2)) - (4π*(30+atan(√(2)))÷360) = √(3) + 2√(2) - (π(30+atan(√(2)))/90) square unit. = 4.56047793232 - 2.95783078745 = 1.60264714487 square units. ≈ 1.60 square units to 2 decimal places.
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