Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2384313

Let the side length of the three congruent regular hexagon be 1 unit. a = ⅙(180(6-2)) a = 120° a is the single interior angle of the three congruent regular hexagon. b² = 2-2cos120 b = √(3) units. tanc = 2√(3)/1 c = atan(2√(3)) c = 73.897886248° d = c-30 d = 73.897886248-30 d = 43.897886248° tan43.897886248 = e/√(3) e = ⅓(5) units. f² = (⅓(5))²+(√(3))² f = ⅓(2√(13)) units. f = 2.4037008503 units. g² = 1²+2/(2√(3))² g = √(13) units. g = 3.6055512755 units. h...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2383929

Calculating r, radius of the smaller semi circle. Radius of the bigger semi circle a, is; a = ½(3+7+2) a = 6 units. 6² = r²+b² 36-r² = b² b = √(36-r²) units. c = 6-3.5-2 c = 0.5 units. b² = d²+0.5² √(36-r²)² = d²+0.25 d² = 36-0.25-r² d² =35.75-r² d = √(35.75-r²) units. r² = d²+3.5² r² = √(35.75-r²)²+3.5² 2r² = 35.75+12.25 2r² = 48 r² = 24 r = √(24) r = 2√(6) units. r = 4.8989794856 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2383108

Notice! AB = AC It implies, triangle ABC is isosceles. a = 180-45-30 a = 105° a is angle ADC. b = ½(180-45) b = ½(135) b = 67.5° b = angle ABC = angle ACB c = 67.5-45 c = 22.5° c is angle DBE. d = 67.5-30 d = 37.5° d is angle BCD. Let BC be 1 unit. Observing Sine Rule. (1/sin(180-45-67.5)) = (e/sin67.5) e = 1 unit. e is BE. Observing Sine Rule. (f/sin(67.5-30)) = (1/sin(180-37.5-67.5)) f = 0.630236207 units. f is BD. Observing Cosine Rule. g² = 0.630236207²+1²-2*0.630236207*1cos22.5 g...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2383046

Area green is; 6(2(area sector with radius 4 units and angle 60°-area triangle with height 4 units and base 4sin60 units.)) = 6(2((60π*4*4/360)-(½*4*4sin60))) = 6(2(⅓(8π)-4√(3))) = 12(⅓(8π)-4√(3)) = 32π-48√(3) = 16(2π-3√(3)) square units. = 17.3925261516 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2383011

Notice. The composite plane shape consist of two regular plane shapes, a square and an equilateral triangle. Let the four congruent units of the square be 1 each. It implies a, the square side is; a = 3 units. b² = 3²+1² b² = 9+1 b = √(10) units. b is a side length of the regular triangle (equilateral triangle). tanc= 1/3 c = atan(⅓)° d = 60-c d = (60-atan(⅓))° e = 90+d e = 90+(60-atan(⅓)) e = (150-atan(⅓))° e = 131.5650511771° It implies; f² = 3²+√(10)²-2*3...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2382942

Let a be the radius of the inscribed circle. Let the circle's centre be X. Calculating a. (3-a)² = √(3)²+a² 9-6a+a² = 3+a² 6 = 6a a = 1 unit. a is the radius of the inscribed circle (XT = XN = XE). tanb = √(3)/1 b = atan(√(3) b = 60° b is angle XTH c = 180-60 c = 120° c is angle EXT. It implies, area grey is; Area sector with radius 1 unit and angle 120° - Area triangle with height 1 unit and base sin120 units. = (120π*1*1/360)-(½*1*1sin120) = ⅓(π)-½*½(√(3)) = ⅓(π)-¼√(3) = (4π-3√(3))/1...

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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/2382365

Let R = 1 unit. sin22.5 = r/(1-r) r = sin22.5-rsin22.5 r+rsin22.5 = sin22.5 r(1+sin22.5) = sin22.5 r = sin22.5/(1+sin22.5) units. r = 0.2767686539 units. It implies; R/r = 1/(sin22.5/(1+sin22.5)) = (1+sin22.5)/sin22.5 = 3.6131259298 units

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2382178

Let BC be 1 unit. a = 180-60-20-40-10 a = 50° a is angle ACB. Observing Sine Rule. (b/sin50) = (1/sin(60+20)) b = 0.7778619134 units. b is AB. c = 180-40-60 c = 80° c is angle ADB. Observing Sine Rule. (0.7778619134/sin80) = (d/sin60) d = 0.6840402867 units. d is BD. Observing Cosine Rule. e² = 0.6840402867²+1²-2*0.6840402867*1cos10 e = 0.3472963553 units. e is CD. Therefore, calculating the required angle alpha (angle BCD). Let it be f. It implies, observing...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2380531

Notice. Radius of the ascribed circle is 7 cm. Calculating a, radius of the big inscribed circle. (7-a)² = 2a² 49-14a+a² = 2a² a²+14a-49 = 0 Resolving the above quadratic equation via completing the square approach. (a+7)² = 49+(7²) (a+7)² = 98 a = -7±√(98) a = -7±7√(2) It implies; a ≠ -7-7√(2) a = -7+7√(2) a = 2.8994949366 cm. b = 2(7)-a b = 14-2.8994949366 b = 11.1005050634 cm. b is the diameter of the orange circle. c = ½(b) c = ½(11.1005050634) c = 5.5502525317 cm. c is the radius...

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