Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2029469

Sir Mike Ambrose is the author of the question. Let the side of the inscribed square be a. Calculating a. ¾ = a/b b = ⅓(4a) It implies; ½*a*((⅓(4a))+3a) = 78 ⅙(13a²) = 78 ⅙(a²) = 6 a = 6 cm. b = ⅓(4*6)+6+6 b = 20 cm. ¾ = c/20 c = 15 cm. Let the side of the regular hexagon be d. Calculating d. e² = 2d²-2d²cos120 e = √(3)d cm. (¾) = √(3)d/f f = ⅓(4√(3)d) cm. It implies; d+e+f = 15 d+√(3)d+⅓(4√(3)d) = 15 3d+3√(3)d+4√(3)d = 45 d = 45/(3+7√(3)) cm It implies; The side of the regular...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2028126

Calculating x. Let the two congruent lengths of the plane shape be 1 unit. (a/sin11) = (1/sin22) a = 0.50935834748 units. (b/sin147) = (1/sin22) b = 1.45389601942 units. c = 1+a c = 1.50935834748 units. d = 1+b d = 2.45389601942 units. It implies; e² = 2.45389601942²+1.50935834748²-2*2.45389601942*1.50935834748cos11 e = 1.01402672537 units. Required angle x is; (1.01402672537/sin11) = (1.50935834748/sinx) x =16.5°

Class with Adesuwa

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Please, browse through the link below to review our teaching and learning activities with super smart and clever Adesuwa. She is in year 6. Thank you. https://www.facebook.com/100085657676431/posts/pfbid0J4K6L27WpGtcJYLU672sE95EwoA4dLyCgMJ7TdfSyBEVZyKYErZNeDWbBVyVTivFl/?app=fbl

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2027792

Sir Mike Ambrose is the author of the question. ⅓(4) = a/9 a = 12 cm. let alpha be b. Calculating b. (2tanb/(1-tan²b)) = (4/3) 2tan²b+3tanb-2= 0 b = atan(½)° ½ = c/9 c = 4.5 cm. d = ½(12-c) d = 3.75 cm. ½ = 3.75/e e = 7.5 cm f² = 8.25²+7.5² f = 11.14955156049 cm. g = ½(90-atan(½))° g = 31.71747441146° h = 90-31.71747441146-atan(8.25/7.5) h = 10.55621459463° Let r be the radius of the inscribed circle. sin31.71747441146 = r/i i = 1.90211303259r j²...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2027786

Notice; AB = AC (6.84/sin40) = (a/sin70) a = 9.99941104856 a = 10 cm. a is AB = AC. (9.99941104856/sin155) = (b/sin15) b = 6.12381965823 cm. (9.99941104856/sin155) = (c/sin10) c = 4.10862393682 cm. d = 70-15 d = 55° Area Green is; 0.5*6.84*4.10862393682sin55 = 11.51030992395 cm² e = 9.99941104856-6.84 e = 3.15941104856 cm. Area Orange is; 0.5*3.15941104856*6.12381965823sin30 = 4.8369158719 cm² It implies; Area Orange ÷ Area Green to 2 decimal places is; 4.8369158719÷11.510309923...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2026898

Let the two congruent longer lengths be 1 unit. (1/sin135) = (b/sin30) b = 0.70710678119 units. (1/sin135) = (c/sin15) c = 0.36602540378 units. d = 1+b d = 1.70710678119 units. e = 1+c e = 1.36602540378 units. f² = 1.36602540378²+1-2*1.36602540378cos30 f = 0.70710678118 units. (0.70710678118sin30) = (1/sing) g = 45° It implies; x = 180-2(g) x = 90° Calculating y. h = 180-15-90 h = 75° y = ½(180-75) y = ½(105) y = 52.5°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2026839

Let the longer pair of congruent lengths be 1 unit. (1/sin60) = (a/sin20) a = 0.39493084363 units. (1/sin60) = (b/sin100) b = 1.1371580426 units. c = 1+a c = 1.39493084363 units. d = 1+b d = 2.1371580426 units. It implies; e² = 1.39493084363²+2.1371580426²-2*1.39493084363*2.1371580426cos20 e = 0.95418889414 units. It implies, the required angle, x is; Let y be the supplementary angle of angle x. Calculating y. (0.95418889414/sin20) = (2.1371580426/siny) y = 50° Therefore; x+y = 180...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2026802

Let the side length of the regular hexagon be 1 unit. sin60 = a/1 a = ½√(3) units. tanb = (a/1) b = atan(0.5√(3))° b = 40.89339464913° b is angle DEH. c = atan(2/√(3)) c = 49.10660535087° d = 120-40.89339464913 d = 79.10660535087° e = 40.89339464913-30 e = 10.89339464913° It implies, the required angle, angle EIH is; d+e = 79.10660535087+10.89339464913 = 90°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2026658

Let r be the radius of the inscribed circle. Calculating r. a = √(2*24²) a = 24√(2) units. b = √(2*r²) b = √(2)r units. It implies; 16+r+√(2)r = 24√(2) r(√(2)+1) = (24√(2)-16) r = (24√(2)-16)/(√(2)+1) r = 8(8-5√(2)) units. r = 7.43145750508 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2025221

Sir Mike Ambrose is the author of the question. Let the side length of the inscribed regular hexagon be 1 unit. Calculating r, radius of the circle. 1+2(sin30) r = 2 units. Area circle is; π(2)² = 4π square units. = 12.56637061436 square units. a = (1800/12) a = 150° b² = 2-2cos150 b = 1.93185165258 units. 2² = c²+1.93185165258²-2*1.93185165258*ccos135 4 = c²+3.73205080758+ 2.73205080757c c²+2.73205080757c-0.26794919242 = 0 c = 0.0947876 unit. (2/sin135) = (1.93185165258/sind) d = 43.0...

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Mathematics Question and Solution

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