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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3113509

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a² = 2²+2² Where; a is radius of the ascribed half circle. 2 is the side length of the inscribed blue square. a² = 8 a = 2√(2) cm. Again, a is the radius of the circle. Therefore, length AC is; AC = a+2 AC = (2√(2)+2) cm. AC = 2(√(2)+1) cm. AC = 4.82842712475 cm. tanb = 2/(2√(2)) b = atan(1/√(2))° b = 35.2643896828° c = 180-2b c = (180-2atan(½√(2)))° Therefore, length AP is; (AP)² = 2(2√(2))²-2(2√(2))²cos(180-2atan(½√(2))) AP = 4.61880215352 cm. It implie...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3113330

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Calculating Area Blue. a = 4+8 a = 12 cm. a is the radius of the ascribed quarter circle. b² = 12²+4² b = √(144+16) b = √(160) b = 4√(10) units. tanc = 12/4 c = atan(3)° (12/sin(60+atan(3))) = (4/sind) d = 14.4420678686° e = 180-60-atan(3)-d e = 33.9928809543° f = 90-e f = 56.0071190457° (12/sin(60+atan(3))) = (g/sin33.9928809543) g = 8.96692386239 cm. Therefore blue area is; (0.5*8.96692386239*4√(10)sin60)+(56.0071190457π*12²/360)-(0.5*12²sin56.0071190457) = 49.1138647086+70.38062...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3112640

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Notice. R-r = 10 cm. It implies; R = (10+r) cm. R is the side length of the ascribed square. Calculating length AB. a = R-r a = 10 cm. b = ½(R) units. It implies; R² = 10²+(½(R))² R²-¼(R²) = 100 ¼(3R²) = 100 3R² = 400 R = √(400/3) R = ⅓(20√(3)) cm. R = 11.5470053838 cm. And R-r = 10 cm. Therefore; ⅓(20√(3))-10 = r r = ⅓(20√(3)-30) cm. r = 1.5470053838 cm. Recall. b = ½(R) units. b = ½(11.5470053838) b = 5.7735026919 cm. tanc = 10/5.77350269...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3112450

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Sir Mike Ambrose is the author of the question. Calculating Yellow Area. It is; Area triangle with height 4sin(atan(⅔)) units and base 8√(13)/9 units - Area triangle with height 1 unit and base ⅔ units - Area triangle with height ¼ units and base ⅔ units. = ½*4sin(atan(⅔))*(8√(13)/9) - ½*1*⅔ - ½*¼*⅔ = (32/9)-⅓-(1/12) = (113/36) square units. Area Orange is; Area triangle with height ⅓(10sin(atan(¾))) units and base (5√(13)/9) units. = ½*⅓(10sin(atan(3/2)))*(5√(13)/9)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3112432

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Sir Mike Ambrose is the author of the question. Let the square aide length be 1 unit. Square Area is; 1² = 1 square unit. Calculating Green Area. a² = 1²+1² a = √(2) units. a is AB. b = ½(a) b = ½√(2) units. sin60 = c/√(2) 2c = √(6) c = ½√(6) units. Or tan60 = c/(½√(2)) √(3) = 2c/√(2) c = ½√(6) units. tan60 = √(2)/d d = ⅓√(6) units. It implies, area green is; Area right-angled trapezium with parallel lengths √(2) units and ½√(2) units respectively an...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3111676

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Let a be the side length of the big inscribed square. Let b be the side length of the bigger inscribed square. Therefore; 27 ~ b a ~ 64 Cross Multiply. ab = 27*64 b = (1728)/a units. Notice. (a+b) units is the side length of the biggest inscribed square. a+b = a+(1728)/a = (a²+1728)/a units. Calculating a. ((a²+1728)/a)² = ((1728)/a)²+64² ((a⁴+3456a²+2985984)/a²) = (2985984/a²)+4096 a⁴+3456a²+2985984 = 2985984+4096a² a⁴ = 4096a²-3456a² a² = 640 a =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3111412

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Sir Mike Ambrose is the author of the question. Let a be the equal lengths inscribing triangle ABC. Calculating a. ½(a)² = 15 a = √(30) units. b = 180-60-45 b = 75° b is angle BAC. c = 180-2b c = 30° d² = 2√(30)² d = √(60) d = 2√(15) units. e = 180-60-45 e = 75° √(30)/sin60 = (f/sin45) 2√(10) = √(2)f f = 2√(5) units. g = d+f g = 2(√(15)+√(5)) units. Area ABC is; ½*2√(30)*2(√(15)+√(5))sin45 = √(30)*(√(15)+√(5))*√(2) = 2√(15)(√(15)+√(5)) = 30+10...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3111075

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Let R=the bigger inscribed circle radius. Let r=the smaller inscribed circle radius. It implies; R+r=6√(3)-6=4.4  R+r=4.4 -------(1). And the radius of the ascribed semi circle is 6 units. Calculating the R and r considering the radius of the ascribed semi circle. R=√((6-R)²-R²) Therefore; R=√(36-12R) ----- (2). r=√((6-R)²-R²) Therefore; r=√(36-12r) ----- (3). Adding (2) and (3). R+r=√(36-12R)+√(36-12r)-----(4) Therefore equating (1) and (4). √(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3111053

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Notice. Radius rod the half circle is 5 units. a = (5-x) units. Calculating x. 5² = 3²+(5-x)² 16 = 25-10x+x² x²-10x+9 = 0 x = 1 unit. It implies; a = 5-x And x = 1 unit. a = 4 units. Therefore, length red is; Let it be y. y² = (5+3)²+a² y = √(64+16) y = √(80) y = 4√(5) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3110613

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a = (2r-14) units. b = ½(a) b = (r-7) units. Calculating r, radius of the half circle. Therefore; 2r ~ 6 6 ~ (r-7) Cross Multiply. 2r(r-7) = 36 r²-7r-18 = 0 Resolving the quadratic equation via factorization approach. r²-9r+2r-18 = 0 (r²-9r)+(2r-18) = 0 r(r-9)+2(r-9) = 0 (r-9)(r+2) = 0 It implies; r ≠ -2 r = 9 units.
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