Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2677566

Calculating x, side length of the regular heptagon. (1/a)+(1/b) = ⅓ 3a+3b = ab ab-3a = 3b a(b-3) = 3b a = 3b/(b-3) --- (1). c = ⅐*180(7-2) c = ⅐(900)° c is the single interior angle of the regular heptagon. d = 90-½*⅐(900) d = ⅐(630-450) d = ⅐(180)° cos(⅐(180)) = ½(b)/x x = 0.5549581321b b = 1.8019377358x --- (2). e = ½(360-⅐x2(900)) e = (180-⅐(900) e = ⅐(1260-900) e = ⅐(360)° cos(⅐(360)) = f/x f = 0.6234898019x units. It implies; a = x+2f a = x+2(0.6234898...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2677371

Let a be the side length of the square. b = (a-3) m. c = (a-5) m. It implies; 10² = (a-3)²+(a-5)² 100 = a²-6a+9+a²-10a+25 2a²-16a+34-100 = 0 2a²-16a-66 = 0 a²-8a-33 = 0 a²-11a+3a-33 = 0 a(a-11)+3(a-11) = 0 (a-11)(a+3) = 0 It implies; a = 11 m. Area square is; a² = 11² = 121 m²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2677223

a² = 64 a = 8 m. a is the side length of the square. Calculating radius of the circle. b = ½(8+c) m. b is the radius of the circle. d = ½(8-c) m. Therefore; (½(8+c))² = (½(8-c))²+4² ¼(64+16c+c²) = ¼(64-16c+c²)+16 64+16c+c² = 64-16c+c²+64 32c = 64 c = 2 m. Recall. b = ½(8+c) And c = 2 m. b = ½(8+2) b = 5 m. Again, b is the radius of the circle. Area circle is; πb² = π*5² = 25π m²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2677209

Let a be the radius of the quarter circle. b = (a-7) units. c = (a-14) units. It implies; a² = (a-7)²+(a-14)² a² = a²-14a+49+a²-28a+196 a²-42a+245 = 0 Resolving the above quadratic equation a via factorization approach to get a. a²-35a-7a+245 = 0 a(a-35)-7(a-35) = 0 (a-35)(a-7) = 0 It implies; a = 35 units. Again, a is the radius of the quarter circle. Recall. b = a-7 And a = 35 units. b = 35-7 b = 28 units. c = a-14 And a = 35 units. c = 35-14 c =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2676476

Please, move the above question left/right one time to review detailed explanation of the solution. Thank you. The derived used equations are; 3y = 5x ------- (1) y = ⅓(5x) ------ (2) (y-5)(x-3) = 5x ------- (3) Substituting (2) in (3) to get x. x = ½(9+3√(5)) units. Therefore; Using (2). y is; ½(15+5√(5)) units. Where x is the length of the pink rectangle, and y the length of the grey rectangle. Therefore; Area rectangle ABCD exact is; Length AD x Len...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2676449

Sir Mike Ambrose is the author of the question. Area red exactly as a single fraction is; Area triangle red with two lengths ⅓(24-8√(3)) cm and ⅓(18-8√(3)) cm, and angle 60° = ½(⅓(24-8√(3)))(⅓(18-8√(3)))sin60 = ((24-8√(3))(18-8√(3))√(3))/36 = (624√(3)-1008)/36 = (104√(3)-168)/6 = ⅓(52√(3)-84) cm² = 2.02221399786 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2675508

Please, move the above question left/right one time to review the solution. Thank you. Area Square OBCD is; ½(3+2√(2)) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2675320

Let a be the radius of the inscribed circle. b = a+2a b = 3a units. c² = 2(3a)² c = 3√(2)a units. d = 180-45 d = 135° Calculating a. 5² = (3√(2)a)²+a²-2*3√(2)a²cos135 25 = 19a²+6a² 25 = 25a² a = 1 unit. Again, a is the radius of the inscribed circle. Calculating length x. x² = 2a² x² = 2 x = √(2) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2675296

Calculating L. Let a be the width of the rectangle. b = (a-x) units. a(a-x) = 4 --- (1). Observing similar plane shape (right-angled) side length ratios. (a-x) - ½(L) L - a Cross Multiply. a(a-x) = ½(L²) --- (2). Therefore, equating (1) and (2) to get L. ½(L²) = 4 L² = 8 L = √(8) L = 2√(2) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2674965

½(a*a) = 10 a² = 20 a = 2√(5) units. a is the equal adjacent lengths of the inscribed purple isosceles right-angled triangle. Let c be the side length of the square. tand = c/(0.5c) d = atan(2)° e = d-45 e = (atan(2)-45)° cos(atan(2)-45) = 0.5c/(2√(5)) c = 8.4852813742 units. c = 6√(2) units. c is the side length of the ascribed square Are ascribed square (ABCD) is; c² = (6√(2))² = 6√(2)*6√(2) = 72 square units.

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