Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2664461

Let the width of the green inscribed rectangle be a. Let the length of the green inscribed rectangle be b. c = (5-a) cm. d = (5-b) cm. It implies. (5-a)(5-b) = 2*5 (5-a)(5-b) = 10 --- (1). 5² = e²+(5-a)² e² = 25-(25-10a+a²) e = √(10a-a²) cm. e+b = 5 b = 5-√(10a-a²) --- (2). Substituting (2) in (1). (5-a)(5-(5-√(10a-a²))) = 10 (5-a)√(10a-a²) = 10 (5-a)²(10a-a²) = 100 a = 0.5279 cm. Again, a is the width of the green inscribed rectangle. Recall. (5-a)(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2664447

Calculating green inscribed area. Let the green inscribed square side length be a. b = (15-a) units. c = (8-a) units. It implies. a² = ½(8(15-a)) 2a² = 120-8a a²+4a-60 = 0 Resolving the above quadratic equation via factorization approach to get a, side length of the inscribed green square. a²+10a-6a-60 = 0 a(a+10)-6(a+10) = 0 (a+10)(a-6) = 0 It implies; a-6 = 0 a = 6 units. Therefore, area inscribed green square plus area inscribed green triangle in total is;...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2663395

Let the radius of sector ABC be 2 units. Therefore; Calculating TP = PA = RP = BR, let it be x. 2x²=(2-x)² x²+4x-4=0 x = (2√(2)-2) units. Calculating RQ = QC, let it be y. 2y+x=2, and x = 2√(2)-2 2y=2-(2√(2)-2) y = (2-√(2)) units. Consider right-angled triangle PQR. Angle RPQ is; atan(1/√(2))° Therefore; Length PQ is; (2-√(2))/(sin(atan(1/√(2))) PQ = 1.01461187235 units. Length PX is; (2√(2)-2)cos(atan(1/√(2))) PX = 0.6764079149 units. Length...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2663292

a² = 11²+8² a = √(121+64) a = √(185) units. a² = 4²+b² √(185)² = 4²+b² b² = 185-16 b = √(169) b = 13 units. b is the length of the 2 congruent rectangles. c = b-11 c = 2 units. Therefore, shaded area is; Area trapezoid with parallel sides 8 units and 4 units, and height 13 units + Area trapezoid with parallel sides 13 units and 2 units, and height 8 units - Area triangle with height 4 units and base 13 units - Area triangle with height 8 units and base 11 units. ½(4+8)*...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2662711

a² = 4²+(5+3)² a = √(80) a = 4√(5) units. a is the hypotenuse of the ascribed right-angled triangle. Observing similar plane shape (right-angled) side length ratios to get ?, the required length. 4 - ? 4√(5) - 3 Cross Multiply. 4√(5)? = 12 ? = 12/(4√(5)) ? = 3/√(5) ? = ⅕(3√(5)) units. ? = 1.3416407865 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2662699

Let a be the radius of the half circle. b = (a-2) cm. It implies, calculating a. a² = 4²+(a-2)² a² = 16+a²-4a+4 4a = 20 a = 5 cm. Again, a is the radius of the half circle. Therefore; b = (a-2) And a = 5 cm. b = 5-2 b = 3 cm. tanc = 4/3 c = atan(4/3)° Area blue is; Area sector with radius 5 cm and angle atan(4/3)° - Area triangle with height and base 4 cm and 3 cm respectively. (atan(4/3)π*5*5/360)-½(4*3) = 11.591190225-6 = 5.591190225 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2662692

Shaded Area is; Area circle with radius 5 cm + 4(¼(area square with single side length 10 cm - area circle with radius 5 cm)) = 25π + 4(¼(100 - 25π)) = 25π + 4(25 - ¼(25π)) = 25π + 100 - 25π = 100 cm² A very smart approach is; Half the area of the rectangle. = ½(20*10) = ½(200) = 100 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2662680

Let AB be 1 unit. a = 180-28-14-64 a = 180-106 a = 74° a is angle BAC. It implies; (1/sin64) = (b/sin74) b = 0.9350149393 units. b is BC. c = 180-14-64-58 c = 180-136 c = 44° c is angle BDC. It implies; (0.9350149393/sin44) = (d/sin(64+58)) d = 1.141478564 units. d is BD. e² = 1.141478564²+1-2*1.141478564cos28 e = 0.5359494485 units. e is AD. (0.5359494485/sin28) = (1.141478564/sinf) f = asin(1) f = 90° f is angle BAD. Therefore, the required angle, ?...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2662675

Calculating yellow area. a = ½(270)-90 a = 135-90 a = 45° Notice. CD = BD = 23 units. Let b be the radius of the quarter circle. c = 2b units. c is the diameter of the circle. d²+24² = (2b)² d = √(4b²-576) units. e = d-23 e = (√(4b²-576)-23) units. f² = 2b² f = √(2)b units. Calculating b. (√(4b²-576)-23)²+23² = (√(2)b)² 4b²-576-46√(4b²-576)+529+529 = 2b² 2b²+482 = 46√(4b²-576) b²+241 = 23√(4b²-576) (b²+241)² = 529(4b²-576) b⁴+482b²+58081 = 2116b²-3047...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2661817

Let a be the radius of the inscribed semi circle. Calculating a. b² = 9²+9² b² = 162 b = 9√(2) cm. b is the diagonal of a square. c = (9+a) cm. It implies: (9+a)² = a²+(9√(2))²-2*9√(2)acos45 (9+a)² = a²+162-18a a²+18a+81 = a²+162-18a 36a = 81 4a = 9 a = ¼(9) cm. Again, a is the radius of the inscribed semi circle. Let d be the radius of the inscribed circle. e = ((9/4)+d) cm. f = ((9/4)-d) cm. ((9/4)+d)² = ((9/4)-d)²+g² ½(9)d+d² = d²-½(9)d+g² 9d = g² g =...

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