Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2660491

Sir Mike Ambrose is the author of the question. Area Square is: √(40)² = 40 square units. Area Green is; Area triangle with two lengths 4√(10) units and 6√(5) units, and angle 45°, the angle they form - Area triangle with height 2√(10) units and base 2√(10) units - Area triangle with height 2√(5) units and base √(5) units. = ½(4√(10)*6√(5))sin(45) - ½(2√(10)*2√(10) - ½(2√(5)*√(5)) = 60 - 20 - 5 = 35 square units. It implies; Area Green ÷ Area Square is; 35 ÷ 40 = 7/8...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2660278

AD = BC = Width of Rectangle ABCD = Radius of the Semi Circle R = √(4-2√(2)) units. AB = CD = Length of Rectangle ABCD = (2/√(4-2√(2))) units. It implies; Area Yellow (Rectangle ABCD) is; AB x AD = (2/√(4-2√(2))) x √(4-2√(2)) = 2 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2659221

Please, move the above question left/right one time to review the solution. Thank you. Length AB is; 2.37 cm to 2 decimal places.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2659018

a² = 36 a = 6 cm. a is the side length of the inscribed big square. b² = a²+(0.5a)² b² = 6²+(0.5*6)² b² = 36+9 b = √(45) b = 3√(5) cm. b is the radius of the ascribed semi circle. Let c be the side length of the inscribed small blue square. d = ½(a)+c d = (3+c) cm. Calculating c. (3√(5))² = (3+c)²+c² 45 = 9+6c+c²+c² 2c²+6c-36 = 0 c²+3c-18 = 0 Resolving the above quadratic equation via factorization approach to get c, side length of the inscribed blue small square....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2658999

Notice. Radius of the ascribed semi circle is 5 units. a = (5-R) units. (5-R)² = b²+R² b² = 25-10R+R²-R² b² = 25-10R b = √(25-10R) units. c = ½(5)+b c = (2.5+√(25-10R)) units. d = (2.5+R) units. Calculating R. (2.5+R)² = R²+(2.5+√(25-10R))² 6.25+5R+R² = R²+6.25+5√(25-10R)+25-10R 15R = 5√(25-10R)+25 3R = √(25-10R)+5 3R-5 = √(25-10R) (3R-5)² = 25-10R 9R²-30R+25 = 25-10R 9R²-20R = 0 9R = 20 R = ⅑(20) units. R = 2.2222222222 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2658995

Sir Mike Ambrose is the author of the question. Area of the curve is; y = -⅓(x³)+2x² Point p coordinate is; (⅓(32), 4) Therefore Area Red is; Area under the curve at the points (4, 0) and (6, 0) - Area triangle with height ⅓(32) units and base 2 units. ⅓(44) - ⅓(32) = ⅓(12) = 4 square units. Area Blue is; Area triangle with two lengths 5.69600024969 units and 10.8525470641 units, and angle (atan(3/8)+atan(3/16))° = ½(5.69600024969)(10.8525470641)sin(atan(3/8)+atan(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2658241

Please, move the above question left/right one time to review the solution. Thank you. Coloured area region is; 22.85 cm² to 2 decimal places.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2657627

Let a be the side length of the inscribed square. Calculating a. b = (6-a) units. Therefore; 6 - (6-a) 8 - a Cross Multiply. 6a = 48-8a 14a = 48 7a = 24 a = (24/7) units. Again, a is the side length of the inscribed square. c = 8-a c = 8-(24/7) c = (32/7) units. d² = 6²+8² d = 10 units. Recall. b = (6-a) units. And a = (24/7) units. b = 6-(24/7) b = (18/7) units. e² = b²+a² e² = (18/7)²+(24/7)² e = √(900/49) e = (30/7) units. f = d-e f = 10-(...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2657338

Let a and b be the height and base of the triangle respectively. It implies; ½(ab) = 64 ab = 128 --- (1). a²+b² = 16² a = √(256-b²) --- (2). Therefore, substituting (2) in (1) to get b. ab = 128 And a = √(256-b²) √(256-b²)b = 128 (256-b²)b² = (128)² b⁴-256b²+16384 = 0 Let p = b² It implies; p²-256p+16384 = 0 Resolving the above quadratic equation via completing the square approach to get p. (p-128)² = -16384+(-128)² (p-128)²= 0 p = 128 And p = b² b² = 1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2657326

Notice. a+b = 28 units. It implies; b = (28-a) units. Calculating a. 20² = a²+b² And b = 28-a 400 = a²+(28-a)² 400 = a²+784-56a+a² 2a²-56a+384 = 0 a²-28a+192 = 0 (a-14)² = -192+(-14)² (a-14)² = -192+196 a-14 = ±√(4) a = 14±2 a ≠ 12 a = 16 units. Recall. a+b = 28 Therefore; b = 28-a And a = 16 b = = 28-16 b = 12 units.

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