answer:1. **Identify the Key Points and Setup:** - Each side of the 12-sided polygon is 5 units, making each angle a right angle. - We need to find the area of quadrilateral ABCM where line segments overline{AF} and overline{CD} intersect at point M. 2. **Calculate the Area of Rectangle ABDE:** - Since each side of the polygon is 5 units, the length of AB and DE are each 10 units (covering two sides of the polygon). - Therefore, the area of rectangle ABDE is: [ text{Area of } ABDE = 5 times 10 = 50. ] 3. **Determine the Areas of Triangles triangle AFE and triangle CDM:** - The area of triangle AFE can be calculated as: [ text{Area of } triangle AFE = frac{1}{2} times AF times FE = frac{1}{2} times 5 times 10 = 25. ] - To find the area of triangle CDM, we use the fact that M falls on line overline{CD}. By symmetry, if M divides overline{AF} and overline{CD} in the same ratio, assume it is the midpoint. The area of triangle CDM is: [ text{Area of } triangle CDM = frac{1}{2} times CD times text{Distance from } M text{ to } overline{CD} = frac{1}{2} times 5 times 2.5 = 6.25. ] 4. **Calculate the Area of Quadrilateral ABCM:** - Subtract the areas of triangle AFE and triangle CDM from the area of rectangle ABDE: [ text{Area of } ABCM = text{Area of } ABDE - (text{Area of } triangle AFE + text{Area of } triangle CDM) = 50 - (25 + 6.25). ] - Simplify the expression: [ text{Area of } ABCM = 50 - 31.25 = 18.75. ] The final answer is boxed{B}.

answer:Given that S_{n} is the sum of the first n terms of a geometric sequence {a_{n}}, and we have two conditions: a_{5}-a_{3}=12 and a_{6}-a_{4}=24. Let's denote the common ratio of the geometric sequence by q. 1. From the condition a_{5}-a_{3}=12, we can express this in terms of the common ratio q as follows: [a_{5} = a_{1}q^{4}, quad a_{3} = a_{1}q^{2}] Substituting these into the given condition gives us: [a_{1}q^{4} - a_{1}q^{2} = 12] 2. Similarly, from a_{6}-a_{4}=24, we can express this as: [a_{6} = a_{1}q^{5}, quad a_{4} = a_{1}q^{3}] Substituting these into the given condition gives us: [a_{1}q^{5} - a_{1}q^{3} = 24] This can be rewritten as q(a_{1}q^{4} - a_{1}q^{2}) = 24, which simplifies to q cdot 12 = 24. 3. Solving for q from the above equation, we find: [q = frac{24}{12} = 2] 4. Knowing q=2, we can substitute back to find a_{1}: [12 = a_{1}(2^{4} - 2^{2})] [12 = 12a_{1}] Thus, a_{1} = 1. 5. The sum of the first n terms of a geometric sequence is given by S_{n} = frac{a_{1}(1-q^{n})}{1-q}. Substituting a_{1}=1 and q=2, we get: [S_{n} = frac{1-2^{n}}{1-2} = 2^{n} - 1] 6. The nth term of the geometric sequence is a_{n} = a_{1}q^{n-1}. Substituting a_{1}=1 and q=2, we find: [a_{n} = 2^{n-1}] 7. Therefore, the ratio frac{S_{n}}{a_{n}} is: [frac{S_{n}}{a_{n}} = frac{2^{n}-1}{2^{n-1}} = 2 - 2^{1-n}] Hence, the correct answer is boxed{text{B: } 2-2^{1-n}}.

answer:Given the problem: In an acute triangle ABC, find a point P such that (BL)^2 + (CM)^2 + (AN)^2 is minimized, where L, M, N are the feet of the perpendiculars from P to BC, CA, AB respectively. To solve this, we use the following steps: 1. Notice that the geometric configuration suggests the use of Euclidean geometry and trigonometric identities. We should express S_{triangle A'B'C} in simpler terms: [ S_{triangle A'B'C'} = 2R^2 cos frac{C}{2} cos frac{A}{2} cos frac{B}{2} ] Here, R is the circumradius of triangle ABC. 2. It is required to prove that [ S_{triangle A'B'C} geq S_{triangle ABC} ] Simplifying S_{triangle A'B'C} (area of the modified triangle with A', B', C') down, we need to verify the inequality: [ sin frac{A}{2} sin frac{B}{2} sin frac{C}{2} leq frac{1}{8} ] 3. The proof of this inequality can be found by invoking the trigonometric identity and standard inequalities that are usually known to hold in acute triangles. Next, we confirm the relationship in the transformed coordinate system: 4. From the given triangles and perpendiculars, evaluate: [ x^2 + h_1^2 = h_3^2 + (c - z)^2 ] [ y^2 + h_2^2 = h_1^2 + (a - x)^2 ] [ z^2 + h_3^2 = h_2^2 + (b - y)^2 ] 5. By adding the equations, we get: [ x^2 + y^2 + z^2 = (a - x)^2 + (b - y)^2 + (c - z)^2 ] 6. Then, simplify: [ 2(x^2 + y^2 + z^2) geq frac{1}{2}[(x + a - x)^2 + (y + b - y)^2 + (z + c - z)^2] ] [ = frac{1}{2}(a^2 + b^2 + c^2) ] Therefore, [ x^2 + y^2 + z^2 geq frac{1}{4}(a^2 + b^2 + c^2) ] 7. The equality holds only when P is the circumcenter of triangle ABC. Thus, the minimum value is achieved when P is at the circumcenter. Conclusion: [ boxed{P text{ is the circumcenter of } triangle ABC} ]

answer:To find the length of the train, we need to calculate the distance it covers in the time it takes to overtake the motorbike. First, we convert the speeds from km/h to m/s, because the time given is in seconds. Speed of train in m/s = 100 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 27.78 m/s Speed of motorbike in m/s = 64 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 17.78 m/s The relative speed at which the train is overtaking the motorbike is the difference in their speeds: Relative speed = Speed of train - Speed of motorbike Relative speed = 27.78 m/s - 17.78 m/s = 10 m/s Now, we can calculate the distance the train covers in the 20 seconds it takes to overtake the motorbike, which will be the length of the train: Distance = Relative speed * Time Distance = 10 m/s * 20 s = 200 m Therefore, the length of the train is boxed{200} meters.