🧠:<think>Okay, so I've got two problems here about a series of nested tetrahedrons in an art gallery exhibition. Let me try to figure them out step by step.Starting with the first problem: Each installation is a series of nested tetrahedra, each scaled down by a factor of 1/3 from the previous one. The largest tetrahedron has vertices at (3, 0, 0), (0, 3, 0), (0, 0, 3), and (1, 1, 1). I need to find the coordinates of the vertices of the smallest tetrahedron in this series.Hmm, okay. So, first, I need to understand how many tetrahedrons are in the series. The problem says "a series of nested tetrahedra," but it doesn't specify how many. Wait, maybe it's an infinite series? But that doesn't make much sense because the smallest would be approaching a point, not a specific tetrahedron. Maybe it's a finite series? But the problem doesn't specify. Hmm.Wait, the problem says "each tetrahedron is scaled down by a factor of 1/3 from the previous one." So, if we start with the largest, each subsequent one is 1/3 the size. But how many times do we scale down? It says "the smallest tetrahedron in this series." Maybe it's a finite series, but without knowing how many, I can't compute the exact scaling factor. Hmm, maybe I need to assume that it's scaled down infinitely, but then the smallest would be a point. That doesn't make sense either.Wait, perhaps the scaling is such that each tetrahedron is scaled by 1/3 from the previous, so each edge is 1/3 the length of the previous. So, if the largest tetrahedron has edge length, let's see, the distance between (3,0,0) and (0,3,0) is sqrt[(3)^2 + (-3)^2 + 0] = sqrt(9 + 9) = sqrt(18) = 3*sqrt(2). So the edge length is 3*sqrt(2). Then the next one would be sqrt(2), then sqrt(2)/3, and so on.But again, without knowing how many times it's scaled, I can't find the exact coordinates. Wait, maybe the series is such that each tetrahedron is scaled down by 1/3, so the scaling factor is 1/3 each time. So, if we have n tetrahedrons, the smallest would be scaled by (1/3)^{n-1}.But since the problem doesn't specify n, maybe it's just one scaling? That is, the next one is 1/3 the size, so the smallest is the second one? But the problem says "a series," which implies more than one. Hmm, I'm confused.Wait, maybe the scaling is such that each tetrahedron is scaled by 1/3 from the previous, so the series is infinite, but the smallest is the limit as n approaches infinity, which is a point. But that can't be, because the problem asks for the coordinates of the vertices, which would be a single point, but a tetrahedron has four vertices.Wait, maybe the scaling is done in such a way that each tetrahedron is scaled down by 1/3 in each dimension, so the coordinates are scaled accordingly. So, starting from the largest tetrahedron, each subsequent one is scaled by 1/3 in x, y, z.So, if the largest tetrahedron has vertices at (3,0,0), (0,3,0), (0,0,3), and (1,1,1), then the next one would be scaled by 1/3. So, each coordinate is multiplied by 1/3.But wait, scaling a tetrahedron by 1/3 from the origin would scale all coordinates by 1/3, but the centroid might not be the origin. Wait, the centroid of the largest tetrahedron is the average of its vertices. Let's compute that.Centroid C = [(3 + 0 + 0 + 1)/4, (0 + 3 + 0 + 1)/4, (0 + 0 + 3 + 1)/4] = (4/4, 4/4, 4/4) = (1,1,1). Oh, interesting, the centroid is at (1,1,1). So, if we scale each tetrahedron about its centroid, then the scaling would be relative to (1,1,1).So, to scale down by 1/3, we can use the formula for scaling about a point. The formula is: for a point P, scaling about point C by factor k is given by P' = C + k*(P - C).So, for each vertex P, we compute P' = (1,1,1) + (1/3)*(P - (1,1,1)).Let me test this with one vertex. Take (3,0,0):P - C = (3-1, 0-1, 0-1) = (2, -1, -1)Multiply by 1/3: (2/3, -1/3, -1/3)Add back C: (1 + 2/3, 1 - 1/3, 1 - 1/3) = (5/3, 2/3, 2/3)Similarly, for (0,3,0):P - C = (-1, 2, -1)Multiply by 1/3: (-1/3, 2/3, -1/3)Add back C: (1 - 1/3, 1 + 2/3, 1 - 1/3) = (2/3, 5/3, 2/3)For (0,0,3):P - C = (-1, -1, 2)Multiply by 1/3: (-1/3, -1/3, 2/3)Add back C: (1 - 1/3, 1 - 1/3, 1 + 2/3) = (2/3, 2/3, 5/3)And for (1,1,1):P - C = (0,0,0)Multiply by 1/3: (0,0,0)Add back C: (1,1,1)So, the scaled-down tetrahedron has vertices at (5/3, 2/3, 2/3), (2/3, 5/3, 2/3), (2/3, 2/3, 5/3), and (1,1,1).Wait, but the problem says "each tetrahedron is scaled down by a factor of 1/3 from the previous one." So, does this mean that each subsequent tetrahedron is scaled by 1/3 from the previous one, which is itself scaled by 1/3 from the one before, and so on?If so, then each scaling is relative to the previous centroid, which is (1,1,1) for all, since scaling about the same centroid each time.So, each time we scale by 1/3, the vertices get closer to (1,1,1). So, the first scaling gives us the second tetrahedron, the second scaling gives us the third, and so on.But the problem says "the smallest tetrahedron in this series." So, is it the limit as the number of scalings approaches infinity? That would be a single point at (1,1,1). But that's not a tetrahedron. So, maybe the series is finite, but the problem doesn't specify how many. Hmm.Wait, maybe the series is such that each tetrahedron is scaled by 1/3, so the edge length is multiplied by 1/3 each time. So, starting from the largest, which has edge length 3*sqrt(2), the next would have sqrt(2), then sqrt(2)/3, etc. But without knowing how many times it's scaled, I can't find the exact coordinates.Wait, perhaps the series is such that it's scaled down infinitely, but the problem asks for the coordinates of the smallest, which would be the limit. But as I thought earlier, that would be a single point, not a tetrahedron. So, maybe the series is finite, and the smallest is the second one, scaled once. So, the coordinates would be as I calculated: (5/3, 2/3, 2/3), (2/3, 5/3, 2/3), (2/3, 2/3, 5/3), and (1,1,1).But the problem says "the smallest tetrahedron in this series." So, maybe it's the second one, scaled once. So, I think that's the answer.So, the vertices of the smallest tetrahedron are (5/3, 2/3, 2/3), (2/3, 5/3, 2/3), (2/3, 2/3, 5/3), and (1,1,1).Wait, but let me double-check. If we scale down by 1/3 each time, starting from the largest, then the first scaling gives us the second tetrahedron, which is smaller. If we scale again, we get the third, even smaller, and so on. So, the "smallest" would be the limit as n approaches infinity, but that's a point. So, maybe the problem assumes that the series is such that it's scaled down once, so the smallest is the second one.Alternatively, maybe the scaling is such that each tetrahedron is scaled by 1/3 relative to the previous one, but the number of tetrahedrons is such that the smallest is the one after scaling once. So, I think the answer is as I calculated.Okay, moving on to the second problem: The critic wants to analyze the visual impact of colors in the installations. Each face of the tetrahedra can be painted with one of three colors: red, blue, or yellow. No two adjacent faces can be the same color. Calculate the number of distinct ways to paint the largest tetrahedron following this rule.Alright, so we have a tetrahedron, which has four triangular faces. Each face can be colored with red, blue, or yellow, and no two adjacent faces can have the same color. We need to find the number of distinct colorings.This is a classic graph coloring problem, where the tetrahedron is a complete graph K4, meaning each face is adjacent to every other face. So, in graph theory terms, we're looking for the number of proper colorings of K4 with three colors.The formula for the number of proper colorings of a complete graph Kn with k colors is k*(k-1)*(k-2)*...*(k - n + 1). So, for K4 with 3 colors, it would be 3*2*1*0, but that can't be right because 0 would make it zero, which doesn't make sense.Wait, no, that formula is for permutations where each vertex is colored, but in our case, it's faces. Wait, actually, in graph coloring, each vertex is assigned a color, but here, each face is a node in the dual graph, which for a tetrahedron is another tetrahedron. So, the dual graph of a tetrahedron is also K4, meaning each face is adjacent to every other face.So, the number of colorings is the number of proper colorings of K4 with 3 colors. The formula for the number of proper colorings is given by the chromatic polynomial. For K4, the chromatic polynomial is k*(k-1)*(k-2)*(k-3). So, for k=3, it's 3*2*1*0=0. That can't be right because we know that with three colors, it's possible to color a tetrahedron.Wait, maybe I'm confusing something. Let me think again.In graph coloring, the chromatic polynomial counts the number of colorings where adjacent vertices have different colors. For K4, which is a complete graph with four vertices, the chromatic polynomial is indeed k*(k-1)*(k-2)*(k-3). So, for k=3, it's 3*2*1*0=0, which suggests that it's impossible to color K4 with three colors such that no two adjacent vertices have the same color. But that's not true because K4 is 4-chromatic, meaning it requires four colors. So, with three colors, it's impossible.But in our case, we're talking about coloring the faces of a tetrahedron, not the vertices. So, the dual graph of the tetrahedron is also K4, so the faces correspond to the vertices in the dual graph. So, the problem reduces to coloring the vertices of K4 with three colors, which is impossible because K4 requires four colors. Therefore, the number of colorings is zero.But that can't be right because the problem states that each face can be painted with one of three colors, and no two adjacent faces can be the same color. So, perhaps I'm misunderstanding the adjacency.Wait, in a tetrahedron, each face is adjacent to the other three faces. So, each face shares an edge with every other face. Therefore, in terms of coloring, each face is adjacent to every other face, meaning that all four faces must have different colors. But we only have three colors, so it's impossible. Therefore, the number of distinct ways is zero.But that seems counterintuitive because the problem is asking to calculate it, implying that there is a positive number. Maybe I'm making a mistake.Wait, perhaps the tetrahedron's faces are not all adjacent in the way I think. Let me visualize a tetrahedron: it has four triangular faces, each face is a triangle, and each face is adjacent to the other three faces. So, yes, each face shares an edge with every other face. Therefore, in terms of coloring, each face must have a different color from all the others. Since we have four faces and only three colors, it's impossible to color them without having at least two adjacent faces sharing the same color. Therefore, the number of distinct colorings is zero.But that seems odd because the problem is presented as a problem to solve, so maybe I'm missing something. Alternatively, perhaps the tetrahedron's dual graph isn't K4? Wait, no, the dual graph of a tetrahedron is indeed another tetrahedron, which is K4.Wait, maybe the problem is considering the tetrahedron as a planar graph, but K4 is planar? No, K4 is planar, but in planar graphs, the four-color theorem applies, but here we're using three colors. So, if K4 is planar, it can be colored with four colors, but with three colors, it's not always possible. In fact, K4 requires four colors because it's a complete graph with four vertices.Therefore, the number of proper colorings with three colors is zero.But let me think again. Maybe the problem is considering the tetrahedron as a 3D object, and the adjacency is different? Wait, no, in 3D, the adjacency of faces is the same as in the dual graph. Each face is adjacent to every other face, so it's still K4.Alternatively, maybe the problem is considering that each face is only adjacent to three others, but in reality, in a tetrahedron, each face is adjacent to three others, but all four faces are mutually adjacent. So, it's a complete graph.Wait, no, each face is adjacent to three others, but not all four. Wait, no, in a tetrahedron, each face is adjacent to the other three faces. So, each face is adjacent to three others, but all four faces are connected in a way that they form a complete graph.Wait, no, in a tetrahedron, each face is a triangle, and each face shares an edge with three other faces. So, each face is adjacent to three others, but not all four. Wait, no, in a tetrahedron, each face is adjacent to the other three faces. So, for example, face 1 is adjacent to faces 2, 3, and 4. Similarly, face 2 is adjacent to 1, 3, and 4, and so on. So, the adjacency graph is K4.Therefore, the number of colorings is zero because we can't color K4 with three colors without adjacent faces having the same color.But the problem says "each face of the tetrahedra can be painted with one of three colors: red, blue, or yellow. No two adjacent faces can be the same color." So, the answer is zero? That seems strange, but mathematically, it's correct.Alternatively, maybe I'm misunderstanding the adjacency. Maybe in the tetrahedron, each face is only adjacent to three others, but not all four. Wait, no, in a tetrahedron, each face is adjacent to the other three. So, it's a complete graph.Wait, let me think differently. Maybe the problem is considering that the tetrahedron is embedded in 3D space, and the adjacency is not as strict as in the graph. But no, in terms of coloring, adjacency is defined as sharing an edge, which in the tetrahedron, each face shares an edge with every other face.Therefore, the conclusion is that it's impossible to color the tetrahedron with three colors such that no two adjacent faces have the same color. Therefore, the number of distinct ways is zero.But let me double-check. Maybe I can try to color it manually. Let's say we have four faces: A, B, C, D.We have three colors: R, B, Y.We need to assign a color to each face such that no two adjacent faces have the same color.Start with face A: color it R.Then face B, adjacent to A, can be B or Y.Let's choose B.Face C, adjacent to A and B, can't be R or B, so it must be Y.Face D, adjacent to A, B, and C. It can't be R (because of A), can't be B (because of B), can't be Y (because of C). So, no color available. Therefore, it's impossible.Therefore, the number of colorings is zero.So, the answer to the second problem is zero.But wait, the problem says "the largest tetrahedron." Maybe the largest tetrahedron is different? Wait, no, the largest tetrahedron is just the size, but the structure is the same. It's still a tetrahedron with four faces, each adjacent to the other three.Therefore, regardless of the size, the number of colorings is zero.So, to summarize:1. The smallest tetrahedron's vertices are (5/3, 2/3, 2/3), (2/3, 5/3, 2/3), (2/3, 2/3, 5/3), and (1,1,1).2. The number of distinct colorings is zero.But wait, for the first problem, I assumed that the series is scaled once, but maybe it's scaled multiple times. If it's scaled infinitely, the smallest would be a point, but that's not a tetrahedron. So, perhaps the problem assumes that the series is scaled once, so the smallest is the second one.Alternatively, maybe the series is such that each tetrahedron is scaled by 1/3 from the previous one, and the smallest is the one after scaling once. So, I think my initial answer is correct.So, final answers:1. The smallest tetrahedron has vertices at (5/3, 2/3, 2/3), (2/3, 5/3, 2/3), (2/3, 2/3, 5/3), and (1,1,1).2. The number of distinct colorings is zero.</think>

🧠:<think>Alright, so I have this problem about a college basketball team's performance over five years. The person wants to analyze their wins, considering home and away games. Let me try to break this down step by step.First, the team played 150 games over five years. Each year, they played an equal number of home and away games. So, let's figure out how many games they played each year. If it's 150 games over five years, that's 150 divided by 5, which is 30 games per year. Since they played an equal number of home and away games each year, that means 15 home games and 15 away games per year.Okay, so each year they have 15 home games with a 0.75 probability of winning each, and 15 away games with a 0.40 probability of winning each. The first part asks for the expected number of total wins over the five-year period. Then, we need to find the variance and standard deviation of the number of wins.Let me recall that expectation is linear, so I can calculate the expected number of wins for each type of game and then sum them up. For home games, each season, the expected number of wins is 15 games multiplied by 0.75. Let me compute that: 15 * 0.75 = 11.25 wins per season. Similarly, for away games, it's 15 * 0.40 = 6 wins per season.So, each season, the expected total wins are 11.25 + 6 = 17.25 wins. Over five seasons, that would be 17.25 * 5. Let me calculate that: 17.25 * 5 = 86.25. So, the expected number of total wins over five years is 86.25.Now, moving on to variance and standard deviation. Since each game is a Bernoulli trial (either win or loss), the variance for each type of game can be calculated separately and then summed up because variance is additive for independent random variables.For home games: Each home game has a variance of p*(1-p) = 0.75*(1 - 0.75) = 0.75*0.25 = 0.1875. Since there are 15 home games per season, the variance per season for home games is 15 * 0.1875. Let me compute that: 15 * 0.1875 = 2.8125.Similarly, for away games: Each away game has a variance of 0.40*(1 - 0.40) = 0.40*0.60 = 0.24. With 15 away games per season, the variance per season for away games is 15 * 0.24 = 3.6.So, the total variance per season is 2.8125 + 3.6 = 6.4125. Over five seasons, since each season is independent, the total variance is 5 * 6.4125. Let me calculate that: 5 * 6.4125 = 32.0625.Therefore, the variance of the number of wins over five years is 32.0625. To find the standard deviation, we take the square root of the variance. So, sqrt(32.0625). Let me compute that: sqrt(32.0625) is 5.6625. Wait, let me verify that because 5.6625 squared is approximately 32.0625. Yes, that seems right.So, summarizing the first part: Expected total wins = 86.25, variance = 32.0625, standard deviation ≈ 5.6625.Moving on to the second part. The player wants to predict the number of wins in the next season using a weighted moving average. The weights are 0.5, 0.3, and 0.2 for the most recent, second most recent, and third most recent seasons, respectively. The number of wins in the last three seasons were 18, 20, and 22.So, to construct the weighted moving average, we multiply each season's wins by their respective weights and sum them up. Let me assign the weights correctly. The most recent season is 18 wins, so that's multiplied by 0.5. The second most recent is 20, multiplied by 0.3. The third most recent is 22, multiplied by 0.2.Calculating each term:- 18 * 0.5 = 9- 20 * 0.3 = 6- 22 * 0.2 = 4.4Adding these up: 9 + 6 + 4.4 = 19.4. So, the weighted moving average is 19.4. Therefore, the prediction for the next season's number of wins is 19.4.Wait, let me double-check the multiplication and addition:18 * 0.5 is indeed 9.20 * 0.3 is 6.22 * 0.2 is 4.4.Adding 9 + 6 is 15, plus 4.4 is 19.4. Yep, that's correct.So, the prediction is 19.4 wins for the next season.Just to recap, for the first part, we calculated the expected number of wins by considering home and away games separately, then found the variance and standard deviation by summing the variances of each category. For the second part, we applied the weighted moving average formula with the given weights and past wins to predict the next season's wins.I think that covers both parts of the problem. Let me just make sure I didn't make any calculation errors.For the expected wins: 15 home games * 0.75 = 11.25, 15 away * 0.4 = 6, total per season 17.25, over five seasons 86.25. That seems right.Variance: home variance per season is 15 * 0.75 * 0.25 = 2.8125, away variance is 15 * 0.4 * 0.6 = 3.6, total per season 6.4125, over five seasons 32.0625. Standard deviation sqrt(32.0625) is 5.6625. Correct.Weighted moving average: 18*0.5 + 20*0.3 + 22*0.2 = 9 + 6 + 4.4 = 19.4. Yep, that's accurate.So, I think I'm confident with these answers.Final Answer1. The expected number of total wins is boxed{86.25}, the variance is boxed{32.0625}, and the standard deviation is boxed{5.6625}.2. The predicted number of wins in the next season is boxed{19.4}.</think>

🧠:<think>Okay, so I have this problem where I need to help a VMware administrator optimize their data center. They want to show that VMware's vSphere is better than Microsoft's Hyper-V in terms of resource utilization and cost efficiency. There are two parts to this problem. Let me tackle them one by one.Starting with the first part: The data center has N physical servers. Each server has C CPU cores and M GB of memory. The administrator wants to deploy V virtual machines (VMs). Each VM needs c CPU cores and m GB of memory. But there's also an overhead for vSphere managing each VM, which is o CPU cores and p GB of memory. I need to derive an inequality that ensures all VMs can be deployed without exceeding the total physical resources available.Hmm, okay. So, each VM requires c CPU cores and m memory, but also adds an overhead of o CPU cores and p memory. So, for each VM, the total resources needed are (c + o) CPU cores and (m + p) GB of memory.Since there are V VMs, the total required CPU cores would be V*(c + o) and the total required memory would be V*(m + p).But these have to fit within the total resources available from the N physical servers. Each server has C CPU cores and M memory, so the total available CPU cores are N*C and the total available memory is N*M.Therefore, the inequalities should be:V*(c + o) ≤ N*CandV*(m + p) ≤ N*MSo, both of these inequalities need to be satisfied to ensure that all VMs can be deployed without exceeding the physical resources.Wait, but the question says to derive an inequality, not inequalities. Maybe they expect a single inequality? Or perhaps they just want the conditions for both resources, so maybe writing both inequalities is acceptable. I think it's better to present both since both CPU and memory are critical resources.Moving on to the second part: The administrator wants to show that the TCO for vSphere is at least α% lower than Hyper-V's TCO. The TCO formulas are given.For vSphere: TCO_VMware = A*N + B*VFor Hyper-V: TCO_HyperV = (A - k)*N + (B + j)*VWe need to derive an inequality that ensures TCO_VMware is at least α% lower than TCO_HyperV.So, mathematically, this means:TCO_VMware ≤ TCO_HyperV - (α/100)*TCO_HyperVWhich simplifies to:TCO_VMware ≤ TCO_HyperV*(1 - α/100)Plugging in the TCO formulas:A*N + B*V ≤ [(A - k)*N + (B + j)*V]*(1 - α/100)Alternatively, we can write this as:(A*N + B*V) / [(A - k)*N + (B + j)*V] ≤ 1 - α/100But the question asks for an inequality involving A, B, k, j, α, N, and V. So, perhaps expanding the right side would be better.Let me compute the right-hand side:[(A - k)*N + (B + j)*V]*(1 - α/100) = (A*N - k*N + B*V + j*V)*(1 - α/100)Expanding this:= (A*N - k*N + B*V + j*V) - (α/100)*(A*N - k*N + B*V + j*V)So, the inequality becomes:A*N + B*V ≤ (A*N - k*N + B*V + j*V) - (α/100)*(A*N - k*N + B*V + j*V)Let me rearrange terms:A*N + B*V - (A*N - k*N + B*V + j*V) ≤ - (α/100)*(A*N - k*N + B*V + j*V)Simplify the left side:A*N - A*N + B*V - B*V + k*N - j*V ≤ - (α/100)*(A*N - k*N + B*V + j*V)Which simplifies to:k*N - j*V ≤ - (α/100)*(A*N - k*N + B*V + j*V)Multiply both sides by -1 (which reverses the inequality):- k*N + j*V ≥ (α/100)*(A*N - k*N + B*V + j*V)So,j*V - k*N ≥ (α/100)*(A*N - k*N + B*V + j*V)Alternatively, we can write this as:(j*V - k*N) ≥ (α/100)*(A*N - k*N + B*V + j*V)But maybe it's better to express it in terms of the original TCO expressions.Alternatively, another approach is to express the percentage difference. The requirement is that TCO_VMware is at least α% lower than TCO_HyperV. So,TCO_VMware ≤ TCO_HyperV * (1 - α/100)Which is the same as:TCO_VMware - TCO_HyperV ≤ - (α/100)*TCO_HyperVSo,(A*N + B*V) - [(A - k)*N + (B + j)*V] ≤ - (α/100)*[(A - k)*N + (B + j)*V]Simplify the left side:A*N + B*V - A*N + k*N - B*V - j*V = k*N - j*VSo,k*N - j*V ≤ - (α/100)*[(A - k)*N + (B + j)*V]Multiply both sides by -1 (reverse inequality):- k*N + j*V ≥ (α/100)*[(A - k)*N + (B + j)*V]Which is the same as:j*V - k*N ≥ (α/100)*(A*N - k*N + B*V + j*V)I think this is the inequality they are looking for.So, summarizing:1. For resource utilization, the inequalities are V*(c + o) ≤ N*C and V*(m + p) ≤ N*M.2. For cost efficiency, the inequality is j*V - k*N ≥ (α/100)*(A*N - k*N + B*V + j*V).I should probably write these neatly.</think>

🧠:<think>Alright, so I need to figure out the total usable area for construction for both Parcel A and Parcel B, considering the restrictions given. Let me take this step by step.Starting with Parcel A. It's a triangular parcel with coordinates (0, 0), (a, 0), and (0, b). I remember that the area of a triangle is given by (base * height) / 2. Here, the base is 'a' and the height is 'b'. So, the area should be (a * b) / 2.Given that a = 500 meters and b = 300 meters, plugging those in: (500 * 300) / 2. Let me calculate that. 500 times 300 is 150,000, and divided by 2 is 75,000 square meters. So, the area of Parcel A is 75,000 m².But wait, the developer needs to reserve 25% of the land for green space and infrastructure. That means only 75% of the area is usable. So, I need to find 75% of 75,000 m². Calculating that: 75,000 * 0.75. Hmm, 75,000 * 0.75 is the same as 75,000 * (3/4), which is 56,250 m². So, the usable area for Parcel A is 56,250 m².Okay, moving on to Parcel B. It's a rectangular parcel with coordinates (0, 0), (0, 600), (400, 0), and (400, 600). So, the length along the x-axis is 400 meters, and the length along the y-axis is 600 meters. The area of a rectangle is length * width, so 400 * 600.Calculating that: 400 * 600 is 240,000 m². So, the total area of Parcel B is 240,000 m².However, there's a circular lake in the middle that can't be developed. The lake has a radius of 100 meters, centered at (200, 300). To find the usable area, I need to subtract the area of the lake from the total area of Parcel B.The area of a circle is π * r². Here, r = 100 meters, so the area is π * (100)². That's π * 10,000, which is approximately 3.1416 * 10,000 = 31,416 m².So, subtracting the lake's area from Parcel B's total area: 240,000 - 31,416. Let me compute that. 240,000 minus 30,000 is 210,000, and then subtract another 1,416, which gives 208,584 m². So, the usable area for Parcel B is approximately 208,584 m².Wait, hold on. Let me double-check that subtraction. 240,000 minus 31,416. Breaking it down: 240,000 - 30,000 = 210,000, and then 210,000 - 1,416 = 208,584. Yeah, that seems right.Now, to find the total usable area combining both parcels, I need to add the usable areas of Parcel A and Parcel B.Parcel A's usable area is 56,250 m², and Parcel B's is 208,584 m². Adding them together: 56,250 + 208,584. Let me add these numbers.56,250 + 200,000 = 256,250. Then, adding the remaining 8,584: 256,250 + 8,584 = 264,834 m².So, the total usable area is 264,834 square meters.Wait, just to make sure I didn't make any calculation errors. Let me verify each step again.For Parcel A:- Area = (500 * 300)/2 = 75,000 m².- Usable area = 75,000 * 0.75 = 56,250 m². That seems correct.For Parcel B:- Total area = 400 * 600 = 240,000 m².- Lake area = π * 100² ≈ 3.1416 * 10,000 ≈ 31,416 m².- Usable area = 240,000 - 31,416 ≈ 208,584 m². That also seems correct.Total usable area: 56,250 + 208,584 = 264,834 m². Yep, that adds up.I think that's all. So, the total usable area is 264,834 square meters.Final AnswerThe total usable area for construction is boxed{264834} square meters.</think>