💡Okay, so I need to find the equation of a plane in three-dimensional space. The plane passes through a specific point, A(1, 2, 3), and has a given normal vector, n = (-1, -2, 1). Hmm, I remember that in two dimensions, the equation of a line can be found using a point and a normal vector, and it involves the dot product. Maybe something similar applies here in three dimensions for planes.Let me recall. In two dimensions, if I have a point A(x₀, y₀) and a normal vector n = (a, b), the equation of the line is a(x - x₀) + b(y - y₀) = 0. So, it's like taking the dot product of the vector from A to any point (x, y) on the line and the normal vector, setting it equal to zero because they are perpendicular.Extending this idea to three dimensions, a plane can be defined similarly. If I have a point A(x₀, y₀, z₀) and a normal vector n = (a, b, c), then any point P(x, y, z) on the plane should satisfy the condition that the vector AP is perpendicular to the normal vector n. So, the dot product of AP and n should be zero.Let me write this down. The vector AP is (x - x₀, y - y₀, z - z₀). The normal vector is (a, b, c). Their dot product is a(x - x₀) + b(y - y₀) + c(z - z₀) = 0. That should give the equation of the plane.Given the point A(1, 2, 3) and normal vector n = (-1, -2, 1), I can plug these into the formula. So, substituting x₀ = 1, y₀ = 2, z₀ = 3, a = -1, b = -2, c = 1.So, the equation becomes:-1*(x - 1) + (-2)*(y - 2) + 1*(z - 3) = 0Let me compute each term step by step.First term: -1*(x - 1) = -x + 1Second term: -2*(y - 2) = -2y + 4Third term: 1*(z - 3) = z - 3Now, adding all these together:(-x + 1) + (-2y + 4) + (z - 3) = 0Combine like terms:- x - 2y + z + (1 + 4 - 3) = 0Calculating the constants: 1 + 4 is 5, minus 3 is 2.So, the equation simplifies to:- x - 2y + z + 2 = 0Hmm, but usually, plane equations are written with the x term positive. Maybe I should multiply both sides by -1 to make it look nicer.Multiplying each term by -1:x + 2y - z - 2 = 0So, the equation of the plane is x + 2y - z - 2 = 0.Let me double-check my steps to make sure I didn't make a mistake. Starting from the vector AP being perpendicular to n, so their dot product is zero. Substituted the coordinates correctly, expanded each term, combined like terms, and simplified. Then, multiplied by -1 to make the x coefficient positive. It seems correct.Alternatively, I can write the equation as x + 2y - z = 2. That might be another way to present it, but both forms are equivalent.I think that's the correct equation for the plane passing through A(1, 2, 3) with normal vector (-1, -2, 1). It makes sense because the normal vector is perpendicular to the plane, so any vector lying on the plane should be orthogonal to n, which is exactly what the equation ensures through the dot product.Just to visualize, the normal vector points in the direction (-1, -2, 1), so the plane is oriented such that it's "tilted" in that direction. The point A(1, 2, 3) lies on the plane, so plugging these coordinates into the equation should satisfy it.Let me verify that. Plugging x=1, y=2, z=3 into x + 2y - z - 2:1 + 2*2 - 3 - 2 = 1 + 4 - 3 - 2 = (1 + 4) - (3 + 2) = 5 - 5 = 0. Perfect, it satisfies the equation.So, I'm confident that the equation x + 2y - z - 2 = 0 is correct for the given plane.**Final Answer**The equation of the plane is boxed{x + 2y - z - 2 = 0}.

💡Okay, so I have this problem here: wz equals 12 minus 8i, and the magnitude of w is the square root of 13. I need to find the magnitude of z. Hmm, let me think about how to approach this.First, I remember that when you multiply two complex numbers, their magnitudes multiply. So, the magnitude of wz should be equal to the magnitude of w times the magnitude of z. That makes sense because if you think of complex numbers as vectors, multiplying them involves scaling and rotating, but the overall length (magnitude) should just be the product of their lengths.So, mathematically, that would be |wz| = |w| * |z|. I know wz is given as 12 - 8i, so I can find its magnitude. Let me calculate that. The magnitude of a complex number a + bi is sqrt(a² + b²). So, for 12 - 8i, that would be sqrt(12² + (-8)²).Calculating that, 12 squared is 144, and (-8) squared is 64. Adding those together gives 144 + 64 = 208. So, the magnitude of wz is sqrt(208). Hmm, sqrt(208) can be simplified. Let me see, 208 divided by 16 is 13, so sqrt(208) is sqrt(16 * 13) which is 4 * sqrt(13). Okay, so |wz| is 4 times the square root of 13.Now, I know that |w| is given as sqrt(13). So, plugging into the equation |wz| = |w| * |z|, I have 4 * sqrt(13) = sqrt(13) * |z|. To solve for |z|, I can divide both sides by sqrt(13). Dividing both sides by sqrt(13), I get |z| = (4 * sqrt(13)) / sqrt(13). The sqrt(13) in the numerator and denominator cancel out, leaving me with |z| = 4. Wait, that seems straightforward. Let me double-check to make sure I didn't make any mistakes. So, wz is 12 - 8i, whose magnitude is sqrt(12² + (-8)²) = sqrt(144 + 64) = sqrt(208) = 4 * sqrt(13). Then, since |w| is sqrt(13), dividing 4 * sqrt(13) by sqrt(13) gives 4. Yep, that checks out.I think I've got it. The magnitude of z is 4.

💡Okay, so I have this problem here with vectors a, b, and c. Vector a is (2, -1, 3), vector b is (-1, 4, -2), and vector c is (3, 2, λ). The question says that vector c equals 2 times vector a plus vector b, and I need to find the real number λ. The options are A: 2, B: 3, C: 4, D: 5.Alright, let me start by understanding what this equation means. Vector c is expressed as a linear combination of vectors a and b. Specifically, it's 2 times vector a plus vector b. So, to find vector c, I need to perform the scalar multiplication and vector addition as given.First, I'll write down the vectors again to make sure I have them correct:- Vector a: (2, -1, 3)- Vector b: (-1, 4, -2)- Vector c: (3, 2, λ)The equation is c = 2a + b. So, I need to compute 2a first. Scalar multiplication of a vector involves multiplying each component of the vector by the scalar. So, 2 times vector a would be:2 * (2, -1, 3) = (2*2, 2*(-1), 2*3) = (4, -2, 6)Okay, so 2a is (4, -2, 6). Now, I need to add vector b to this result. Vector addition is done component-wise, so I'll add each corresponding component of 2a and b.Vector b is (-1, 4, -2). So, adding this to 2a:(4, -2, 6) + (-1, 4, -2) = (4 + (-1), -2 + 4, 6 + (-2)) = (3, 2, 4)Hmm, so when I compute 2a + b, I get the vector (3, 2, 4). But vector c is given as (3, 2, λ). That means that the components of c must match the components of 2a + b. So, comparing the two vectors:c = (3, 2, λ) and 2a + b = (3, 2, 4)So, the first two components already match: 3 and 2. The third component is λ in vector c and 4 in the computed vector. Therefore, λ must be equal to 4.Let me double-check my calculations to make sure I didn't make a mistake. Sometimes when dealing with negative numbers, it's easy to slip up.First, 2a:2 * 2 = 42 * (-1) = -22 * 3 = 6So, 2a is indeed (4, -2, 6).Then, adding vector b:4 + (-1) = 3-2 + 4 = 26 + (-2) = 4Yes, that's correct. So, 2a + b is (3, 2, 4). Therefore, vector c must be (3, 2, 4), which means λ is 4.Looking back at the options, C is 4, so that should be the correct answer.I think that's solid. I don't see any mistakes in my reasoning. It's straightforward vector operations: scalar multiplication followed by vector addition. Each step checks out, and the components match up except for λ, which we've determined is 4.Just to recap:1. Compute 2a: (4, -2, 6)2. Add vector b: (4 -1, -2 +4, 6 -2) = (3, 2, 4)3. Compare with c: (3, 2, λ) => λ = 4Yep, that's it. I feel confident that λ is 4, which is option C.**Final Answer**boxed{C}

💡First, Melanie picked a total of 7 plums from the orchard.She then gave 3 plums to Sam.To find out how many plums Melanie has now, subtract the number of plums she gave away from the total number she picked.So, 7 plums minus 3 plums equals 4 plums.Therefore, Melanie has 4 plums remaining.

💡Okay, so I have this problem where I need to find two positive integers, both less than 1000. Their arithmetic mean and geometric mean are consecutive odd integers. I need to find the maximum possible value of the difference between these two integers. Hmm, let me break this down step by step.First, let me recall what arithmetic mean (AM) and geometric mean (GM) are. The arithmetic mean of two numbers ( a ) and ( b ) is ( frac{a + b}{2} ), and the geometric mean is ( sqrt{ab} ). The problem says these two means are consecutive odd integers. So, if one is an odd integer, the other is the next odd integer. Let me denote the arithmetic mean as ( 2k + 1 ) and the geometric mean as ( 2k - 1 ) for some integer ( k ). This is because consecutive odd integers differ by 2, so if one is ( 2k + 1 ), the next one is ( 2k - 1 ) or ( 2k + 3 ). Wait, actually, depending on which one is larger. Since the arithmetic mean is always greater than or equal to the geometric mean for non-negative numbers, the arithmetic mean should be the larger one. So, if the arithmetic mean is ( 2k + 1 ), then the geometric mean should be ( 2k - 1 ). That makes sense.So, writing down the equations:1. ( frac{a + b}{2} = 2k + 1 )2. ( sqrt{ab} = 2k - 1 )From the first equation, multiplying both sides by 2 gives:( a + b = 4k + 2 )From the second equation, squaring both sides gives:( ab = (2k - 1)^2 = 4k^2 - 4k + 1 )So now I have two equations:1. ( a + b = 4k + 2 )2. ( ab = 4k^2 - 4k + 1 )I need to find integers ( a ) and ( b ) that satisfy these equations. Since ( a ) and ( b ) are roots of the quadratic equation ( x^2 - (a + b)x + ab = 0 ), substituting the known values:( x^2 - (4k + 2)x + (4k^2 - 4k + 1) = 0 )Let me try to solve this quadratic equation for ( x ). Using the quadratic formula:( x = frac{(4k + 2) pm sqrt{(4k + 2)^2 - 4(4k^2 - 4k + 1)}}{2} )Calculating the discriminant:( D = (4k + 2)^2 - 4(4k^2 - 4k + 1) )Expanding ( (4k + 2)^2 ):( 16k^2 + 16k + 4 )Subtracting ( 4(4k^2 - 4k + 1) ):( 16k^2 + 16k + 4 - 16k^2 + 16k - 4 )Simplifying:( 32k )So, the discriminant is ( 32k ). Therefore, the roots are:( x = frac{4k + 2 pm sqrt{32k}}{2} )Simplify ( sqrt{32k} ):( sqrt{32k} = sqrt{16 times 2k} = 4sqrt{2k} )So, the roots become:( x = frac{4k + 2 pm 4sqrt{2k}}{2} = 2k + 1 pm 2sqrt{2k} )Thus, the two numbers are:( a = 2k + 1 + 2sqrt{2k} )( b = 2k + 1 - 2sqrt{2k} )Since ( a ) and ( b ) must be integers, ( 2sqrt{2k} ) must also be an integer. Let me denote ( m = sqrt{2k} ), so ( m ) must be an integer because ( 2k ) is a perfect square. Therefore, ( 2k = m^2 ), which implies ( k = frac{m^2}{2} ). But ( k ) must be an integer, so ( m^2 ) must be even, meaning ( m ) must be even. Let me set ( m = 2n ), so ( m^2 = 4n^2 ), and ( k = frac{4n^2}{2} = 2n^2 ).Substituting back into ( a ) and ( b ):( a = 2(2n^2) + 1 + 2(2n) = 4n^2 + 1 + 4n = 4n^2 + 4n + 1 )( b = 2(2n^2) + 1 - 2(2n) = 4n^2 + 1 - 4n = 4n^2 - 4n + 1 )So, ( a = 4n^2 + 4n + 1 ) and ( b = 4n^2 - 4n + 1 ). Let me compute the difference ( a - b ):( a - b = (4n^2 + 4n + 1) - (4n^2 - 4n + 1) = 8n )So, the difference is ( 8n ). To maximize this difference, I need to maximize ( n ), but keeping both ( a ) and ( b ) less than 1000.So, let's write the conditions:1. ( a = 4n^2 + 4n + 1 < 1000 )2. ( b = 4n^2 - 4n + 1 < 1000 )Since ( a ) is larger than ( b ), it's sufficient to ensure ( a < 1000 ).So, ( 4n^2 + 4n + 1 < 1000 )Let me solve this inequality for ( n ):( 4n^2 + 4n + 1 < 1000 )Subtract 1000:( 4n^2 + 4n - 999 < 0 )This is a quadratic inequality. Let's find the roots of the equation ( 4n^2 + 4n - 999 = 0 ).Using the quadratic formula:( n = frac{-4 pm sqrt{16 + 4 times 4 times 999}}{8} )Calculating discriminant:( D = 16 + 16 times 999 = 16(1 + 999) = 16 times 1000 = 16000 )So,( n = frac{-4 pm sqrt{16000}}{8} )( sqrt{16000} = sqrt{16 times 1000} = 4 times sqrt{1000} approx 4 times 31.6227766 = 126.4911064 )So,( n = frac{-4 + 126.4911064}{8} approx frac{122.4911064}{8} approx 15.3113883 )And the other root is negative, which we can ignore since ( n ) must be positive.So, ( n ) must be less than approximately 15.31. Since ( n ) is an integer, the maximum possible value is ( n = 15 ).But let me check if ( n = 15 ) satisfies ( a < 1000 ):Compute ( a = 4(15)^2 + 4(15) + 1 = 4(225) + 60 + 1 = 900 + 60 + 1 = 961 ). That's less than 1000.What about ( n = 16 ):( a = 4(16)^2 + 4(16) + 1 = 4(256) + 64 + 1 = 1024 + 64 + 1 = 1089 ). That's more than 1000, so ( n = 16 ) is invalid.Therefore, the maximum ( n ) is 15, giving a difference of ( 8n = 8 times 15 = 120 ).Wait, but hold on, earlier I thought the difference was ( 8n ), but in my initial steps, I had ( a - b = 4sqrt{2k} ). Let me double-check that.Wait, when I had ( a = 2k + 1 + 2sqrt{2k} ) and ( b = 2k + 1 - 2sqrt{2k} ), so ( a - b = 4sqrt{2k} ). But later, when I substituted ( k = 2n^2 ), I got ( a - b = 8n ). Let me verify this.Given ( k = 2n^2 ), then ( sqrt{2k} = sqrt{4n^2} = 2n ). Therefore, ( a - b = 4 times 2n = 8n ). So, yes, that's correct.So, with ( n = 15 ), the difference is 120. But wait, let me check the values of ( a ) and ( b ) when ( n = 15 ):( a = 4(15)^2 + 4(15) + 1 = 900 + 60 + 1 = 961 )( b = 4(15)^2 - 4(15) + 1 = 900 - 60 + 1 = 841 )Difference: ( 961 - 841 = 120 ). So, that's correct.But wait, the problem says both numbers must be less than 1000. ( a = 961 < 1000 ), ( b = 841 < 1000 ). So, that's okay.But hold on, earlier when I tried ( n = 16 ), ( a = 1089 ), which is over 1000, so ( n = 16 ) is invalid. So, 15 is the maximum.But wait, let me think again. Maybe I can get a larger difference by choosing a different ( k ) where ( sqrt{8k} ) is an integer but not necessarily ( m = 2n ). Wait, in my earlier steps, I assumed ( m = sqrt{2k} ) is integer, which led me to ( k = 2n^2 ). But perhaps there are other values of ( k ) where ( sqrt{8k} ) is integer without ( k ) being twice a square.Wait, ( sqrt{8k} ) must be integer because ( a ) and ( b ) are integers. So, ( 8k ) must be a perfect square. Let me denote ( sqrt{8k} = m ), so ( 8k = m^2 ), which implies ( k = frac{m^2}{8} ). Since ( k ) must be integer, ( m^2 ) must be divisible by 8, so ( m ) must be divisible by 2. Let me set ( m = 2n ), so ( m^2 = 4n^2 ), and ( k = frac{4n^2}{8} = frac{n^2}{2} ). Therefore, ( n^2 ) must be even, so ( n ) must be even. Let me set ( n = 2p ), so ( n^2 = 4p^2 ), and ( k = frac{4p^2}{2} = 2p^2 ).Wait, so this brings me back to ( k = 2p^2 ), which is similar to my earlier substitution. So, it seems that ( k ) must be twice a square, which leads to ( a ) and ( b ) as above.Therefore, my initial approach was correct, and the maximum difference is 120 when ( n = 15 ). But wait, in the initial problem, the user had a different approach and got 62 as the answer. Let me check where I might have gone wrong.Wait, in my initial steps, I set the arithmetic mean as ( 2k + 1 ) and geometric mean as ( 2k - 1 ). But maybe the other way around? If the geometric mean is larger, but that's impossible because AM ≥ GM. So, arithmetic mean must be larger. So, my initial setup is correct.Wait, but in the user's solution, they set AM as ( 2k + 1 ) and GM as ( 2k - 1 ), leading to a difference of ( 4sqrt{2k} ). Then, they found ( k = 120 ), leading to a difference of 62. Wait, but in my approach, I got a difference of 120. So, there's a discrepancy here.Let me check the user's solution again. They had:1. ( a + b = 4k + 2 )2. ( ab = 4k^2 - 4k + 1 )3. Then, solving the quadratic, they got ( a = 2k + 1 + sqrt{8k} ) and ( b = 2k + 1 - sqrt{8k} )4. Then, the difference is ( 2sqrt{8k} = 4sqrt{2k} )5. Then, they set ( 2k + 1 + sqrt{8k} < 1000 )6. Solving for ( k ), they found ( k = 120 ), leading to ( a = 272 ), ( b = 210 ), difference 62.Wait, so in their solution, they didn't substitute ( k ) as twice a square, but rather directly solved for ( k ) such that ( a ) and ( b ) are integers. So, perhaps my approach was too restrictive by assuming ( k = 2n^2 ). Maybe there are other values of ( k ) where ( sqrt{8k} ) is integer without ( k ) being twice a square.Wait, let's think again. For ( a ) and ( b ) to be integers, ( sqrt{8k} ) must be integer because ( a = 2k + 1 + sqrt{8k} ) and ( b = 2k + 1 - sqrt{8k} ). So, ( sqrt{8k} ) must be integer, say ( m ). Therefore, ( 8k = m^2 ), so ( k = frac{m^2}{8} ). For ( k ) to be integer, ( m^2 ) must be divisible by 8, which implies ( m ) must be divisible by 2. Let ( m = 2n ), so ( m^2 = 4n^2 ), and ( k = frac{4n^2}{8} = frac{n^2}{2} ). Therefore, ( n^2 ) must be even, so ( n ) must be even. Let ( n = 2p ), so ( n^2 = 4p^2 ), and ( k = frac{4p^2}{2} = 2p^2 ). So, this brings us back to ( k = 2p^2 ), which is the same as my initial substitution.Therefore, the only possible ( k ) are those where ( k = 2p^2 ), leading to ( a = 4p^2 + 4p + 1 ) and ( b = 4p^2 - 4p + 1 ), with difference ( 8p ). So, in this case, the maximum ( p ) is 15, giving a difference of 120.But in the user's solution, they got a difference of 62. So, perhaps there's a misunderstanding here. Let me check the user's solution again.Wait, in the user's solution, they set ( a = 2k + 1 + sqrt{8k} ) and ( b = 2k + 1 - sqrt{8k} ), then set ( 2k + 1 + sqrt{8k} < 1000 ), leading to ( k = 120 ), giving ( a = 272 ), ( b = 210 ), difference 62.But according to my approach, ( k = 2p^2 ), so ( p = sqrt{k/2} ). If ( k = 120 ), then ( p = sqrt{60} ), which is not integer. Therefore, ( k = 120 ) is not of the form ( 2p^2 ), so ( a ) and ( b ) would not be integers in that case. Wait, but in the user's solution, they got integer values for ( a ) and ( b ). So, how is that possible?Wait, let me check the user's solution again. They set ( k = 120 ), then ( a = 2(120) + 1 + sqrt{8(120)} = 241 + 31 = 272 ), and ( b = 241 - 31 = 210 ). So, ( sqrt{8k} = sqrt{960} = 30.98... ), but they rounded it to 31. Wait, but 31 squared is 961, which is not 960. So, that's a problem.Wait, actually, ( sqrt{960} ) is approximately 30.98, which is not an integer. Therefore, ( a ) and ( b ) would not be integers in this case. So, the user's solution might have an error here.Wait, but in their solution, they wrote ( sqrt{8k} = 31 ), which would imply ( 8k = 961 ), so ( k = 961 / 8 = 120.125 ), which is not integer. Therefore, their solution is incorrect because ( k ) must be integer, and ( sqrt{8k} ) must be integer as well.Therefore, my initial approach is correct, and the maximum difference is 120 when ( n = 15 ), giving ( a = 961 ) and ( b = 841 ), both less than 1000.Wait, but let me double-check if ( a = 961 ) and ( b = 841 ) satisfy the conditions.Compute AM: ( (961 + 841)/2 = 1802/2 = 901 ), which is an odd integer.Compute GM: ( sqrt{961 times 841} ). Let me compute 961 × 841. 961 is 31², and 841 is 29². So, ( sqrt{31^2 times 29^2} = 31 times 29 = 899 ). So, GM is 899, which is also an odd integer. And 901 and 899 are consecutive odd integers. So, yes, this satisfies the conditions.Therefore, the maximum difference is 120.But wait, in the user's solution, they got 62 as the answer, but that seems incorrect because their ( sqrt{8k} ) was not integer, leading to non-integer ( a ) and ( b ). Therefore, the correct maximum difference is 120.But let me check if there are other possible ( k ) values where ( sqrt{8k} ) is integer, leading to larger differences.Wait, for ( k = 2p^2 ), the maximum ( p ) is 15, giving ( k = 2(15)^2 = 450 ). Then, ( a = 4(15)^2 + 4(15) + 1 = 900 + 60 + 1 = 961 ), and ( b = 4(15)^2 - 4(15) + 1 = 900 - 60 + 1 = 841 ). Difference is 120.If I try ( p = 16 ), ( k = 2(16)^2 = 512 ), then ( a = 4(16)^2 + 4(16) + 1 = 1024 + 64 + 1 = 1089 ), which is over 1000, so invalid.Therefore, 120 is indeed the maximum difference.Wait, but let me think again. Maybe there are other ( k ) values where ( sqrt{8k} ) is integer, but ( k ) is not twice a square. For example, ( k = 8 times 1 = 8 ), ( sqrt{8k} = sqrt{64} = 8 ), which is integer. So, ( k = 8 ), ( a = 2(8) + 1 + 8 = 16 + 1 + 8 = 25 ), ( b = 16 + 1 - 8 = 9 ). Difference is 16.Similarly, ( k = 18 ), ( sqrt{8 times 18} = sqrt{144} = 12 ), so ( a = 37 + 12 = 49 ), ( b = 37 - 12 = 25 ). Difference is 24.Wait, but in these cases, ( k ) is not twice a square. For ( k = 8 ), ( 8 = 2 times 2^2 ), which fits my earlier substitution ( k = 2p^2 ) with ( p = 2 ). Similarly, ( k = 18 = 2 times 3^2 ), so ( p = 3 ). So, all such ( k ) are indeed twice a square.Therefore, my initial approach was correct, and the maximum difference is 120.But wait, in the user's solution, they got 62 as the answer, which is less than 120. So, perhaps they made a mistake in their approach.Wait, let me check the user's solution again. They set ( a = 2k + 1 + sqrt{8k} ) and ( b = 2k + 1 - sqrt{8k} ), then set ( 2k + 1 + sqrt{8k} < 1000 ). They solved for ( k ) and found ( k = 120 ), leading to ( a = 272 ), ( b = 210 ), difference 62.But as I saw earlier, ( sqrt{8k} = sqrt{960} approx 30.98 ), which is not integer, so ( a ) and ( b ) would not be integers. Therefore, their solution is incorrect because ( a ) and ( b ) must be integers.Therefore, the correct maximum difference is 120, achieved when ( n = 15 ), ( a = 961 ), ( b = 841 ).Wait, but let me double-check if ( a = 961 ) and ( b = 841 ) are both less than 1000. Yes, 961 < 1000 and 841 < 1000. So, that's valid.Therefore, the maximum possible difference is 120.

💡First, I recognize that the largest square that can be inscribed in a circle will have its diagonal equal to the diameter of the circle. Given that the diameter of the circle is 8 decimeters, the diagonal of the square is also 8 decimeters.Next, I recall that for a square, the relationship between the side length (s) and the diagonal (d) is given by the formula:[ d = s times sqrt{2} ]Substituting the known diagonal:[ 8 = s times sqrt{2} ]To find the side length, I solve for ( s ):[ s = frac{8}{sqrt{2}} ][ s = 4sqrt{2} text{ decimeters} ]Finally, to find the area of the square, I use the formula:[ text{Area} = s^2 ]Substituting the value of ( s ):[ text{Area} = (4sqrt{2})^2 ][ text{Area} = 16 times 2 ][ text{Area} = 32 text{ square decimeters} ]