💡First, I need to determine the total number of letters in the word "BANANA." Counting each letter, I find there are 6 letters in total.Next, I observe that some letters are repeated. Specifically, the letter 'A' appears 3 times, the letter 'N' appears 2 times, and the letter 'B' appears once.If all the letters were unique, the total number of ways to arrange them would be 6 factorial (6!), which equals 720.However, since there are repeated letters, I need to adjust for these repetitions to avoid overcounting. I do this by dividing the total number of arrangements by the factorial of the number of times each repeated letter occurs. In this case, I divide by 3! for the three 'A's and 2! for the two 'N's.Calculating the adjusted number of arrangements: 720 divided by (3! multiplied by 2!) equals 720 divided by 12, which results in 60.Therefore, the number of distinct ways to arrange the letters of the word "BANANA" is 60.

💡First, I need to evaluate the expression (left(frac{1}{3}right)^{10} cdot left(frac{2}{5}right)^{-4}).I notice that there is a negative exponent in the second term, (left(frac{2}{5}right)^{-4}). To simplify this, I can use the property of exponents that states (left(frac{a}{b}right)^{-n} = left(frac{b}{a}right)^{n}). Applying this property, I can rewrite the term as (left(frac{5}{2}right)^{4}).Now, I have two separate terms to evaluate: (left(frac{1}{3}right)^{10}) and (left(frac{5}{2}right)^{4}).For the first term, (left(frac{1}{3}right)^{10}), I can apply the rule (left(frac{a}{b}right)^{n} = frac{a^{n}}{b^{n}}). This gives me (frac{1^{10}}{3^{10}} = frac{1}{3^{10}}).For the second term, (left(frac{5}{2}right)^{4}), using the same exponent rule, I get (frac{5^{4}}{2^{4}} = frac{625}{16}).Next, I need to multiply these two results together: (frac{1}{3^{10}} cdot frac{625}{16}).Multiplying the numerators and denominators separately, I have (frac{1 cdot 625}{3^{10} cdot 16} = frac{625}{59049 cdot 16}).Calculating the denominator, (3^{10} = 59049), so the expression becomes (frac{625}{59049 cdot 16}).Finally, multiplying (59049) by (16) gives (943744), resulting in the simplified fraction (frac{625}{943744}).

💡Okay, so I have this problem where numbers from 1 to 20 are written on a board. I can erase any two numbers, say a and b, and replace them with ab + a + b. I need to figure out what number will be left after doing this 19 times. Hmm, let me think about how to approach this.First, I notice that each operation reduces the total count of numbers by one. So starting with 20 numbers, after 19 operations, I should have just one number left. That makes sense.Now, the operation is to replace a and b with ab + a + b. Let me see if I can find a pattern or an invariant here. Maybe if I can find something that doesn't change when I perform this operation, it could help me figure out the final number.Let me try to rewrite the operation ab + a + b. If I factor this, it becomes (a + 1)(b + 1) - 1. Oh, that's interesting! So replacing a and b with ab + a + b is the same as taking (a + 1)(b + 1) and then subtracting 1.So, if I consider each number on the board as (number + 1), then the operation is just multiplying those two numbers together. That seems promising because multiplication is associative and commutative, so the order might not matter in the end.Let me define a new set where each number is one more than the original numbers. So instead of 1, 2, ..., 20, I have 2, 3, ..., 21. Now, each time I perform the operation on two numbers a and b, I'm effectively multiplying (a + 1) and (b + 1) and replacing them with that product.If I keep doing this, eventually, I'll end up with just one number, which is the product of all the numbers from 2 to 21. That product is 21 factorial, written as 21!.But wait, in the original problem, we're subtracting 1 each time we perform the operation. So, the final number on the board is 21! - 1.Let me double-check this. If I start with numbers 1 to 20, and each operation replaces two numbers a and b with (a + 1)(b + 1) - 1, then the product of all (number + 1) remains the same throughout the operations. So, the product of 2 to 21 is 21!, and the final number is 21! - 1.Yeah, that seems to make sense. I don't think I made any mistakes here. The key was noticing that the operation can be rewritten in terms of multiplication, which preserves the product, leading to the final result.

💡Okay, so I have this problem about an ellipse. Let me try to understand what it's asking.First, part (I) says that the ellipse C is centered at the origin, its focus is on the x-axis, and its eccentricity is 1/2. It also passes through the point M(1, 3/2). I need to find the equation of this ellipse.Alright, I remember that the standard equation of an ellipse centered at the origin with a horizontal major axis is (x²/a²) + (y²/b²) = 1, where a is the semi-major axis, b is the semi-minor axis, and the relationship between a, b, and the eccentricity e is e = c/a, where c is the distance from the center to each focus. Also, c² = a² - b².Given that the eccentricity e is 1/2, so e = c/a = 1/2, which means c = a/2. Then, c² = a²/4, so a² - b² = a²/4. That simplifies to b² = a² - a²/4 = (3/4)a². So, b = (√3/2)a.Now, the ellipse passes through the point M(1, 3/2). So, plugging this point into the ellipse equation:(1²)/a² + ( (3/2)² ) / b² = 1Which is 1/a² + (9/4)/b² = 1But we know that b² = (3/4)a², so substitute that in:1/a² + (9/4)/( (3/4)a² ) = 1Simplify the second term: (9/4) divided by (3/4) is (9/4)*(4/3) = 3. So, the equation becomes:1/a² + 3/a² = 1Which is 4/a² = 1, so a² = 4. Therefore, a = 2.Then, b² = (3/4)*a² = (3/4)*4 = 3. So, b = √3.Therefore, the equation of the ellipse is (x²)/4 + (y²)/3 = 1.Okay, that seems straightforward. Let me just double-check my steps:1. Eccentricity e = 1/2, so c = a/2.2. c² = a² - b² => (a²)/4 = a² - b² => b² = (3/4)a².3. Plugging point M(1, 3/2) into the ellipse equation.4. Substituted b² and solved for a², got a²=4, so a=2.5. Then, b²=3, so the equation is x²/4 + y²/3 =1.Yep, that seems correct.Now, part (II) is a bit more complex. It asks whether there exists a line l₁ passing through point P(2,1) that intersects the ellipse C at two distinct points A and B such that the dot product of vectors PA and PB equals the square of vector PM. If such a line exists, find its equation; otherwise, explain why not.First, let's parse the condition: PA · PB = |PM|².Point M is (1, 3/2), so vector PM is M - P, which is (1-2, 3/2 -1) = (-1, 1/2). So, |PM|² is (-1)² + (1/2)² = 1 + 1/4 = 5/4.So, the condition is PA · PB = 5/4.Let me think about how to approach this. Since the line passes through P(2,1), let's denote the line as l₁. Let me assume that l₁ has a slope k, so its equation is y - 1 = k(x - 2), or y = kx - 2k +1.We need to find if such a line intersects the ellipse at two distinct points A and B, and then check if PA · PB = 5/4.So, first, let's find the points of intersection between the line and the ellipse.Substitute y = kx - 2k +1 into the ellipse equation:x²/4 + (kx - 2k +1)² /3 =1.Let me expand this:x²/4 + [k²x² - 4k²x + 4k² + 2kx - 4k +1]/3 =1.Wait, let me do it step by step.First, compute (kx - 2k +1)²:= (kx)^2 + (-2k +1)^2 + 2*(kx)*(-2k +1)= k²x² + (4k² -4k +1) + (-4k²x + 2kx)So, expanding:= k²x² + 4k² -4k +1 -4k²x + 2kxSo, putting it back into the ellipse equation:x²/4 + [k²x² + 4k² -4k +1 -4k²x + 2kx]/3 =1.Multiply through by 12 to eliminate denominators:3x² + 4[k²x² + 4k² -4k +1 -4k²x + 2kx] =12.Expand the 4:3x² + 4k²x² +16k² -16k +4 -16k²x +8kx =12.Combine like terms:(3 + 4k²)x² + (-16k² +8k)x + (16k² -16k +4) =12.Bring 12 to the left:(3 + 4k²)x² + (-16k² +8k)x + (16k² -16k +4 -12) =0.Simplify constants:16k² -16k +4 -12 =16k² -16k -8.So, the quadratic equation is:(3 + 4k²)x² + (-16k² +8k)x + (16k² -16k -8) =0.Let me denote this as Ax² + Bx + C =0, where:A = 3 +4k²,B = -16k² +8k,C =16k² -16k -8.Now, for the line to intersect the ellipse at two distinct points, the discriminant must be positive.Discriminant D = B² -4AC.Compute D:= (-16k² +8k)^2 -4*(3 +4k²)*(16k² -16k -8).First, compute (-16k² +8k)^2:= (16k² -8k)^2 = (16k²)^2 - 2*16k²*8k + (8k)^2 =256k^4 -256k^3 +64k².Now, compute 4AC:4*(3 +4k²)*(16k² -16k -8).First, compute (3 +4k²)*(16k² -16k -8):Multiply term by term:3*16k² =48k²,3*(-16k)= -48k,3*(-8)= -24,4k²*16k²=64k^4,4k²*(-16k)= -64k^3,4k²*(-8)= -32k².So, adding all together:64k^4 -64k^3 +48k² -48k -24 -32k².Combine like terms:64k^4 -64k^3 + (48k² -32k²) + (-48k) + (-24)=64k^4 -64k^3 +16k² -48k -24.Multiply by 4:4AC =4*(64k^4 -64k^3 +16k² -48k -24)=256k^4 -256k^3 +64k² -192k -96.Now, D = [256k^4 -256k^3 +64k²] - [256k^4 -256k^3 +64k² -192k -96]Subtract term by term:256k^4 -256k^4 =0,-256k^3 +256k^3=0,64k² -64k²=0,0 - (-192k)= +192k,0 - (-96)= +96.So, D=192k +96.We need D>0 for two distinct real roots.So, 192k +96 >0 => 192k > -96 => k > -96/192 => k > -0.5.So, the slope k must be greater than -0.5.Alright, so any line with slope k > -0.5 will intersect the ellipse at two distinct points.Now, moving on to the condition PA · PB =5/4.Let me recall that PA and PB are vectors from P to A and P to B.Given that A and B lie on the line l₁, which passes through P(2,1). So, points A and B can be parametrized as points on the line.Alternatively, since we have the quadratic equation in x, whose roots are x₁ and x₂, the x-coordinates of A and B.Let me denote A as (x₁, y₁) and B as (x₂, y₂). Then, vectors PA and PB are (x₁ -2, y₁ -1) and (x₂ -2, y₂ -1), respectively.Their dot product is (x₁ -2)(x₂ -2) + (y₁ -1)(y₂ -1).Given that y =k(x -2) +1, so y₁ =k(x₁ -2)+1, y₂ =k(x₂ -2)+1.So, y₁ -1 =k(x₁ -2), y₂ -1 =k(x₂ -2).Therefore, the dot product becomes:(x₁ -2)(x₂ -2) + [k(x₁ -2)][k(x₂ -2)] = (x₁ -2)(x₂ -2)(1 +k²).This is equal to 5/4.So, (x₁ -2)(x₂ -2)(1 +k²) =5/4.Let me compute (x₁ -2)(x₂ -2). Since x₁ and x₂ are roots of the quadratic equation Ax² + Bx + C=0, we can use Vieta's formula.From quadratic equation, x₁ +x₂ = -B/A, and x₁x₂ = C/A.Compute (x₁ -2)(x₂ -2)=x₁x₂ -2(x₁ +x₂) +4.So, substituting:= (C/A) -2*(-B/A) +4= (C/A) + (2B)/A +4= (C + 2B)/A +4.So, let's compute C + 2B:C =16k² -16k -8,2B=2*(-16k² +8k)= -32k² +16k.So, C +2B=16k² -16k -8 -32k² +16k= (-16k²) +0k -8.Therefore, (x₁ -2)(x₂ -2)= (-16k² -8)/A +4.But A=3 +4k².So,= (-16k² -8)/(3 +4k²) +4.Let me write 4 as 4*(3 +4k²)/(3 +4k²):= [(-16k² -8) +4*(3 +4k²)] / (3 +4k²)Compute numerator:-16k² -8 +12 +16k²= (-16k² +16k²) + (-8 +12)=0 +4=4.Therefore, (x₁ -2)(x₂ -2)=4/(3 +4k²).So, going back to the dot product:(x₁ -2)(x₂ -2)(1 +k²)= [4/(3 +4k²)]*(1 +k²)=4(1 +k²)/(3 +4k²).Set this equal to 5/4:4(1 +k²)/(3 +4k²)=5/4.Multiply both sides by (3 +4k²):4(1 +k²)= (5/4)(3 +4k²).Multiply both sides by 4 to eliminate denominators:16(1 +k²)=5(3 +4k²).Expand:16 +16k²=15 +20k².Bring all terms to left:16 +16k² -15 -20k²=0 => (16 -15) + (16k² -20k²)=1 -4k²=0.So, 1 -4k²=0 => 4k²=1 =>k²=1/4 =>k=±1/2.But earlier, we found that k must be greater than -0.5 for the discriminant to be positive. So, k=1/2 is acceptable, but k=-1/2 is not, since -1/2 is not greater than -0.5; it's equal. Wait, actually, k > -0.5, so k=-0.5 is the boundary case where discriminant is zero, meaning the line is tangent to the ellipse. But in our case, we need two distinct points, so k must be greater than -0.5. Therefore, k=-1/2 is not acceptable because it would make discriminant zero, leading to only one point of contact, not two.Therefore, only k=1/2 is acceptable.Therefore, the slope k=1/2.So, the equation of the line is y= (1/2)(x -2) +1.Simplify:y= (1/2)x -1 +1= (1/2)x.So, the equation is y=(1/2)x.Therefore, such a line exists, and its equation is y=(1/2)x.Let me just recap:1. Found the ellipse equation in part (I): x²/4 + y²/3=1.2. For part (II), assumed a line through P(2,1) with slope k, found its equation.3. Substituted into ellipse equation, found quadratic in x.4. Used discriminant condition to find k > -0.5.5. Expressed the dot product condition in terms of x₁ and x₂, used Vieta's formulas.6. Derived an equation in k, solved it to get k=±1/2.7. Discarded k=-1/2 because it doesn't satisfy the discriminant condition for two distinct points.8. Concluded that k=1/2 is the valid slope, giving the line y=(1/2)x.I think that's thorough. I don't see any mistakes in the calculations.**Final Answer**(I) The equation of the ellipse is boxed{dfrac{x^2}{4} + dfrac{y^2}{3} = 1}.(II) The equation of the line ( l_1 ) is boxed{y = dfrac{1}{2}x}.

💡Okay, so I have this geometry problem here. It says there are six right triangles in the figure, and all the angles at point A are equal. They also give me that AB is 24 cm and AC is 54 cm. I need to find the length of AD.Hmm, right triangles with equal angles at A. That probably means they're all similar triangles because they have the same angles, right? So, if they're similar, their sides should be in proportion.Let me try to visualize this. There's a point A, and from there, there are six right triangles. Maybe they're arranged in a way where each triangle is connected to the next, forming some sort of chain from AB to AC? So, AD would be somewhere in the middle of this chain.Since all the triangles are similar, the ratios of their corresponding sides should be equal. Let's denote the lengths of the sides of these triangles as a, b, x, c, d, and so on. But I'm not sure exactly how they're connected. Maybe each triangle's hypotenuse is a side of the next triangle?Wait, the problem mentions that the angles at A are equal. So, all these triangles share the same angle at A, which is a right angle? Or is it another angle? Wait, no, they're right triangles, so one of the angles is 90 degrees, and the other two angles are acute. Since all the angles at A are equal, that means the angle at A is the same for all six triangles.So, if they're all right triangles with a common angle at A, they must all be similar to each other. That makes sense. So, the sides should be in proportion.Let me think about the proportions. If AB is 24 cm and AC is 54 cm, and AD is somewhere in between, maybe AD is the geometric mean of AB and AC? Because in similar triangles, sometimes the geometric mean comes into play.Wait, if I consider AB and AC as corresponding sides of the first and last triangles, then AD could be the geometric mean. So, AD squared equals AB times AC. Let me check that.So, AD^2 = AB * AC = 24 * 54. Let me calculate that. 24 times 54. Hmm, 24 times 50 is 1200, and 24 times 4 is 96, so 1200 + 96 is 1296. So, AD squared is 1296, which means AD is the square root of 1296.What's the square root of 1296? Let me think. 36 times 36 is 1296, right? Because 30 squared is 900, and 35 squared is 1225, and 36 squared is 1296. Yeah, so AD is 36 cm.Wait, but let me make sure I didn't skip any steps. So, if all the triangles are similar, the ratio of their sides should be consistent. If I have six triangles, each subsequent triangle's sides are scaled by some factor. So, starting from AB, each triangle's sides are multiplied by a common ratio until I get to AC.So, if I denote the common ratio as r, then after six multiplications, I should get from AB to AC. So, AB * r^6 = AC. Let's see, 24 * r^6 = 54. Then, r^6 = 54/24, which simplifies to 9/4. So, r^6 = (3/2)^2. Hmm, that's interesting.Wait, but if r^6 is (3/2)^2, then r^3 is 3/2. So, r is the cube root of 3/2. But I'm not sure if that helps me directly. Maybe I need to find AD, which is somewhere in the middle.If I think of AD as the third triangle from AB, then AD = AB * r^3. Since r^3 is 3/2, then AD = 24 * (3/2) = 36 cm. Yeah, that matches what I got earlier.So, both methods give me AD as 36 cm. That must be the answer. I think I'm confident with that.**Final Answer**boxed{C}

💡Okay, so I have this problem about a parallelogram ABCD. The vertices A, B, C, and D are connected to the midpoints of the sides CD, AD, AB, and BC respectively. I need to prove that the area of the quadrilateral formed by these lines is 1/5 of the area of the parallelogram. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, let me visualize the parallelogram. Parallelogram ABCD has opposite sides equal and parallel. Let me label the vertices in order: A, B, C, D. So sides AB and CD are equal and parallel, and sides AD and BC are equal and parallel. Now, the midpoints of these sides: let's denote them as M, N, P, Q. Specifically, M is the midpoint of CD, N is the midpoint of AD, P is the midpoint of AB, and Q is the midpoint of BC.So, according to the problem, we connect each vertex to the midpoint of the opposite side. That means connecting A to M, B to N, C to P, and D to Q. These four lines should intersect each other, forming a smaller quadrilateral inside the parallelogram. The goal is to find the area of this inner quadrilateral.I think a good approach would be to use coordinate geometry. If I assign coordinates to the vertices of the parallelogram, I can find the equations of the lines connecting the vertices to the midpoints and then find the points of intersection. Once I have the coordinates of the vertices of the inner quadrilateral, I can calculate its area and compare it to the area of the original parallelogram.Let me set up a coordinate system. Let's place point A at the origin (0,0). Since it's a parallelogram, I can denote point B as (a,0), point D as (0,b), and consequently, point C would be at (a,b). This way, sides AB and DC are both length a, and sides AD and BC are both length b. The midpoints can then be calculated easily.Calculating the midpoints:- Midpoint M of CD: Since C is (a,b) and D is (0,b), the midpoint M would be ((a+0)/2, (b+b)/2) = (a/2, b).Wait, that doesn't seem right. If CD is from (a,b) to (0,b), the midpoint should be halfway along the x-axis. So, yes, (a/2, b) is correct.- Midpoint N of AD: A is (0,0) and D is (0,b), so midpoint N is ((0+0)/2, (0+b)/2) = (0, b/2).- Midpoint P of AB: A is (0,0) and B is (a,0), so midpoint P is ((0+a)/2, (0+0)/2) = (a/2, 0).- Midpoint Q of BC: B is (a,0) and C is (a,b), so midpoint Q is ((a+a)/2, (0+b)/2) = (a, b/2).Okay, so now I have the coordinates of all midpoints:- M: (a/2, b)- N: (0, b/2)- P: (a/2, 0)- Q: (a, b/2)Now, I need to find the equations of the lines connecting each vertex to the midpoint of the opposite side.Let's start with line AM: connecting A(0,0) to M(a/2, b). The slope of this line is (b - 0)/(a/2 - 0) = (b)/(a/2) = 2b/a. So the equation is y = (2b/a)x.Next, line BN: connecting B(a,0) to N(0, b/2). The slope is (b/2 - 0)/(0 - a) = (b/2)/(-a) = -b/(2a). The equation can be written using point-slope form from point B: y - 0 = (-b/(2a))(x - a), which simplifies to y = (-b/(2a))x + b/2.Then, line CP: connecting C(a,b) to P(a/2, 0). The slope is (0 - b)/(a/2 - a) = (-b)/(-a/2) = 2b/a. The equation using point C is y - b = (2b/a)(x - a). Simplifying: y = (2b/a)x - 2b + b = (2b/a)x - b.Lastly, line DQ: connecting D(0,b) to Q(a, b/2). The slope is (b/2 - b)/(a - 0) = (-b/2)/a = -b/(2a). The equation using point D is y - b = (-b/(2a))(x - 0), which simplifies to y = (-b/(2a))x + b.Now, I have the equations of all four lines:1. AM: y = (2b/a)x2. BN: y = (-b/(2a))x + b/23. CP: y = (2b/a)x - b4. DQ: y = (-b/(2a))x + bThe inner quadrilateral is formed by the intersections of these lines. Let's find the points of intersection.First, find the intersection of AM and BN:Set y = (2b/a)x equal to y = (-b/(2a))x + b/2.So, (2b/a)x = (-b/(2a))x + b/2.Multiply both sides by 2a to eliminate denominators:4b x = -b x + a bBring terms with x to one side:4b x + b x = a b5b x = a bx = a/5Substitute back into y = (2b/a)x:y = (2b/a)(a/5) = 2b/5So, the intersection point is (a/5, 2b/5).Next, find the intersection of BN and CP:Set y = (-b/(2a))x + b/2 equal to y = (2b/a)x - b.So, (-b/(2a))x + b/2 = (2b/a)x - bMultiply both sides by 2a to eliminate denominators:-b x + a b = 4b x - 2a bBring all terms to one side:-b x - 4b x + a b + 2a b = 0-5b x + 3a b = 0-5b x = -3a bx = (3a)/5Substitute back into y = (-b/(2a))x + b/2:y = (-b/(2a))(3a/5) + b/2 = (-3b/10) + (5b/10) = (2b)/10 = b/5So, the intersection point is (3a/5, b/5).Now, find the intersection of CP and DQ:Set y = (2b/a)x - b equal to y = (-b/(2a))x + b.So, (2b/a)x - b = (-b/(2a))x + bMultiply both sides by 2a:4b x - 2a b = -b x + 2a bBring all terms to one side:4b x + b x - 2a b - 2a b = 05b x - 4a b = 05b x = 4a bx = (4a)/5Substitute back into y = (2b/a)x - b:y = (2b/a)(4a/5) - b = (8b/5) - b = (8b - 5b)/5 = 3b/5So, the intersection point is (4a/5, 3b/5).Finally, find the intersection of DQ and AM:Set y = (-b/(2a))x + b equal to y = (2b/a)x.So, (-b/(2a))x + b = (2b/a)xMultiply both sides by 2a:-b x + 2a b = 4b xBring terms with x to one side:-b x - 4b x + 2a b = 0-5b x + 2a b = 0-5b x = -2a bx = (2a)/5Substitute back into y = (2b/a)x:y = (2b/a)(2a/5) = 4b/5So, the intersection point is (2a/5, 4b/5).Now, I have all four vertices of the inner quadrilateral:1. (a/5, 2b/5)2. (3a/5, b/5)3. (4a/5, 3b/5)4. (2a/5, 4b/5)To find the area of this quadrilateral, I can use the shoelace formula. Let me list the coordinates in order:1. (a/5, 2b/5)2. (3a/5, b/5)3. (4a/5, 3b/5)4. (2a/5, 4b/5)Let's apply the shoelace formula:Area = 1/2 |(x1y2 + x2y3 + x3y4 + x4y1) - (y1x2 + y2x3 + y3x4 + y4x1)|Plugging in the values:x1 = a/5, y1 = 2b/5x2 = 3a/5, y2 = b/5x3 = 4a/5, y3 = 3b/5x4 = 2a/5, y4 = 4b/5Compute the terms:x1y2 = (a/5)(b/5) = ab/25x2y3 = (3a/5)(3b/5) = 9ab/25x3y4 = (4a/5)(4b/5) = 16ab/25x4y1 = (2a/5)(2b/5) = 4ab/25Sum of these terms: ab/25 + 9ab/25 + 16ab/25 + 4ab/25 = (1 + 9 + 16 + 4)ab/25 = 30ab/25 = 6ab/5Now the other part:y1x2 = (2b/5)(3a/5) = 6ab/25y2x3 = (b/5)(4a/5) = 4ab/25y3x4 = (3b/5)(2a/5) = 6ab/25y4x1 = (4b/5)(a/5) = 4ab/25Sum of these terms: 6ab/25 + 4ab/25 + 6ab/25 + 4ab/25 = (6 + 4 + 6 + 4)ab/25 = 20ab/25 = 4ab/5Now, subtract the two sums:6ab/5 - 4ab/5 = 2ab/5Take the absolute value and multiply by 1/2:Area = (1/2)(2ab/5) = ab/5So, the area of the inner quadrilateral is ab/5.Now, what's the area of the original parallelogram ABCD? Since it's a parallelogram with base a and height b (assuming the sides are along the axes), the area is base times height, which is ab.Therefore, the area of the inner quadrilateral is ab/5, which is 1/5 of the area of the parallelogram ABCD.Wait, that seems straightforward. But let me double-check my calculations to make sure I didn't make any mistakes.Looking back at the shoelace formula:First sum: x1y2 + x2y3 + x3y4 + x4y1= (a/5)(b/5) + (3a/5)(3b/5) + (4a/5)(4b/5) + (2a/5)(2b/5)= ab/25 + 9ab/25 + 16ab/25 + 4ab/25= (1 + 9 + 16 + 4)ab/25= 30ab/25= 6ab/5Second sum: y1x2 + y2x3 + y3x4 + y4x1= (2b/5)(3a/5) + (b/5)(4a/5) + (3b/5)(2a/5) + (4b/5)(a/5)= 6ab/25 + 4ab/25 + 6ab/25 + 4ab/25= (6 + 4 + 6 + 4)ab/25= 20ab/25= 4ab/5Subtracting: 6ab/5 - 4ab/5 = 2ab/5Multiply by 1/2: (1/2)(2ab/5) = ab/5Yes, that seems correct. So the area of the inner quadrilateral is indeed ab/5, which is 1/5 of the area of the parallelogram ABCD, whose area is ab.Therefore, the proof is complete. The area of the quadrilateral formed by connecting the vertices to the midpoints of the opposite sides in a parallelogram is 1/5 of the area of the parallelogram.