answer：Okay, so I have this problem about a triangle with integer side lengths. One side is four times as long as the second side, and the third side is 20. I need to find the greatest possible perimeter of this triangle. Hmm, let me think about how to approach this.First, I know that in any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side. This is called the triangle inequality theorem. So, whatever sides I come up with, they have to satisfy this condition.Let me denote the sides of the triangle. Let's say the shorter side is x. Then, the longer side that's four times as long would be 4x. The third side is given as 20. So, the sides are x, 4x, and 20.Now, I need to apply the triangle inequality theorem to these sides. There are three conditions to check:1. The sum of the two shorter sides must be greater than the longest side.2. The sum of the shorter side and the third side must be greater than the longer side.3. The sum of the longer side and the third side must be greater than the shorter side.Wait, actually, since the sides are x, 4x, and 20, I need to figure out which one is the longest. It depends on the value of x. If 4x is greater than 20, then 4x is the longest side. If 4x is less than 20, then 20 is the longest side. So, I need to consider both possibilities.Let me first assume that 4x is the longest side. That would mean 4x > 20, so x > 5. Since x has to be an integer, x would be at least 6.Now, applying the triangle inequality conditions:1. x + 20 > 4x2. x + 4x > 203. 4x + 20 > xLet me solve each inequality.Starting with the first inequality: x + 20 > 4xSubtract x from both sides: 20 > 3xDivide both sides by 3: 20/3 > xWhich is approximately 6.666... > xSo, x must be less than 6.666...But earlier, I assumed that x > 5 because 4x > 20. So, combining these two, x must be greater than 5 and less than 6.666...Since x is an integer, the possible values are x = 6.Let me check x = 6.Then, the sides would be 6, 24, and 20.Check the triangle inequalities:1. 6 + 20 > 24? 26 > 24, which is true.2. 6 + 24 > 20? 30 > 20, which is true.3. 24 + 20 > 6? 44 > 6, which is true.So, x = 6 works, and the perimeter would be 6 + 24 + 20 = 50.But wait, is this the maximum possible perimeter? Maybe I should check if 4x could be less than 20, making 20 the longest side.If 4x < 20, then x < 5. Since x is an integer, x can be 1, 2, 3, or 4.Let me check these values.Starting with x = 4:Sides would be 4, 16, and 20.Check the triangle inequalities:1. 4 + 16 > 20? 20 > 20? No, it's equal, which doesn't satisfy the inequality. So, x = 4 doesn't work.Next, x = 3:Sides would be 3, 12, and 20.Check the inequalities:1. 3 + 12 > 20? 15 > 20? No, that's false.So, x = 3 doesn't work.x = 2:Sides would be 2, 8, and 20.Check the inequalities:1. 2 + 8 > 20? 10 > 20? No.x = 2 doesn't work.x = 1:Sides would be 1, 4, and 20.Check the inequalities:1. 1 + 4 > 20? 5 > 20? No.So, none of the x values less than 5 work when 20 is the longest side.Therefore, the only valid integer value for x is 6, giving sides of 6, 24, and 20, with a perimeter of 50.But wait, could there be a larger perimeter? Maybe if I consider different configurations or if I misunderstood the problem.Wait, the problem says one side is four times as long as the second side. It doesn't specify which side is which. So, maybe I should consider the case where the third side, 20, is four times one of the other sides.Let me explore that possibility.Case 1: 4x is one side, x is another, and 20 is the third.Case 2: 20 is four times one of the sides, say 20 = 4y, so y = 5. Then, the sides would be y = 5, 20, and another side, let's say z.Wait, but in this case, the sides would be 5, 20, and z. But then, z has to satisfy the triangle inequality as well.But in this case, z could be either the longer side or the shorter side. Let me see.If z is the longer side, then z < 5 + 20 = 25. So, z < 25.If z is the shorter side, then 5 + z > 20, so z > 15.So, z must be greater than 15 and less than 25.Since z has to be an integer, z can be from 16 to 24.But wait, in this case, the sides would be 5, 20, and z, where z is between 16 and 24.But the problem states that one side is four times as long as the second side. So, if 20 is four times 5, then that's satisfied. But we need to ensure that the other side, z, doesn't violate the triangle inequality.But in this case, z can be up to 24, which would make the perimeter 5 + 20 + 24 = 49, which is less than 50.Wait, so in this case, the perimeter is 49, which is less than the previous case where the perimeter was 50.So, the maximum perimeter is still 50.But let me double-check.If I take z = 24, then the sides are 5, 20, 24.Check the triangle inequalities:1. 5 + 20 > 24? 25 > 24, which is true.2. 5 + 24 > 20? 29 > 20, which is true.3. 20 + 24 > 5? 44 > 5, which is true.So, that works, but the perimeter is 49, which is less than 50.Alternatively, if z is 16, the sides would be 5, 20, 16.Check the inequalities:1. 5 + 16 > 20? 21 > 20, which is true.2. 5 + 20 > 16? 25 > 16, which is true.3. 20 + 16 > 5? 36 > 5, which is true.So, that works too, but the perimeter is 5 + 20 + 16 = 41, which is even smaller.Therefore, the maximum perimeter in this case is 49, which is still less than 50.So, the maximum perimeter is 50.Wait, but let me think again. Maybe I can have a larger perimeter if I consider that the third side, 20, is not necessarily the middle side.Wait, in the first case, when x = 6, the sides are 6, 24, and 20. So, 24 is the longest side, 20 is the middle, and 6 is the shortest.In the second case, when 20 is four times 5, the sides are 5, 20, and z, where z can be up to 24, making the perimeter 49.So, 50 is larger than 49, so 50 is the maximum.But let me check if there are any other possibilities.Wait, what if the third side, 20, is not four times any of the other sides, but one of the other sides is four times 20? That would make one side 80, which seems too large, but let's see.If one side is four times 20, that would be 80. So, the sides would be 20, 80, and x.But then, applying the triangle inequality:1. 20 + x > 80? So, x > 60.But also, 20 + 80 > x? So, x < 100.And 80 + x > 20? Which is always true if x is positive.So, x must be greater than 60 and less than 100.But since x has to be an integer, x can be from 61 to 99.But wait, in this case, the sides would be 20, 80, and x, where x is between 61 and 99.But the problem states that one side is four times as long as the second side. So, if 80 is four times 20, that's satisfied.But then, the perimeter would be 20 + 80 + x, where x is between 61 and 99.So, the perimeter would be between 161 and 199.Wait, that's way larger than 50.But hold on, does this satisfy the triangle inequality?Let me take x = 61:Sides are 20, 80, 61.Check the inequalities:1. 20 + 61 > 80? 81 > 80, which is true.2. 20 + 80 > 61? 100 > 61, which is true.3. 80 + 61 > 20? 141 > 20, which is true.So, x = 61 works, and the perimeter is 20 + 80 + 61 = 161.Similarly, x = 99:Sides are 20, 80, 99.Check the inequalities:1. 20 + 80 > 99? 100 > 99, which is true.2. 20 + 99 > 80? 119 > 80, which is true.3. 80 + 99 > 20? 179 > 20, which is true.So, x = 99 works, and the perimeter is 20 + 80 + 99 = 199.Wait, so this seems possible. But the problem states that one side is four times as long as the second side, and the third side is 20.So, in this case, 80 is four times 20, and the third side is x, which can be up to 99, making the perimeter up to 199.But wait, the problem says "one side is four times as long as the second side, and the length of the third side is 20."So, does that mean that the third side is fixed at 20, and the other two sides are x and 4x?Or does it mean that one side is four times another, and the third side is 20, but not necessarily that 20 is one of the sides being multiplied by four?I think the problem is a bit ambiguous, but I think the intended interpretation is that the sides are x, 4x, and 20, with 20 being the third side, not necessarily being four times any of the others.But in my earlier analysis, I considered both cases: 4x being the longest side and 20 being the longest side.But in this new case, where 80 is four times 20, and the third side is x, which can be up to 99, making the perimeter much larger.But wait, the problem says "one side is four times as long as the second side, and the length of the third side is 20."So, it could be interpreted as:Either:- The sides are x, 4x, and 20, with 20 being the third side, not necessarily four times any of the others.Or:- One side is four times another, and the third side is 20, which could be four times one of the sides.So, both interpretations are possible.But in the first interpretation, the maximum perimeter is 50.In the second interpretation, where 20 is four times one of the sides, making that side 5, and the third side can be up to 24, making the perimeter 49.But in the third interpretation, where one side is four times 20, making it 80, and the third side can be up to 99, making the perimeter 199.But wait, the problem says "one side is four times as long as the second side, and the length of the third side is 20."So, it's saying that among the three sides, one is four times another, and the third is 20.So, it could be that 20 is four times one of the sides, or one of the other sides is four times another, with 20 being the third.So, both cases are possible.Therefore, to find the maximum perimeter, I need to consider both possibilities.Case 1: The sides are x, 4x, and 20, with 4x being the longest side.As before, x = 6, perimeter = 50.Case 2: The sides are 5, 20, and z, where z can be up to 24, making the perimeter up to 49.Case 3: The sides are 20, 80, and z, where z can be up to 99, making the perimeter up to 199.Wait, but in case 3, is 80 considered as four times 20? Yes, 80 = 4 * 20.So, that satisfies the condition that one side is four times as long as the second side.Therefore, the maximum perimeter would be when z is as large as possible, which is 99, making the perimeter 20 + 80 + 99 = 199.But wait, is there a restriction on z? Because z has to satisfy the triangle inequality.In this case, z must be less than 20 + 80 = 100, so z < 100. Since z is an integer, z can be up to 99.So, yes, z = 99 is allowed.Therefore, the maximum perimeter is 199.But wait, the problem says "one side is four times as long as the second side, and the length of the third side is 20."So, in this case, the third side is 20, and one side is four times another.So, if 80 is four times 20, then 80 is one side, 20 is another, and the third side is z.But in this case, the third side is z, not 20. Wait, no, the third side is 20.Wait, hold on, I think I made a mistake here.If one side is four times another, and the third side is 20, then the sides are either:- x, 4x, and 20Or- 20, 4*20=80, and xBut in the second case, the third side is x, not 20.Wait, the problem says "the length of the third side is 20."So, if we consider the sides as 20, 80, and x, then the third side is x, not 20.But the problem says the third side is 20, so x must be 20.Wait, no, the problem says "the length of the third side is 20."So, in this case, the sides would be 20, 80, and 20?Wait, that can't be, because then two sides are 20 and 80, and the third is 20.But then, the triangle would have sides 20, 20, 80.But that doesn't satisfy the triangle inequality because 20 + 20 = 40, which is less than 80.So, that's not a valid triangle.Wait, so maybe I misinterpreted.If the third side is 20, then the sides are x, 4x, and 20.Alternatively, if one side is four times another, and the third side is 20, it could be that 20 is four times one of the sides, making that side 5, and the third side is x.But in that case, the sides would be 5, 20, and x, with x being the third side.But the problem says "the length of the third side is 20," so x would have to be 20, making the sides 5, 20, 20.But that's a valid triangle because 5 + 20 > 20, 20 + 20 > 5, and 5 + 20 > 20.Wait, but in this case, the third side is 20, and one side is four times another (20 is four times 5).So, that works, and the perimeter is 5 + 20 + 20 = 45.But earlier, when I considered x = 6, the perimeter was 50, which is larger.So, in that case, 50 is larger.But wait, earlier I thought about making one side four times 20, which would be 80, but then the third side would have to be 20, which doesn't work because 20 + 20 is not greater than 80.So, that's invalid.Therefore, the only valid cases are:1. Sides are x, 4x, and 20, with x = 6, perimeter = 50.2. Sides are 5, 20, and 20, perimeter = 45.So, the maximum perimeter is 50.Wait, but earlier I thought about sides being 20, 80, and x, but that leads to the third side being x, not 20, which contradicts the problem statement.So, I think the correct interpretation is that the sides are x, 4x, and 20, with 20 being the third side, not necessarily four times any of the others.Therefore, the maximum perimeter is 50.But let me double-check.If I take x = 6, sides are 6, 24, 20.Perimeter = 50.If I try x = 7, sides would be 7, 28, 20.Check triangle inequalities:1. 7 + 20 > 28? 27 > 28? No, that's false.So, x = 7 doesn't work.x = 5:Sides would be 5, 20, 20.Perimeter = 45.x = 4:Sides would be 4, 16, 20.Check inequalities:1. 4 + 16 > 20? 20 > 20? No.So, x = 4 doesn't work.x = 3:Sides would be 3, 12, 20.1. 3 + 12 > 20? 15 > 20? No.x = 2:2, 8, 20.1. 2 + 8 > 20? 10 > 20? No.x = 1:1, 4, 20.1. 1 + 4 > 20? 5 > 20? No.So, only x = 5 and x = 6 work.x = 5 gives perimeter 45, x = 6 gives perimeter 50.Therefore, the maximum perimeter is 50.So, I think the answer is 50.

answer：First, I need to evaluate the square root of 16. The square root of 16 is 4 because 4 multiplied by itself equals 16.Next, I need to calculate 2 squared. Squaring 2 means multiplying 2 by itself, which equals 4.Finally, I subtract the result of 2 squared from the square root of 16: 4 minus 4 equals 0.Therefore, the final answer is 0.

answer：First, I need to identify the given information and what is being asked. The problem states that the simple interest on a certain sum of money for 5 years is one-fifth of the sum. I need to find the rate of interest per annum.I'll start by recalling the formula for simple interest:[ text{SI} = frac{P times R times T}{100} ]where:- ( text{SI} ) is the simple interest,- ( P ) is the principal amount,- ( R ) is the rate of interest per annum,- ( T ) is the time in years.According to the problem, the simple interest for 5 years is one-fifth of the principal:[ text{SI} = frac{P}{5} ]Substituting the values into the simple interest formula:[ frac{P}{5} = frac{P times R times 5}{100} ]To solve for ( R ), I'll first simplify the equation by canceling out ( P ) from both sides:[ frac{1}{5} = frac{R times 5}{100} ]Next, I'll multiply both sides by 100 to eliminate the denominator:[ 20 = 5R ]Finally, I'll divide both sides by 5 to find the value of ( R ):[ R = 4 ]Therefore, the rate of interest per annum is 4%.

answer：Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. The problem gives me two pieces of information: first, that the product of sides a and c is equal to one fourth of b squared, so ac = (1/4)b². Second, it tells me that the sum of the sines of angles A and C is equal to t times the sine of angle B, so sin A + sin C = t sin B. Also, angle B is acute, which means it's less than 90 degrees.I need to find the range of the real number t. Hmm, okay. Let me think about how to approach this.First, I remember the Law of Sines, which states that in any triangle, the ratio of a side to the sine of its opposite angle is constant. So, that gives me a/sin A = b/sin B = c/sin C. Maybe I can use this to relate the sides and the sines of the angles.Given that ac = (1/4)b², maybe I can express a and c in terms of b and the sines of their opposite angles. Let me denote the common ratio from the Law of Sines as 2R, where R is the radius of the circumscribed circle. So, a = 2R sin A, b = 2R sin B, and c = 2R sin C.Substituting these into the equation ac = (1/4)b², we get:(2R sin A)(2R sin C) = (1/4)(2R sin B)²Simplify this:4R² sin A sin C = (1/4)(4R² sin² B)Simplify both sides:4R² sin A sin C = R² sin² BDivide both sides by R²:4 sin A sin C = sin² BSo, 4 sin A sin C = sin² B. Hmm, that's one equation.Now, the other given is sin A + sin C = t sin B. So, I have two equations:1) 4 sin A sin C = sin² B2) sin A + sin C = t sin BI need to find t, so maybe I can express sin A and sin C in terms of t and sin B, and then substitute into the first equation.Let me denote S = sin A + sin C = t sin B, and P = sin A sin C. From the first equation, 4P = sin² B, so P = (1/4) sin² B.Now, I can think of sin A and sin C as the roots of a quadratic equation. Let me consider x² - Sx + P = 0, where x represents sin A or sin C.So, substituting S and P:x² - (t sin B) x + (1/4 sin² B) = 0For this quadratic to have real roots, the discriminant must be non-negative. The discriminant D is:D = (t sin B)^2 - 4 * 1 * (1/4 sin² B) = t² sin² B - sin² BSo, D = sin² B (t² - 1) ≥ 0Since sin² B is always non-negative (because it's a square), the discriminant is non-negative when t² - 1 ≥ 0, which implies t² ≥ 1, so |t| ≥ 1.But since angle B is acute, sin B is positive, and sin A and sin C are also positive because angles A and C are between 0 and 180 degrees in a triangle. Therefore, sin A + sin C is positive, so t must be positive. Hence, t ≥ 1.But wait, that's just from the discriminant condition. I need more constraints because the range of t can't be just t ≥ 1. There must be an upper limit as well.Let me think about the triangle's angle sum. In triangle ABC, angles A + B + C = 180 degrees. Since angle B is acute, it's less than 90 degrees, so angles A and C must add up to more than 90 degrees.Also, from the Law of Sines, a/sin A = b/sin B = c/sin C. So, if I denote this common ratio as 2R, then a = 2R sin A, b = 2R sin B, c = 2R sin C.Given that ac = (1/4) b², substituting:(2R sin A)(2R sin C) = (1/4)(2R sin B)^2Which simplifies to:4R² sin A sin C = (1/4)(4R² sin² B)So, 4 sin A sin C = sin² B, as before.So, sin A sin C = (1/4) sin² B.Also, sin A + sin C = t sin B.Let me denote sin A = x and sin C = y. So, x + y = t sin B and xy = (1/4) sin² B.From these, I can write that x and y are roots of the quadratic equation:z² - (t sin B) z + (1/4 sin² B) = 0Which is the same as before.We already considered the discriminant, but perhaps I can use another approach. Maybe express sin A and sin C in terms of t and sin B, and then use the fact that in a triangle, the sum of angles is 180 degrees.Alternatively, maybe use the Law of Cosines. Let me try that.From the Law of Cosines, for angle B:b² = a² + c² - 2ac cos BWe know that ac = (1/4) b², so substitute:b² = a² + c² - 2*(1/4 b²)*cos BSimplify:b² = a² + c² - (1/2) b² cos BBring the (1/2) b² cos B to the left:b² + (1/2) b² cos B = a² + c²Factor b²:b² (1 + (1/2) cos B) = a² + c²But from the Law of Sines, a = 2R sin A, c = 2R sin C, so a² + c² = 4R² (sin² A + sin² C)Similarly, b² = 4R² sin² BSo, substituting back:4R² sin² B (1 + (1/2) cos B) = 4R² (sin² A + sin² C)Divide both sides by 4R²:sin² B (1 + (1/2) cos B) = sin² A + sin² CSo, sin² A + sin² C = sin² B (1 + (1/2) cos B)But we also know that sin A + sin C = t sin B. Let me square both sides:(sin A + sin C)^2 = t² sin² BWhich gives:sin² A + 2 sin A sin C + sin² C = t² sin² BBut from earlier, sin² A + sin² C = sin² B (1 + (1/2) cos B), and sin A sin C = (1/4) sin² B.So, substituting into the squared equation:sin² B (1 + (1/2) cos B) + 2*(1/4 sin² B) = t² sin² BSimplify:sin² B (1 + (1/2) cos B) + (1/2) sin² B = t² sin² BCombine like terms:sin² B [1 + (1/2) cos B + (1/2)] = t² sin² BSimplify inside the brackets:1 + (1/2) + (1/2) cos B = (3/2) + (1/2) cos BSo, sin² B [ (3/2) + (1/2) cos B ] = t² sin² BDivide both sides by sin² B (assuming sin B ≠ 0, which it isn't since B is acute):(3/2) + (1/2) cos B = t²So, t² = (3/2) + (1/2) cos BSince angle B is acute, cos B is positive and less than 1. So, cos B ∈ (0, 1). Therefore, (1/2) cos B ∈ (0, 1/2). Thus, t² ∈ (3/2, 3/2 + 1/2) = (3/2, 2).So, t² is between 3/2 and 2, which means t is between sqrt(3/2) and sqrt(2). Since t is positive, as established earlier, the range of t is (sqrt(6)/2, sqrt(2)).Wait, sqrt(3/2) is equal to sqrt(6)/2, right? Because sqrt(3/2) = sqrt(6)/2. Yes, because sqrt(3)/sqrt(2) = sqrt(6)/2.So, t is between sqrt(6)/2 and sqrt(2).Therefore, the range of t is (sqrt(6)/2, sqrt(2)).I think that makes sense. Let me just recap to make sure I didn't make any mistakes.1. Started with given ac = (1/4) b² and sin A + sin C = t sin B.2. Used Law of Sines to express a, b, c in terms of sines of angles and a common ratio.3. Substituted into ac = (1/4) b² to get 4 sin A sin C = sin² B.4. Let sin A + sin C = t sin B and sin A sin C = (1/4) sin² B.5. Formed quadratic equation with roots sin A and sin C, found discriminant condition leading to t ≥ 1.6. Then used Law of Cosines to relate sides and angles, leading to an expression for t² in terms of cos B.7. Since B is acute, cos B is between 0 and 1, so t² is between 3/2 and 2, hence t is between sqrt(6)/2 and sqrt(2).Seems solid. I don't see any errors in the steps. So, the range of t is (sqrt(6)/2, sqrt(2)).**Final Answer**The range of the real number ( t ) is boxed{left( dfrac{sqrt{6}}{2}, sqrt{2} right)}.

answer：Okay, so I have this problem about a farmer's right-angled triangular farm. The legs are 3 and 4 units long. At the right-angle corner, there's an unplanted square area S. The shortest distance from this square to the hypotenuse is 2 units. I need to find the ratio of the planted area to the total area of the farm. The options are given, so I can check my answer against them.First, let me visualize the farm. It's a right-angled triangle with legs 3 and 4. So, the right angle is at the origin, one leg along the x-axis from (0,0) to (4,0), and the other leg along the y-axis from (0,0) to (0,3). The hypotenuse connects (4,0) to (0,3). Now, there's a square at the right-angle corner, so that square is in the corner where the two legs meet. The square has side length s, which I need to find. The square is unplanted, so the planted area will be the total area of the triangle minus the area of the square.The key piece of information is that the shortest distance from the square to the hypotenuse is 2 units. So, I need to relate the side length of the square to this distance.Let me recall the formula for the distance from a point to a line. The distance from a point (x0, y0) to the line ax + by + c = 0 is |ax0 + by0 + c| / sqrt(a^2 + b^2).First, I need the equation of the hypotenuse. The hypotenuse connects (4,0) to (0,3). Let me find its equation.The slope of the hypotenuse is (3 - 0)/(0 - 4) = -3/4. So, the equation in point-slope form is y - 0 = (-3/4)(x - 4), which simplifies to y = (-3/4)x + 3. To write it in standard form, I can rearrange it: (3/4)x + y - 3 = 0. Multiplying both sides by 4 to eliminate the fraction: 3x + 4y - 12 = 0.So, the equation of the hypotenuse is 3x + 4y - 12 = 0.Now, the square is at the origin, so the closest point on the square to the hypotenuse would be one of its corners. Wait, actually, the square is at the right-angle corner, so its sides are along the x and y axes. The square extends from (0,0) to (s,0) along the x-axis and from (0,0) to (0,s) along the y-axis. So, the square itself is from (0,0) to (s,s). The closest point on the square to the hypotenuse would be the corner (s,s), right? Because as we move away from the origin, the point (s,s) is the farthest corner of the square from the origin, and it's the closest point on the square to the hypotenuse.Wait, actually, is that correct? The closest point from the square to the hypotenuse might not necessarily be the corner. It could be somewhere on the edge. Hmm, maybe I need to think about this more carefully.Alternatively, perhaps the distance from the square to the hypotenuse is the same as the distance from the point (s,s) to the hypotenuse. Since the square is in the corner, the point (s,s) is the farthest point on the square from the origin, and it's the closest point on the square to the hypotenuse. So, maybe the distance from (s,s) to the hypotenuse is 2.Let me test this idea. If I compute the distance from (s,s) to the hypotenuse, which is 3x + 4y - 12 = 0, using the distance formula:Distance = |3s + 4s - 12| / sqrt(3^2 + 4^2) = |7s - 12| / 5.This distance is given as 2. So, |7s - 12| / 5 = 2.Solving for s:|7s - 12| = 10So, 7s - 12 = 10 or 7s - 12 = -10Case 1: 7s - 12 = 10 => 7s = 22 => s = 22/7 ≈ 3.14Case 2: 7s - 12 = -10 => 7s = 2 => s = 2/7 ≈ 0.2857Now, considering the triangle has legs of 3 and 4, the square cannot have a side longer than 3 or 4. So, s = 22/7 ≈ 3.14 is longer than 3, which is not possible because the leg along the y-axis is only 3 units. Therefore, s = 22/7 is invalid. So, the correct value is s = 2/7.Wait, but 2/7 is approximately 0.2857, which is less than both 3 and 4, so that's feasible.So, the side length of the square is 2/7.Therefore, the area of the square is (2/7)^2 = 4/49.Now, the total area of the triangle is (3*4)/2 = 6.So, the planted area is the total area minus the square area: 6 - 4/49.Calculating that: 6 is 294/49, so 294/49 - 4/49 = 290/49.Therefore, the ratio of planted area to total area is (290/49)/6 = (290/49)*(1/6) = 290/294 = 145/147.Wait, that's one of the options, option D: 145/147.But let me double-check my reasoning because I might have made a mistake.First, I assumed that the closest point on the square to the hypotenuse is (s,s). Is that correct?Alternatively, maybe the closest point is not (s,s) but somewhere else on the square. Let me think.The square is from (0,0) to (s,0) to (s,s) to (0,s). The hypotenuse is the line 3x + 4y = 12. The distance from the square to the hypotenuse is the minimum distance from any point on the square to the hypotenuse.The square is in the corner, so the closest point on the square to the hypotenuse is indeed the point (s,s). Because as we move along the square towards (s,s), we get closer to the hypotenuse.Alternatively, we can think of the distance from the square to the hypotenuse as the distance from the hypotenuse to the origin minus the distance from the origin to the square.Wait, that might not be accurate. Let me think differently.The distance from the origin (0,0) to the hypotenuse is |3*0 + 4*0 - 12| / sqrt(3^2 + 4^2) = | -12 | / 5 = 12/5 = 2.4.So, the distance from the origin to the hypotenuse is 2.4 units.The distance from the square to the hypotenuse is given as 2 units. So, the square is located such that the distance from the square to the hypotenuse is 2, which is less than 2.4. So, the square is somewhere between the origin and the hypotenuse.Wait, actually, the square is at the origin, so the distance from the square to the hypotenuse is the same as the distance from the origin to the hypotenuse minus the distance from the origin to the square.But I'm not sure if that's the right way to think about it.Alternatively, maybe the distance from the square to the hypotenuse is the distance from the hypotenuse to the point (s,s). So, as I calculated earlier, |7s - 12| / 5 = 2, leading to s = 2/7.But let me verify this.If s = 2/7, then the square is from (0,0) to (2/7, 0) to (2/7, 2/7) to (0, 2/7). The point (2/7, 2/7) is on the square, and the distance from this point to the hypotenuse is 2.But wait, the distance from (2/7, 2/7) to the hypotenuse is |3*(2/7) + 4*(2/7) - 12| / 5 = |6/7 + 8/7 - 12| / 5 = |14/7 - 12| / 5 = |2 - 12| / 5 = | -10 | / 5 = 10/5 = 2. So, yes, that checks out.Therefore, s = 2/7 is correct.So, the area of the square is (2/7)^2 = 4/49.Total area of the triangle is 6.Planted area is 6 - 4/49 = (294 - 4)/49 = 290/49.So, the ratio is 290/49 divided by 6, which is 290/(49*6) = 290/294.Simplify 290/294: divide numerator and denominator by 2: 145/147.So, the ratio is 145/147, which is option D.Wait, but let me think again. Is the area of the square 4/49? Because s = 2/7, so area is (2/7)^2 = 4/49. Yes, that's correct.Total area is 6, which is 294/49. So, 294/49 - 4/49 = 290/49. Then, 290/49 divided by 6 is 290/(49*6) = 290/294 = 145/147. Yes, that seems correct.Alternatively, maybe I should consider the distance from the square to the hypotenuse differently. Perhaps the square is such that the distance from its side to the hypotenuse is 2, not from a point.Wait, the problem says the shortest distance from area S to the hypotenuse is 2. So, the shortest distance from the square to the hypotenuse is 2. That would be the perpendicular distance from the square to the hypotenuse.But since the square is at the origin, the closest point on the square to the hypotenuse is indeed (s,s), as I thought earlier. So, the distance from (s,s) to the hypotenuse is 2, leading to s = 2/7.Therefore, I think my calculation is correct, and the answer is 145/147, which is option D.

answer：First, I recognize that the number is a two-digit number, which can be represented as (10a + b), where (a) is the tens digit and (b) is the units digit.According to the problem, the units digit exceeds the tens digit by 2. This gives me the equation:[b = a + 2]I'm also told that the number is 24, so:[10a + b = 24]Substituting the expression for (b) from the first equation into the second equation:[10a + (a + 2) = 24][11a + 2 = 24][11a = 22][a = 2]Now that I have the value of (a), I can find (b):[b = 2 + 2 = 4]The sum of the digits is:[a + b = 2 + 4 = 6]Finally, the product of the number and the sum of its digits is:[24 times 6 = 144]