Explanation:
Okay, so I need to find the area of triangle ACD in triangle ABC where AB is 17, AC is 8, and BC is 15. D is the foot of the altitude from C to AB. Hmm, let me visualize this first. Triangle ABC with sides AB=17, AC=8, BC=15. Let me sketch it mentally. Point A, point B, point C. AB is the base, which is 17 units long. AC is 8 units, BC is 15 units. D is where the altitude from C meets AB. So, CD is perpendicular to AB.

First, maybe I should find the coordinates of the points? Or maybe use some area formulas? Wait, since we need the area of ACD, which is part of triangle ABC. Maybe if I can find the area of ABC first, and then figure out the ratio of areas between ACD and ABC? But how?

Alternatively, since D is the foot of the altitude from C to AB, maybe I can find the length of CD first, and then find AD? Because once I have AD and CD, the area of ACD would be (AD * CD)/2. Similarly, if I can find AD, then I can compute the area.

But how do I find AD? Well, maybe using the Pythagorean theorem? If I can find the length of CD, then in triangle ACD, which is a right triangle, with right angle at D, so AC² = AD² + CD². Similarly, in triangle BCD, BC² = BD² + CD². Also, since D is on AB, AD + DB = AB = 17.

So maybe I can set up equations. Let me denote AD as x, then DB would be 17 - x. Then, from triangle ACD: AC² = AD² + CD² => 8² = x² + CD² => 64 = x² + CD². From triangle BCD: BC² = DB² + CD² => 15² = (17 - x)² + CD² => 225 = (289 - 34x + x²) + CD².

So now I have two equations:

1) 64 = x² + CD²

2) 225 = (289 - 34x + x²) + CD²

If I subtract equation 1 from equation 2, I can eliminate CD². Let me try that:

225 - 64 = [289 - 34x + x² + CD²] - [x² + CD²]

Which simplifies to:

161 = 289 - 34x

Then, solving for x:

161 - 289 = -34x

-128 = -34x

Divide both sides by -34:

x = (-128)/(-34) = 128/34. Let me simplify that. Both numerator and denominator are divisible by 2: 64/17. So x = 64/17. Therefore, AD is 64/17.

Wait, so AD is 64/17? Let me check my calculations again.

Original equations:

From ACD: 64 = x² + CD²

From BCD: 225 = (17 - x)^2 + CD²

Subtracting first equation from second:

225 - 64 = (17 - x)^2 + CD² - x² - CD²

Which is 161 = (17 - x)^2 - x²

Expanding (17 - x)^2: 289 - 34x + x²

So 161 = 289 - 34x + x² - x² = 289 - 34x

So 161 = 289 - 34x

Then, 34x = 289 - 161 = 128

Therefore, x = 128 / 34 = 64 / 17. Yes, that's correct. So AD is 64/17.

So then, once I have AD, I can find CD from equation 1: CD² = 64 - x². Let me compute that.

First, x = 64/17. So x² = (64)^2 / (17)^2. Let me compute 64 squared: 4096. 17 squared is 289. So CD² = 64 - 4096/289. Let me convert 64 to over 289: 64 = 64 * 289 / 289. Let me compute 64 * 289. Hmm, 64 * 289. Let me compute 64 * 289. Well, 289 is 17², which is 289. Let me compute 64 * 289. Let me break it down:

First, 60 * 289 = 17,340

Then 4 * 289 = 1,156

Adding them together: 17,340 + 1,156 = 18,496. Therefore, 64 = 18,496 / 289. Therefore, CD² = 18,496 / 289 - 4,096 / 289 = (18,496 - 4,096)/289 = 14,400 / 289. Therefore, CD = sqrt(14,400 / 289) = 120 / 17.

So CD is 120/17.

Therefore, the area of triangle ACD is (AD * CD)/2 = (64/17 * 120/17)/2. Let me compute that.

First, multiply 64 * 120: 64 * 120. Well, 60*120=7,200 and 4*120=480, so total is 7,200 + 480 = 7,680. Then divide by 17*17=289. Then divide by 2. So total area is 7,680 / (289 * 2) = 7,680 / 578. Let me simplify that fraction.

First, divide numerator and denominator by 2: 3,840 / 289. Let me check if 289 divides into 3,840. 289 * 13 = 3,757. Because 289*10=2,890, 289*3=867, so 2,890 + 867 = 3,757. Then 3,840 - 3,757 = 83. So 3,840 / 289 = 13 + 83/289. Wait, but maybe there's a common factor? Let me check if 83 and 289 have common factors. 289 is 17². 83 is a prime number? Let me check. 83 divided by 2, no. 3? 8+3=11, not divisible by 3. 5? No. 7? 7*11=77, 7*12=84. So yes, 83 is prime. So 83 and 289 share no common factors. Therefore, the simplified fraction is 3,840 / 289.

But wait, maybe I made a mistake in calculation? Let me check again.

Alternatively, maybe there's another way to compute the area of triangle ACD. Since we know the area of triangle ABC, maybe we can find the ratio of AD to AB, and then multiply by the area of ABC? Let me try that approach as a check.

First, let me compute the area of triangle ABC. Since we have all three sides, we can use Heron's formula.

The sides are AB=17, BC=15, AC=8. Let me check if this is a valid triangle. The sum of AC + BC = 8 + 15 = 23, which is greater than AB=17. Similarly, other combinations. So yes, valid triangle.

Compute semi-perimeter, s = (17 + 15 + 8)/2 = (40)/2 = 20.

Then area = sqrt[s(s - AB)(s - BC)(s - AC)] = sqrt[20*(20 - 17)*(20 - 15)*(20 - 8)] = sqrt[20*3*5*12]. Let me compute that.

First, 20*3=60, 5*12=60. So sqrt[60*60] = sqrt[3600] = 60. So area of ABC is 60.

So area of ABC is 60. Then, if I can find the ratio of AD to AB, then area of ACD would be (AD / AB) * area of ABC. Wait, is that correct? Because the area of ACD is proportional to the base AD if they share the same height. Since both ACD and ABC share the same height from C to AB. Wait, but ABC's area is (AB * CD)/2, and ACD's area is (AD * CD)/2. Therefore, the ratio of areas is AD / AB. Therefore, area of ACD = (AD / AB) * area ABC.

Since AD is 64/17 and AB is 17, then AD / AB = (64/17)/17 = 64 / 289. Therefore, area of ACD = (64 / 289) * 60 = (64 * 60)/289. Let me compute that.

64 * 60 = 3,840. So same as before, 3,840 / 289. Which is the same result as before. So that's a good check.

But let me check if 3,840 divided by 289 is the simplest form. Since 289 is 17², and 3,840 divided by 17? Let me check if 17 divides into 3,840. 17*200=3,400. 3,840 - 3,400 = 440. 17*25=425. 440 - 425 = 15. So 17*225= 3,825. Then 3,840 - 3,825 = 15. Therefore, 3,840 / 17 = 225 + 15/17. Therefore, 3,840 / 289 = (3,840 / 17)/17 = (225 + 15/17)/17 = 225/17 + 15/(17*17). But maybe it's better to leave it as 3,840 / 289. However, maybe we can simplify it? Wait, 3,840 and 289. Since 289 is 17², and 3,840 divided by 17 is 225.88... which is not integer. So no, the fraction is reduced.

But let me check if 3,840 and 289 have any common factors. 289 is 17². Let me check if 17 divides into 3,840. As above, 17*225 = 3,825, and 3,840 - 3,825 = 15, so remainder 15. Therefore, 17 does not divide into 3,840. Therefore, the fraction is reduced.

But maybe the problem expects a fractional answer? Or maybe I made a mistake in calculation? Let me check again.

Alternatively, maybe there's another way to compute the area of ACD. Since we know CD is 120/17 and AD is 64/17, then area is (64/17 * 120/17)/2 = (7680)/578. Let me divide numerator and denominator by 2: 3840 / 289. Which is same as before.

Alternatively, maybe I can check with coordinates. Let me try coordinate geometry. Let me place point A at (0, 0), point B at (17, 0). Then point C is somewhere in the plane. Let me find coordinates of C.

We know AC = 8, BC = 15, AB = 17. Coordinates of A(0,0), B(17,0). Let coordinates of C be (x, y). Then distance from A to C is sqrt(x² + y²) = 8, so x² + y² = 64. Distance from B to C is sqrt((x - 17)^2 + y²) = 15, so (x - 17)^2 + y² = 225.

Subtracting the first equation from the second: (x - 17)^2 + y² - x² - y² = 225 - 64 => (x² - 34x + 289) - x² = 161 => -34x + 289 = 161 => -34x = 161 - 289 = -128 => x = (-128)/(-34) = 64/17. Which matches the previous result. So x = 64/17. Then, from x² + y² = 64, we can find y.

So y² = 64 - (64/17)^2. Let me compute that.

First, 64 is 64/1, so common denominator is 17² = 289.

So y² = (64 * 289 - 64²)/289.

Factor out 64: 64*(289 - 64)/289.

Compute 289 - 64 = 225. Therefore, y² = 64 * 225 / 289. Therefore, y = sqrt(64*225)/sqrt(289) = (8*15)/17 = 120/17. Which is the same as CD. So coordinates of C are (64/17, 120/17).

Now, D is the foot of the altitude from C to AB. Since AB is on the x-axis from (0,0) to (17,0), the foot D has the same x-coordinate as C? Wait, no. Wait, the foot of the altitude from C to AB. Since AB is horizontal, the altitude from C to AB is vertical? Wait, no. Wait, AB is horizontal, so the altitude from C to AB is vertical only if AB is horizontal and the altitude is vertical. Wait, but AB is horizontal, so the altitude from C to AB is vertical? Wait, no. Wait, if AB is horizontal, then the altitude from C to AB is vertical only if AB is horizontal and the altitude is perpendicular to AB. Since AB is horizontal, its slope is 0, so the altitude is vertical, which has an undefined slope. Therefore, the foot D has the same x-coordinate as C? Wait, no. Wait, if AB is horizontal, then the altitude from C to AB is vertical, so yes, the x-coordinate of D is same as that of C? Wait, no. Wait, if AB is horizontal, then the altitude from C to AB is vertical, meaning that D has the same x-coordinate as C? Wait, no. Wait, suppose AB is along x-axis. Then, the altitude from C to AB is a vertical line? Wait, no. Wait, if AB is horizontal, then the slope of AB is 0, so the slope of altitude is undefined (vertical) only if AB is horizontal. Wait, no. Wait, the altitude is perpendicular to AB. Since AB is horizontal, the altitude is vertical. Therefore, yes, the altitude is vertical, so D has same x-coordinate as C? Wait, no. Wait, if you drop a vertical line from C to AB, which is horizontal, then yes, the foot D would have same x-coordinate as C, and y-coordinate 0. Wait, but in our case, coordinates of C are (64/17, 120/17). Therefore, the foot D would be (64/17, 0). Therefore, AD is the distance from A(0,0) to D(64/17, 0), which is 64/17, which matches our previous result. Therefore, coordinates of D are (64/17, 0).

Therefore, triangle ACD has coordinates A(0,0), C(64/17, 120/17), D(64/17, 0). So we can compute its area.

Since A is (0,0), D is (64/17, 0), and C is (64/17, 120/17). So triangle ACD is a right triangle with base AD = 64/17 and height CD = 120/17. Therefore, area is (base * height)/2 = (64/17 * 120/17)/2 = same as before, 3840/289.

Alternatively, since coordinates are known, we can use the formula for area with coordinates. The coordinates are A(0,0), D(64/17, 0), C(64/17, 120/17). Using the formula for area of a triangle with coordinates:

Area = |(x_A(y_D - y_C) + x_D(y_C - y_A) + x_C(y_A - y_D))/2|

Plugging in:

x_A = 0, y_A = 0

x_D = 64/17, y_D = 0

x_C = 64/17, y_C = 120/17

So:

Area = |0*(0 - 120/17) + (64/17)*(120/17 - 0) + (64/17)*(0 - 0)| / 2

Simplify:

= |0 + (64/17)*(120/17) + 0| / 2

= |(7680)/289| / 2

= 7680 / 578

= 3840 / 289

Same result. So that's consistent.

Alternatively, since we know the area of ABC is 60, and AD is 64/17, AB is 17, so ratio is 64/17 divided by 17 is 64/289, so area of ACD is 60*(64/289) = 3840/289.

So all methods lead to the same answer. Therefore, the area of triangle ACD is 3840/289.

But let me check if this fraction can be simplified further. As before, 289 is 17². Let me check if 17 divides into 3840. 17*200 = 3400, 3840 - 3400 = 440. 17*25 = 425, 440 - 425 = 15. So remainder 15. Therefore, 17 does not divide into 3840. Therefore, 3840/289 is in simplest terms.

But maybe the problem expects a mixed number or decimal? But since it's a competition problem, likely expects a reduced fraction. So 3840/289. Let me check if that reduces. Let me check GCD of 3840 and 289.

Find GCD(3840, 289).

Using Euclidean algorithm:

GCD(3840, 289)

Divide 3840 by 289:

289*13 = 3757, 3840 - 3757 = 83. So GCD(289, 83)

Now GCD(289, 83)

Divide 289 by 83: 83*3 = 249, 289 - 249 = 40. GCD(83, 40)

GCD(83, 40)

Divide 83 by 40: 40*2 = 80, remainder 3. GCD(40, 3)

GCD(40, 3)

Divide 40 by 3: 3*13=39, remainder 1. GCD(3,1)

GCD(3,1) is 1. Therefore, GCD is 1. Therefore, 3840/289 is reduced.

Therefore, the answer is 3840⁄289. But let me check if that's correct. Wait, but maybe I made a mistake in calculation earlier? Let me check with another approach.

Alternatively, since we found CD = 120/17, and AD = 64/17, then area is ½ * AD * CD = ½ * (64/17) * (120/17) = (64 * 120) / (2 * 17²) = 7680 / 578 = 3840 / 289. Yes, same result.

Alternatively, maybe the problem expects an integer? But given the sides are 17, 8, 15, which don't seem to form a right triangle. Wait, let me check if triangle ABC is a right triangle. Let me check if 8² + 15² equals 17²? 64 + 225 = 289. 17² is 289. Oh! Wait, 8² + 15² = 64 + 225 = 289 = 17². So triangle ABC is a right triangle with right angle at C? Wait, hold on. Wait, if AC = 8, BC = 15, AB = 17, then by Pythagoras, AC² + BC² = AB². Therefore, triangle ABC is right-angled at C. Wait, but then CD is the altitude from C to AB. But in a right-angled triangle, the altitude from the right angle to the hypotenuse has a known formula.

Wait, this seems conflicting with previous calculations. Wait, if ABC is right-angled at C, then AB is the hypotenuse, and CD is the altitude from C to AB. In a right-angled triangle, the altitude to hypotenuse is given by (AC * BC)/AB. Let me check that.

Yes, in a right-angled triangle, the length of the altitude to hypotenuse is (product of legs)/hypotenuse. So here, AC = 8, BC = 15, AB = 17. Therefore, CD = (8 * 15)/17 = 120/17. Which matches our previous result.

Also, in a right-angled triangle, the length of AD can be found by (AC²)/AB. Wait, let me recall. In a right-angled triangle, the length of the segments into which the altitude divides the hypotenuse are given by AD = (AC²)/AB and BD = (BC²)/AB.

Yes, because in a right triangle, the altitude to hypotenuse creates two similar triangles. Therefore, AD / AC = AC / AB => AD = AC² / AB. Similarly, BD = BC² / AB.

So let me check that. AD = AC² / AB = 8² / 17 = 64 / 17. Which matches our previous result. Therefore, AD = 64/17, CD = 120/17. Therefore, area of ACD is ½ * AD * CD = ½ * (64/17) * (120/17) = 3840 / 289.

But since ABC is a right-angled triangle, maybe there's a simpler way? For example, area of ABC is (AC * BC)/2 = (8 * 15)/2 = 60, which matches our Heron's formula result.

Therefore, since ABC is right-angled at C, then coordinates of C are (0, 0)? Wait, no. Wait, if ABC is right-angled at C, then coordinates can be set with C at (0, 0), A at (0, 8), and B at (15, 0). Wait, but AB would then be sqrt(15² + 8²) = 17, which matches. But earlier, we had coordinates with A at (0,0), B at (17, 0), and C at (64/17, 120/17). But that's also consistent with ABC being right-angled at C? Wait, if ABC is right-angled at C, then vectors AC and BC should be perpendicular. Let me check with coordinates.

In our coordinate system, A(0,0), B(17,0), C(64/17, 120/17). Then vector AC is (64/17, 120/17), vector BC is (64/17 - 17, 120/17 - 0) = (64/17 - 289/17, 120/17) = (-225/17, 120/17). Then the dot product of AC and BC is (64/17)(-225/17) + (120/17)(120/17) = (-64*225 + 120²)/17². Let me compute numerator: -64*225 + 14,400.

Compute 64*225: 64*200=12,800; 64*25=1,600; total 14,400. Therefore, numerator is -14,400 + 14,400 = 0. Therefore, the dot product is zero, so vectors AC and BC are perpendicular. Therefore, triangle ABC is right-angled at C.

Therefore, that confirms that ABC is right-angled at C. Therefore, CD is the altitude from right angle to hypotenuse AB. Therefore, in this case, area of ACD can be calculated as (AD * CD)/2, which we did.

Alternatively, since ABC is right-angled at C, then area of ACD can be found by similar triangles. Since triangle ACD is similar to triangle ABC? Wait, triangle ACD is similar to triangle ABC? Let me check.

In triangle ABC, right-angled at C, and D is foot of altitude from C to AB. Then triangles ABC, ACD, and CBD are all similar.

Therefore, triangle ACD ~ triangle ABC. Therefore, ratio of areas is (AD / AB)^2. Therefore, area of ACD = (AD / AB)^2 * area ABC.

AD / AB = (64/17)/17 = 64/289. Therefore, area ACD = (64/289)^2 * 60? Wait, no. Wait, if they are similar, the ratio of areas is square of ratio of sides. But if triangle ACD ~ triangle ABC, then ratio of areas is (AD / AB)^2. Wait, but earlier we had area ACD = (AD / AB) * area ABC. Which contradicts? Wait, no. Wait, if they are similar, then the ratio of areas is (AD / AB)^2. But earlier, we had area ACD = (AD / AB) * area ABC. Which is different. Therefore, there must be a mistake here.

Wait, let me think. If triangle ACD ~ triangle ABC, then the ratio of areas would be (AD / AB)^2. But according to previous calculation, area ACD = (AD / AB) * area ABC. Which is different. Therefore, which one is correct?

Wait, let me check with numbers. If ABC is right-angled at C, then area ABC is 60. If AD is 64/17, then (AD / AB) * area ABC is (64/17 / 17) * 60 = (64/289)*60 = 3840/289 ≈ 13.3.

Alternatively, if triangle ACD ~ triangle ABC, then area ratio is (AD / AB)^2. So area ACD = (AD / AB)^2 * area ABC = (64/289)^2 * 60 ≈ (0.222)^2 * 60 ≈ 0.049 * 60 ≈ 2.94. Which is different. Therefore, clearly, there is a mistake in assuming similarity.

Wait, why? Because triangle ACD is similar to triangle ABC? Let me check angles. In triangle ABC, right-angled at C. In triangle ACD, angle at A is common with triangle ABC. Angle at D is right angle. Therefore, triangle ACD is similar to triangle ABC by AA similarity. Because angle A is common, and both have a right angle. Therefore, triangle ACD ~ triangle ABC. Therefore, ratio of sides is AD / AB = AC / AC? Wait, no. Wait, in triangle ACD and triangle ABC:

Angle A is common.

Angle ACD is right angle, angle ABC is right angle? Wait, no. Wait, triangle ABC has right angle at C. Triangle ACD has right angle at D. Therefore, angle at D is right angle. Therefore, angle ACD corresponds to angle ABC? Wait, maybe I need to write down the correspondence.

In triangle ACD and triangle ABC:

Angle A is common.

Angle ADC is right angle, angle ACB is right angle. Therefore, the correspondence is ACD ~ ABC with right angles at D and C respectively. Therefore, the ratio of sides would be:

AD / AC = AC / AB.

Wait, because in similar triangles, corresponding sides are proportional. So:

AD / AC = AC / AB => AD = AC² / AB, which matches our previous result. Therefore, the ratio of areas would be (AD / AC)^2.

Wait, area of ACD / area of ABC = (AD / AC)^2.

Wait, let me check. If triangle ACD ~ triangle ABC, then the ratio of areas is (ratio of sides)^2. The ratio of sides is AD / AB? Or AD / AC?

Wait, let me think. In triangle ACD and triangle ABC:

Angle A is common.

Angle ADC = angle ACB = 90 degrees.

Therefore, triangle ACD ~ triangle ABC by AA similarity. Therefore, the correspondence is ACD ~ ABC. Therefore, side AD corresponds to side AB, side AC corresponds to side AC? Wait, no. Wait, in triangle ACD, sides are AD, AC, CD. In triangle ABC, sides are AB, AC, BC.

Wait, maybe better to write the correspondence:

In triangle ACD and triangle ABC:

Angle A is common.

Angle ADC corresponds to angle ACB (both right angles).

Therefore, triangle ACD ~ triangle ABC with correspondence A -> A, D -> C, C -> B.

Therefore, sides:

AD / AB = AC / AC = CD / BC.

Wait, that can't be. Wait, if triangle ACD ~ triangle ABC, then:

AD / AB = AC / AC = CD / BC? Wait, that would mean AD / AB = 1, which is not true. Therefore, my correspondence is wrong.

Wait, maybe triangle ACD ~ triangle ACB? Let me check.

If angle A is common, and angle ADC = angle ACB = 90 degrees. Therefore, triangle ACD ~ triangle ACB.

Yes, because angle A is common, and angle ADC = angle ACB. Therefore, triangle ACD ~ triangle ACB.

Therefore, the correspondence is ACD ~ ACB. Therefore, sides:

AD / AC = AC / AB = CD / CB.

Therefore, AD / AC = AC / AB => AD = AC² / AB, which is correct.

Therefore, the ratio of areas of ACD to ACB is (AD / AC)^2.

But triangle ACB is the same as triangle ABC. Therefore, area ACD / area ABC = (AD / AC)^2.

Therefore, area ACD = area ABC * (AD / AC)^2.

Let me compute that.

AD = 64/17, AC = 8, area ABC = 60.

Therefore, area ACD = 60 * ( (64/17)/8 )² = 60 * ( (64)/(17*8) )² = 60 * (8/17)² = 60 * 64 / 289 = 3840 / 289. Which matches our previous result.

Therefore, this confirms the area is indeed 3840/289.

So, after multiple methods and checks, I can be confident that the area of triangle ACD is 3840⁄289.

But let me check if this fraction can be simplified or if there's a calculation mistake. Wait, 3840 divided by 289. Let me check decimal value. 289*13 = 3757, 3840 - 3757 = 83. So 3840/289 ≈ 13.3.

Alternatively, since area of ABC is 60, and area of ACD is approx 13.3, which is roughly 1/4.5 of the total area. Given that AD is 64/17 ≈ 3.76, which is less than AB=17, so it's a reasonable proportion.

Alternatively, maybe the problem expects the answer as a reduced fraction, which is 3840⁄289. However, let me check if 3840 and 289 have any common factors. As before, GCD is 1, so it's reduced.

Alternatively, maybe I made a mistake in assuming ABC is right-angled? Wait, earlier we saw that AC² + BC² = 8² + 15² = 64 + 225 = 289 = 17² = AB². Therefore, ABC is right-angled at C. Therefore, that is correct.

Therefore, all steps are correct, and answer is 3840⁄289.

But let me check if that's the answer expected. Maybe the problem expects it as a mixed number or something else? But in math competitions, usually, improper fraction is acceptable, and since it's reduced, that's the answer.

Alternatively, maybe I can write it as a decimal? But 3840 divided by 289. Let me compute that.

Well, 289*13 = 3757, as before. 3840 - 3757 = 83. So 83/289 ≈ 0.287. Therefore, total is approximately 13.287. But since the problem says "Find the area", and given the exact values, fractional form is better.

Therefore, I think the answer is 3840⁄289.

But let me check once more with another approach. Since ABC is right-angled at C, coordinates of C are (0, 0), A(0, 8), B(15, 0). Then AB is from (0,8) to (15, 0). Wait, no, if right-angled at C, then coordinates can be set as C(0,0), A(0, 8), B(15, 0). Then AB is hypotenuse from (0,8) to (15, 0). Then foot of altitude from C(0,0) to AB. Wait, but in this coordinate system, the foot of the altitude from C to AB is D. Wait, but in this case, since C is already the right angle, the altitude from C to AB is the same as the vertex C? No, wait, no. Wait, in this coordinate system, C is at (0,0), and AB is the hypotenuse. The altitude from C to AB is the same as the line from C to AB, which is the same as the altitude. Wait, but in this case, since ABC is right-angled at C, the altitude from C to AB is the same as the vertex C? No, that doesn't make sense. Wait, no. Wait, in a right-angled triangle, the altitude from the right angle to hypotenuse is the same as the height.

Wait, in this coordinate system, AB is hypotenuse from A(0,8) to B(15,0). The foot of the altitude from C(0,0) to AB is point D. Let me compute coordinates of D.

Equation of AB: passing through (0,8) and (15,0). Slope of AB is (0 - 8)/(15 - 0) = -8/15. Therefore, equation of AB is y - 8 = (-8/15)(x - 0), so y = (-8/15)x + 8.

The altitude from C(0,0) to AB is perpendicular to AB. Slope of AB is -8/15, so slope of altitude is 15/8. Equation of altitude is y = (15/8)x.

Find intersection point D between AB and altitude.

Set equations equal:

(15/8)x = (-8/15)x + 8

Multiply both sides by 120 to eliminate denominators:

120*(15/8)x = 120*(-8/15)x + 120*8

Simplify:

(15*15)x = (-8*8)x + 960

225x = -64x + 960

225x + 64x = 960

289x = 960

x = 960 / 289

Then y = (15/8)x = (15/8)*(960 / 289) = (15 * 120)/289 = 1800 / 289.

Wait, but in this coordinate system, point D is (960/289, 1800/289). Then, coordinates of A are (0,8), which is (0, 2336/289). Wait, 8 = 2336/289? Let me check: 289*8 = 2312. Wait, no. Wait, 289*8 = 2312. But 2336 - 2312 = 24. So 8 is 2312/289. Therefore, coordinates of A are (0, 2312/289). Coordinates of D are (960/289, 1800/289).

Then, area of triangle ACD can be calculated using coordinates. Points A(0, 2312/289), C(0,0), D(960/289, 1800/289).

Using the formula for area with coordinates:

Area = |(x_A(y_C - y_D) + x_C(y_D - y_A) + x_D(y_A - y_C))/2|

Plugging in:

x_A = 0, y_A = 2312/289

x_C = 0, y_C = 0

x_D = 960/289, y_D = 1800/289

So:

Area = |0*(0 - 1800/289) + 0*(1800/289 - 2312/289) + (960/289)*(2312/289 - 0)| / 2

Simplify:

= |0 + 0 + (960/289)*(2312/289)| / 2

= |(960 * 2312) / (289²)| / 2

Compute numerator: 960 * 2312. This seems complex, but let me see if it relates to previous results.

But wait, earlier we had area of ACD as 3840/289. Let me check if this is equal.

Let me compute (960 * 2312) / (289²) / 2.

First, note that 289² = 289*289.

But maybe there's a relation between 960 * 2312 and 3840 * 289? Let me check:

Let me compute 960 * 2312. Let me note that 2312 = 8 * 289. Because 289 * 8 = 2312. Yes! Because 289*8 = 2312.

Therefore, 960 * 2312 = 960 * 8 * 289 = 7680 * 289.

Therefore, Area = |7680 * 289 / 289²| / 2 = |7680 / 289| / 2 = 7680 / (2 * 289) = 3840 / 289. Which matches our previous result.

Therefore, this confirms again that the area is 3840⁄289.

Therefore, after multiple methods and checks, I can be confident that the area of triangle ACD is 3840⁄289.

**Final Answer**
\boxed{\dfrac{3840}{289}}

To find the area of triangle $ ACD $, we are given triangle $ ABC $ with side lengths $ AB = 17 $, $ AC = 8 $, and $ BC = 15 $. Point $ D $ is the foot of the perpendicular from $ C $ to $ AB $, making triangle $ ACD $ a right triangle with right angle at $ D $.

---

### Step 1: Recognize the Triangle Type

We first check if triangle $ ABC $ is a right triangle:

$$
AC^2 + BC^2 = 8^2 + 15^2 = 64 + 225 = 289 = 17^2 = AB^2
$$

So, triangle $ ABC $ is a **right triangle** with the right angle at $ C $.

---

### Step 2: Use Altitude Properties in a Right Triangle

In a right triangle, the **altitude from the right angle to the hypotenuse** divides the triangle into two smaller, similar right triangles. The length of the altitude $ CD $ from the right angle $ C $ to the hypotenuse $ AB $ is given by:

$$
CD = \frac{AC \cdot BC}{AB} = \frac{8 \cdot 15}{17} = \frac{120}{17}
$$

Also, the segment $ AD $, which is the projection of $ AC $ onto $ AB $, is given by:

$$
AD = \frac{AC^2}{AB} = \frac{8^2}{17} = \frac{64}{17}
$$

---

### Step 3: Compute the Area of Triangle $ ACD $

Since triangle $ ACD $ is a right triangle with legs $ AD $ and $ CD $, its area is:

$$
\text{Area of } \triangle ACD = \frac{1}{2} \cdot AD \cdot CD = \frac{1}{2} \cdot \frac{64}{17} \cdot \frac{120}{17}
$$

$$
= \frac{1}{2} \cdot \frac{7680}{289} = \frac{3840}{289}
$$

---

### Final Answer

$$
\boxed{\dfrac{3840}{289}}
$$

[Assistant]
Explanation:
Okay, so I need to find the area of triangle ACD in triangle ABC where AB is 17, AC is 8, and BC is 15. D is the foot of the altitude from C to AB. Hmm, let me visualize this first. Triangle ABC with sides AB=17, AC=8, BC=15. Let me sketch it mentally. Point A, point B, point C. AB is the base, which is 17 units long. AC is 8 units, BC is 15 units. D is where the altitude from C meets AB. So, CD is perpendicular to AB.

First, maybe I should find the coordinates of the points? Or maybe use some area formulas? Wait, since we need the area of ACD, which is part of triangle ABC. Maybe if I can find the area of ABC first, and then figure out the ratio of areas between ACD and ABC? But how?

Alternatively, since D is the foot of the altitude from C to AB, maybe I can find the length of CD first, and then find AD? Because once I have AD and CD, the area of ACD would be (AD * CD)/2. Similarly, if I can find AD, then I can compute the area.

But how do I find AD? Well, maybe using the Pythagorean theorem? If I can find the length of CD, then in triangle ACD, which is a right triangle, with right angle at D, so AC² = AD² + CD². Similarly, in triangle BCD, BC² = BD² + CD². Also, since D is on AB, AD + DB = AB = 17.

So maybe I can set up equations. Let me denote AD as x, then DB would be 17 - x. Then, from triangle ACD: AC² = AD² + CD² => 8² = x² + CD² => 64 = x² + CD². From triangle BCD: BC² = DB² + CD² => 15² = (17 - x)² + CD² => 225 = (289 - 34x + x²) + CD².

So now I have two equations:

1) 64 = x² + CD²

2) 225 = (289 - 34x + x²) + CD²

If I subtract equation 1 from equation 2, I can eliminate CD². Let me try that:

225 - 64 = [289 - 34x + x² + CD²] - [x² + CD²]

Which simplifies to:

161 = 289 - 34x

Then, solving for x:

161 - 289 = -34x

-128 = -34x

Divide both sides by -34:

x = (-128)/(-34) = 128/34. Let me simplify that. Both numerator and denominator are divisible by 2: 64/17. So x = 64/17. Therefore, AD is 64/17.

Wait, so AD is 64/17? Let me check my calculations again.

Original equations:

From ACD: 64 = x² + CD²

From BCD: 225 = (17 - x)^2 + CD²

Subtracting first equation from second:

225 - 64 = (17 - x)^2 + CD² - x² - CD²

Which is 161 = (17 - x)^2 - x²

Expanding (17 - x)^2: 289 - 34x + x²

So 161 = 289 - 34x + x² - x² = 289 - 34x

So 161 = 289 - 34x

Then, 34x = 289 - 161 = 128

Therefore, x = 128 / 34 = 64 / 17. Yes, that's correct. So AD is 64/17.

So then, once I have AD, I can find CD from equation 1: CD² = 64 - x². Let me compute that.

First, x = 64/17. So x² = (64)^2 / (17)^2. Let me compute 64 squared: 4096. 17 squared is 289. So CD² = 64 - 4096/289. Let me convert 64 to over 289: 64 = 64 * 289 / 289. Let me compute 64 * 289. Hmm, 64 * 289. Let me compute 64 * 289. Well, 289 is 17², which is 289. Let me compute 64 * 289. Let me break it down:

First, 60 * 289 = 17,340

Then 4 * 289 = 1,156

Adding them together: 17,340 + 1,156 = 18,496. Therefore, 64 = 18,496 / 289. Therefore, CD² = 18,496 / 289 - 4,096 / 289 = (18,496 - 4,096)/289 = 14,400 / 289. Therefore, CD = sqrt(14,400 / 289) = 120 / 17.

So CD is 120/17.

Therefore, the area of triangle ACD is (AD * CD)/2 = (64/17 * 120/17)/2. Let me compute that.

First, multiply 64 * 120: 64 * 120. Well, 60*120=7,200 and 4*120=480, so total is 7,200 + 480 = 7,680. Then divide by 17*17=289. Then divide by 2. So total area is 7,680 / (289 * 2) = 7,680 / 578. Let me simplify that fraction.

First, divide numerator and denominator by 2: 3,840 / 289. Let me check if 289 divides into 3,840. 289 * 13 = 3,757. Because 289*10=2,890, 289*3=867, so 2,890 + 867 = 3,757. Then 3,840 - 3,757 = 83. So 3,840 / 289 = 13 + 83/289. Wait, but maybe there's a common factor? Let me check if 83 and 289 have common factors. 289 is 17². 83 is a prime number? Let me check. 83 divided by 2, no. 3? 8+3=11, not divisible by 3. 5? No. 7? 7*11=77, 7*12=84. So yes, 83 is prime. So 83 and 289 share no common factors. Therefore, the simplified fraction is 3,840 / 289.

But wait, maybe I made a mistake in calculation? Let me check again.

Alternatively, maybe there's another way to compute the area of triangle ACD. Since we know the area of triangle ABC, maybe we can find the ratio of AD to AB, and then multiply by the area of ABC? Let me try that approach as a check.

First, let me compute the area of triangle ABC. Since we have all three sides, we can use Heron's formula.

The sides are AB=17, BC=15, AC=8. Let me check if this is a valid triangle. The sum of AC + BC = 8 + 15 = 23, which is greater than AB=17. Similarly, other combinations. So yes, valid triangle.

Compute semi-perimeter, s = (17 + 15 + 8)/2 = (40)/2 = 20.

Then area = sqrt[s(s - AB)(s - BC)(s - AC)] = sqrt[20*(20 - 17)*(20 - 15)*(20 - 8)] = sqrt[20*3*5*12]. Let me compute that.

First, 20*3=60, 5*12=60. So sqrt[60*60] = sqrt[3600] = 60. So area of ABC is 60.

So area of ABC is 60. Then, if I can find the ratio of AD to AB, then area of ACD would be (AD / AB) * area of ABC. Wait, is that correct? Because the area of ACD is proportional to the base AD if they share the same height. Since both ACD and ABC share the same height from C to AB. Wait, but ABC's area is (AB * CD)/2, and ACD's area is (AD * CD)/2. Therefore, the ratio of areas is AD / AB. Therefore, area of ACD = (AD / AB) * area ABC.

Since AD is 64/17 and AB is 17, then AD / AB = (64/17)/17 = 64 / 289. Therefore, area of ACD = (64 / 289) * 60 = (64 * 60)/289. Let me compute that.

64 * 60 = 3,840. So same as before, 3,840 / 289. Which is the same result as before. So that's a good check.

But let me check if 3,840 divided by 289 is the simplest form. Since 289 is 17², and 3,840 divided by 17? Let me check if 17 divides into 3,840. 17*200=3,400. 3,840 - 3,400 = 440. 17*25=425. 440 - 425 = 15. So 17*225= 3,825. Then 3,840 - 3,825 = 15. Therefore, 3,840 / 17 = 225 + 15/17. Therefore, 3,840 / 289 = (3,840 / 17)/17 = (225 + 15/17)/17 = 225/17 + 15/(17*17). But maybe it's better to leave it as 3,840 / 289. However, maybe we can simplify it? Wait, 3,840 and 289. Since 289 is 17², and 3,840 divided by 17 is 225.88... which is not integer. So no, the fraction is reduced.

But let me check if 3,840 and 289 have any common factors. 289 is 17². Let me check if 17 divides into 3,840. As above, 17*225 = 3,825, and 3,840 - 3,825 = 15, so remainder 15. Therefore, 17 does not divide into 3,840. Therefore, the fraction is reduced.

But maybe the problem expects a fractional answer? Or maybe I made a mistake in calculation? Let me check again.

Alternatively, maybe there's another way to compute the area of ACD. Since we know CD is 120/17 and AD is 64/17, then area is (64/17 * 120/17)/2 = (7680)/578. Let me divide numerator and denominator by 2: 3840 / 289. Which is same as before.

Alternatively, maybe I can check with coordinates. Let me try coordinate geometry. Let me place point A at (0, 0), point B at (17, 0). Then point C is somewhere in the plane. Let me find coordinates of C.

We know AC = 8, BC = 15, AB = 17. Coordinates of A(0,0), B(17,0). Let coordinates of C be (x, y). Then distance from A to C is sqrt(x² + y²) = 8, so x² + y² = 64. Distance from B to C is sqrt((x - 17)^2 + y²) = 15, so (x - 17)^2 + y² = 225.

Subtracting the first equation from the second: (x - 17)^2 + y² - x² - y² = 225 - 64 => (x² - 34x + 289) - x² = 161 => -34x + 289 = 161 => -34x = 161 - 289 = -128 => x = (-128)/(-34) = 64/17. Which matches the previous result. So x = 64/17. Then, from x² + y² = 64, we can find y.

So y² = 64 - (64/17)^2. Let me compute that.

First, 64 is 64/1, so common denominator is 17² = 289.

So y² = (64 * 289 - 64²)/289.

Factor out 64: 64*(289 - 64)/289.

Compute 289 - 64 = 225. Therefore, y² = 64 * 225 / 289. Therefore, y = sqrt(64*225)/sqrt(289) = (8*15)/17 = 120/17. Which is the same as CD. So coordinates of C are (64/17, 120/17).

Now, D is the foot of the altitude from C to AB. Since AB is on the x-axis from (0,0) to (17,0), the foot D has the same x-coordinate as C? Wait, no. Wait, the foot of the altitude from C to AB. Since AB is horizontal, the altitude from C to AB is vertical? Wait, no. Wait, AB is horizontal, so the altitude from C to AB is vertical only if AB is horizontal and the altitude is vertical. Wait, but AB is horizontal, so the altitude from C to AB is vertical? Wait, no. Wait, if AB is horizontal, then the altitude from C to AB is vertical only if AB is horizontal and the altitude is perpendicular to AB. Since AB is horizontal, its slope is 0, so the altitude is vertical, which has an undefined slope. Therefore, the foot D has the same x-coordinate as C? Wait, no. Wait, if AB is horizontal, then the altitude from C to AB is vertical, so yes, the x-coordinate of D is same as that of C? Wait, no. Wait, if AB is horizontal, then the altitude from C to AB is vertical, meaning that D has the same x-coordinate as C? Wait, no. Wait, suppose AB is along x-axis. Then, the altitude from C to AB is a vertical line? Wait, no. Wait, if AB is horizontal, then the slope of AB is 0, so the slope of altitude is undefined (vertical) only if AB is horizontal. Wait, no. Wait, the altitude is perpendicular to AB. Since AB is horizontal, the altitude is vertical. Therefore, yes, the altitude is vertical, so D has same x-coordinate as C? Wait, no. Wait, if you drop a vertical line from C to AB, which is horizontal, then yes, the foot D would have same x-coordinate as C, and y-coordinate 0. Wait, but in our case, coordinates of C are (64/17, 120/17). Therefore, the foot D would be (64/17, 0). Therefore, AD is the distance from A(0,0) to D(64/17, 0), which is 64/17, which matches our previous result. Therefore, coordinates of D are (64/17, 0).

Therefore, triangle ACD has coordinates A(0,0), C(64/17, 120/17), D(64/17, 0). So we can compute its area.

Since A is (0,0), D is (64/17, 0), and C is (64/17, 120/17). So triangle ACD is a right triangle with base AD = 64/17 and height CD = 120/17. Therefore, area is (base * height)/2 = (64/17 * 120/17)/2 = same as before, 3840/289.

Alternatively, since coordinates are known, we can use the formula for area with coordinates. The coordinates are A(0,0), D(64/17, 0), C(64/17, 120/17). Using the formula for area of a triangle with coordinates:

Area = |(x_A(y_D - y_C) + x_D(y_C - y_A) + x_C(y_A - y_D))/2|

Plugging in:

x_A = 0, y_A = 0

x_D = 64/17, y_D = 0

x_C = 64/17, y_C = 120/17

So:

Area = |0*(0 - 120/17) + (64/17)*(120/17 - 0) + (64/17)*(0 - 0)| / 2

Simplify:

= |0 + (64/17)*(120/17) + 0| / 2

= |(7680)/289| / 2

= 7680 / 578

= 3840 / 289

Same result. So that's consistent.

Alternatively, since we know the area of ABC is 60, and AD is 64/17, AB is 17, so ratio is 64/17 divided by 17 is 64/289, so area of ACD is 60*(64/289) = 3840/289.

So all methods lead to the same answer. Therefore, the area of triangle ACD is 3840/289.

But let me check if this fraction can be simplified further. As before, 289 is 17². Let me check if 17 divides into 3840. 17*200 = 3400, 3840 - 3400 = 440. 17*25 = 425, 440 - 425 = 15. So remainder 15. Therefore, 17 does not divide into 3840. Therefore, 3840/289 is in simplest terms.

But maybe the problem expects a mixed number or decimal? But since it's a competition problem, likely expects a reduced fraction. So 3840/289. Let me check if that reduces. Let me check GCD of 3840 and 289.

Find GCD(3840, 289).

Using Euclidean algorithm:

GCD(3840, 289)

Divide 3840 by 289:

289*13 = 3757, 3840 - 3757 = 83. So GCD(289, 83)

Now GCD(289, 83)

Divide 289 by 83: 83*3 = 249, 289 - 249 = 40. GCD(83, 40)

GCD(83, 40)

Divide 83 by 40: 40*2 = 80, remainder 3. GCD(40, 3)

GCD(40, 3)

Divide 40 by 3: 3*13=39, remainder 1. GCD(3,1)

GCD(3,1) is 1. Therefore, GCD is 1. Therefore, 3840/289 is reduced.

Therefore, the answer is 3840⁄289. But let me check if that's correct. Wait, but maybe I made a mistake in calculation earlier? Let me check with another approach.

Alternatively, since we found CD = 120/17, and AD = 64/17, then area is ½ * AD * CD = ½ * (64/17) * (120/17) = (64 * 120) / (2 * 17²) = 7680 / 578 = 3840 / 289. Yes, same result.

Alternatively, maybe the problem expects an integer? But given the sides are 17, 8, 15, which don't seem to form a right triangle. Wait, let me check if triangle ABC is a right triangle. Let me check if 8² + 15² equals 17²? 64 + 225 = 289. 17² is 289. Oh! Wait, 8² + 15² = 64 + 225 = 289 = 17². So triangle ABC is a right triangle with right angle at C? Wait, hold on. Wait, if AC = 8, BC = 15, AB = 17, then by Pythagoras, AC² + BC² = AB². Therefore, triangle ABC is right-angled at C. Wait, but then CD is the altitude from C to AB. But in a right-angled triangle, the altitude from the right angle to the hypotenuse has a known formula.

Wait, this seems conflicting with previous calculations. Wait, if ABC is right-angled at C, then AB is the hypotenuse, and CD is the altitude from C to AB. In a right-angled triangle, the altitude to hypotenuse is given by (AC * BC)/AB. Let me check that.

Yes, in a right-angled triangle, the length of the altitude to hypotenuse is (product of legs)/hypotenuse. So here, AC = 8, BC = 15, AB = 17. Therefore, CD = (8 * 15)/17 = 120/17. Which matches our previous result.

Also, in a right-angled triangle, the length of AD can be found by (AC²)/AB. Wait, let me recall. In a right-angled triangle, the length of the segments into which the altitude divides the hypotenuse are given by AD = (AC²)/AB and BD = (BC²)/AB.

Yes, because in a right triangle, the altitude to hypotenuse creates two similar triangles. Therefore, AD / AC = AC / AB => AD = AC² / AB. Similarly, BD = BC² / AB.

So let me check that. AD = AC² / AB = 8² / 17 = 64 / 17. Which matches our previous result. Therefore, AD = 64/17, CD = 120/17. Therefore, area of ACD is ½ * AD * CD = ½ * (64/17) * (120/17) = 3840 / 289.

But since ABC is a right-angled triangle, maybe there's a simpler way? For example, area of ABC is (AC * BC)/2 = (8 * 15)/2 = 60, which matches our Heron's formula result.

Therefore, since ABC is right-angled at C, then coordinates of C are (0, 0)? Wait, no. Wait, if ABC is right-angled at C, then coordinates can be set with C at (0, 0), A at (0, 8), and B at (15, 0). Wait, but AB would then be sqrt(15² + 8²) = 17, which matches. But earlier, we had coordinates with A at (0,0), B at (17, 0), and C at (64/17, 120/17). But that's also consistent with ABC being right-angled at C? Wait, if ABC is right-angled at C, then vectors AC and BC should be perpendicular. Let me check with coordinates.

In our coordinate system, A(0,0), B(17,0), C(64/17, 120/17). Then vector AC is (64/17, 120/17), vector BC is (64/17 - 17, 120/17 - 0) = (64/17 - 289/17, 120/17) = (-225/17, 120/17). Then the dot product of AC and BC is (64/17)(-225/17) + (120/17)(120/17) = (-64*225 + 120²)/17². Let me compute numerator: -64*225 + 14,400.

Compute 64*225: 64*200=12,800; 64*25=1,600; total 14,400. Therefore, numerator is -14,400 + 14,400 = 0. Therefore, the dot product is zero, so vectors AC and BC are perpendicular. Therefore, triangle ABC is right-angled at C.

Therefore, that confirms that ABC is right-angled at C. Therefore, CD is the altitude from right angle to hypotenuse AB. Therefore, in this case, area of ACD can be calculated as (AD * CD)/2, which we did.

Alternatively, since ABC is right-angled at C, then area of ACD can be found by similar triangles. Since triangle ACD is similar to triangle ABC? Wait, triangle ACD is similar to triangle ABC? Let me check.

In triangle ABC, right-angled at C, and D is foot of altitude from C to AB. Then triangles ABC, ACD, and CBD are all similar.

Therefore, triangle ACD ~ triangle ABC. Therefore, ratio of areas is (AD / AB)^2. Therefore, area of ACD = (AD / AB)^2 * area ABC.

AD / AB = (64/17)/17 = 64/289. Therefore, area ACD = (64/289)^2 * 60? Wait, no. Wait, if they are similar, the ratio of areas is square of ratio of sides. But if triangle ACD ~ triangle ABC, then ratio of areas is (AD / AB)^2. Wait, but earlier we had area ACD = (AD / AB) * area ABC. Which contradicts? Wait, no. Wait, if they are similar, then the ratio of areas is (AD / AB)^2. But earlier, we had area ACD = (AD / AB) * area ABC. Which is different. Therefore, there must be a mistake here.

Wait, let me think. If triangle ACD ~ triangle ABC, then the ratio of areas would be (AD / AB)^2. But according to previous calculation, area ACD = (AD / AB) * area ABC. Which is different. Therefore, which one is correct?

Wait, let me check with numbers. If ABC is right-angled at C, then area ABC is 60. If AD is 64/17, then (AD / AB) * area ABC is (64/17 / 17) * 60 = (64/289)*60 = 3840/289 ≈ 13.3.

Alternatively, if triangle ACD ~ triangle ABC, then area ratio is (AD / AB)^2. So area ACD = (AD / AB)^2 * area ABC = (64/289)^2 * 60 ≈ (0.222)^2 * 60 ≈ 0.049 * 60 ≈ 2.94. Which is different. Therefore, clearly, there is a mistake in assuming similarity.

Wait, why? Because triangle ACD is similar to triangle ABC? Let me check angles. In triangle ABC, right-angled at C. In triangle ACD, angle at A is common with triangle ABC. Angle at D is right angle. Therefore, triangle ACD is similar to triangle ABC by AA similarity. Because angle A is common, and both have a right angle. Therefore, triangle ACD ~ triangle ABC. Therefore, ratio of sides is AD / AB = AC / AC? Wait, no. Wait, in triangle ACD and triangle ABC:

Angle A is common.

Angle ACD is right angle, angle ABC is right angle? Wait, no. Wait, triangle ABC has right angle at C. Triangle ACD has right angle at D. Therefore, angle at D is right angle. Therefore, angle ACD corresponds to angle ABC? Wait, maybe I need to write down the correspondence.

In triangle ACD and triangle ABC:

Angle A is common.

Angle ADC is right angle, angle ACB is right angle. Therefore, the correspondence is ACD ~ ABC with right angles at D and C respectively. Therefore, the ratio of sides would be:

AD / AC = AC / AB.

Wait, because in similar triangles, corresponding sides are proportional. So:

AD / AC = AC / AB => AD = AC² / AB, which matches our previous result. Therefore, the ratio of areas would be (AD / AC)^2.

Wait, area of ACD / area of ABC = (AD / AC)^2.

Wait, let me check. If triangle ACD ~ triangle ABC, then the ratio of areas is (ratio of sides)^2. The ratio of sides is AD / AB? Or AD / AC?

Wait, let me think. In triangle ACD and triangle ABC:

Angle A is common.

Angle ADC = angle ACB = 90 degrees.

Therefore, triangle ACD ~ triangle ABC by AA similarity. Therefore, the correspondence is ACD ~ ABC. Therefore, side AD corresponds to side AB, side AC corresponds to side AC? Wait, no. Wait, in triangle ACD, sides are AD, AC, CD. In triangle ABC, sides are AB, AC, BC.

Wait, maybe better to write the correspondence:

In triangle ACD and triangle ABC:

Angle A is common.

Angle ADC corresponds to angle ACB (both right angles).

Therefore, triangle ACD ~ triangle ABC with correspondence A -> A, D -> C, C -> B.

Therefore, sides:

AD / AB = AC / AC = CD / BC.

Wait, that can't be. Wait, if triangle ACD ~ triangle ABC, then:

AD / AB = AC / AC = CD / BC? Wait, that would mean AD / AB = 1, which is not true. Therefore, my correspondence is wrong.

Wait, maybe triangle ACD ~ triangle ACB? Let me check.

If angle A is common, and angle ADC = angle ACB = 90 degrees. Therefore, triangle ACD ~ triangle ACB.

Yes, because angle A is common, and angle ADC = angle ACB. Therefore, triangle ACD ~ triangle ACB.

Therefore, the correspondence is ACD ~ ACB. Therefore, sides:

AD / AC = AC / AB = CD / CB.

Therefore, AD / AC = AC / AB => AD = AC² / AB, which is correct.

Therefore, the ratio of areas of ACD to ACB is (AD / AC)^2.

But triangle ACB is the same as triangle ABC. Therefore, area ACD / area ABC = (AD / AC)^2.

Therefore, area ACD = area ABC * (AD / AC)^2.

Let me compute that.

AD = 64/17, AC = 8, area ABC = 60.

Therefore, area ACD = 60 * ( (64/17)/8 )² = 60 * ( (64)/(17*8) )² = 60 * (8/17)² = 60 * 64 / 289 = 3840 / 289. Which matches our previous result.

Therefore, this confirms the area is indeed 3840/289.

So, after multiple methods and checks, I can be confident that the area of triangle ACD is 3840⁄289.

But let me check if this fraction can be simplified or if there's a calculation mistake. Wait, 3840 divided by 289. Let me check decimal value. 289*13 = 3757, 3840 - 3757 = 83. So 3840/289 ≈ 13.3.

Alternatively, since area of ABC is 60, and area of ACD is approx 13.3, which is roughly 1/4.5 of the total area. Given that AD is 64/17 ≈ 3.76, which is less than AB=17, so it's a reasonable proportion.

Alternatively, maybe the problem expects the answer as a reduced fraction, which is 3840⁄289. However, let me check if 3840 and 289 have any common factors. As before, GCD is 1, so it's reduced.

Alternatively, maybe I made a mistake in assuming ABC is right-angled? Wait, earlier we saw that AC² + BC² = 8² + 15² = 64 + 225 = 289 = 17² = AB². Therefore, ABC is right-angled at C. Therefore, that is correct.

Therefore, all steps are correct, and answer is 3840⁄289.

But let me check if that's the answer expected. Maybe the problem expects it as a mixed number or something else? But in math competitions, usually, improper fraction is acceptable, and since it's reduced, that's the answer.

Alternatively, maybe I can write it as a decimal? But 3840 divided by 289. Let me compute that.

Well, 289*13 = 3757, as before. 3840 - 3757 = 83. So 83/289 ≈ 0.287. Therefore, total is approximately 13.287. But since the problem says "Find the area", and given the exact values, fractional form is better.

Therefore, I think the answer is 3840⁄289.

But let me check once more with another approach. Since ABC is right-angled at C, coordinates of C are (0, 0), A(0, 8), B(15, 0). Then AB is from (0,8) to (15, 0). Wait, no, if right-angled at C, then coordinates can be set as C(0,0), A(0, 8), B(15, 0). Then AB is hypotenuse from (0,8) to (15, 0). Then foot of altitude from C(0,0) to AB. Wait, but in this coordinate system, the foot of the altitude from C to AB is D. Wait, but in this case, since C is already the right angle, the altitude from C to AB is the same as the vertex C? No, wait, no. Wait, in this coordinate system, C is at (0,0), and AB is the hypotenuse. The altitude from C to AB is the same as the line from C to AB, which is the same as the altitude. Wait, but in this case, since ABC is right-angled at C, the altitude from C to AB is the same as the vertex C? No, that doesn't make sense. Wait, no. Wait, in a right-angled triangle, the altitude from the right angle to hypotenuse is the same as the height.

Wait, in this coordinate system, AB is hypotenuse from A(0,8) to B(15,0). The foot of the altitude from C(0,0) to AB is point D. Let me compute coordinates of D.

Equation of AB: passing through (0,8) and (15,0). Slope of AB is (0 - 8)/(15 - 0) = -8/15. Therefore, equation of AB is y - 8 = (-8/15)(x - 0), so y = (-8/15)x + 8.

The altitude from C(0,0) to AB is perpendicular to AB. Slope of AB is -8/15, so slope of altitude is 15/8. Equation of altitude is y = (15/8)x.

Find intersection point D between AB and altitude.

Set equations equal:

(15/8)x = (-8/15)x + 8

Multiply both sides by 120 to eliminate denominators:

120*(15/8)x = 120*(-8/15)x + 120*8

Simplify:

(15*15)x = (-8*8)x + 960

225x = -64x + 960

225x + 64x = 960

289x = 960

x = 960 / 289

Then y = (15/8)x = (15/8)*(960 / 289) = (15 * 120)/289 = 1800 / 289.

Wait, but in this coordinate system, point D is (960/289, 1800/289). Then, coordinates of A are (0,8), which is (0, 2336/289). Wait, 8 = 2336/289? Let me check: 289*8 = 2312. Wait, no. Wait, 289*8 = 2312. But 2336 - 2312 = 24. So 8 is 2312/289. Therefore, coordinates of A are (0, 2312/289). Coordinates of D are (960/289, 1800/289).

Then, area of triangle ACD can be calculated using coordinates. Points A(0, 2312/289), C(0,0), D(960/289, 1800/289).

Using the formula for area with coordinates:

Area = |(x_A(y_C - y_D) + x_C(y_D - y_A) + x_D(y_A - y_C))/2|

Plugging in:

x_A = 0, y_A = 2312/289

x_C = 0, y_C = 0

x_D = 960/289, y_D = 1800/289

So:

Area = |0*(0 - 1800/289) + 0*(1800/289 - 2312/289) + (960/289)*(2312/289 - 0)| / 2

Simplify:

= |0 + 0 + (960/289)*(2312/289)| / 2

= |(960 * 2312) / (289²)| / 2

Compute numerator: 960 * 2312. This seems complex, but let me see if it relates to previous results.

But wait, earlier we had area of ACD as 3840/289. Let me check if this is equal.

Let me compute (960 * 2312) / (289²) / 2.

First, note that 289² = 289*289.

But maybe there's a relation between 960 * 2312 and 3840 * 289? Let me check:

Let me compute 960 * 2312. Let me note that 2312 = 8 * 289. Because 289 * 8 = 2312. Yes! Because 289*8 = 2312.

Therefore, 960 * 2312 = 960 * 8 * 289 = 7680 * 289.

Therefore, Area = |7680 * 289 / 289²| / 2 = |7680 / 289| / 2 = 7680 / (2 * 289) = 3840 / 289. Which matches our previous result.

Therefore, this confirms again that the area is 3840⁄289.

Therefore, after multiple methods and checks, I can be confident that the area of triangle ACD is 3840⁄289.

**Final Answer**
\boxed{\dfrac{3840}{289}}

To find the area of triangle $ ACD $, we are given triangle $ ABC $ with side lengths $ AB = 17 $, $ AC = 8 $, and $ BC = 15 $. Point $ D $ is the foot of the perpendicular from $ C $ to $ AB $, making triangle $ ACD $ a right triangle with right angle at $ D $.

---

### Step 1: Recognize the Triangle Type

We first check if triangle $ ABC $ is a right triangle:

$$
AC^2 + BC^2 = 8^2 + 15^2 = 64 + 225 = 289 = 17^2 = AB^2
$$

So, triangle $ ABC $ is a **right triangle** with the right angle at $ C $.

---

### Step 2: Use Altitude Properties in a Right Triangle

In a right triangle, the **altitude from the right angle to the hypotenuse** divides the triangle into two smaller, similar right triangles. The length of the altitude $ CD $ from the right angle $ C $ to the hypotenuse $ AB $ is given by:

$$
CD = \frac{AC \cdot BC}{AB} = \frac{8 \cdot 15}{17} = \frac{120}{17}
$$

Also, the segment $ AD $, which is the projection of $ AC $ onto $ AB $, is given by:

$$
AD = \frac{AC^2}{AB} = \frac{8^2}{17} = \frac{64}{17}
$$

---

### Step 3: Compute the Area of Triangle $ ACD $

Since triangle $ ACD $ is a right triangle with legs $ AD $ and $ CD $, its area is:

$$
\text{Area of } \triangle ACD = \frac{1}{2} \cdot AD \cdot CD = \frac{1}{2} \cdot \frac{64}{17} \cdot \frac{120}{17}
$$

$$
= \frac{1}{2} \cdot \frac{7680}{289} = \frac{3840}{289}
$$

---

### Final Answer

$$
\boxed{\dfrac{3840}{289}}
$$

Solution:
By Pythagoras, $\angle C = 90^\circ$.  Triangles $ACD$ and $ABC$ are similar, so \[CD = BC \cdot \frac{AC}{AB} = 15 \cdot \frac{8}{17} = \frac{120}{17},\]and \[AD = AC \cdot \frac{AC}{AB} = 8 \cdot \frac{8}{17} = \frac{64}{17}.\][asy]
unitsize(0.4 cm);

pair A, B, C, D;

A = (0,8);
B = (15,0);
C = (0,0);
D = (C + reflect(A,B)*(C))/2;

draw(A--B--C--cycle);
draw(C--D);

label("$A$", A, NW);
label("$B$", B, SE);
label("$C$", C, SW);
label("$D$", D, NE);
[/asy]

Therefore, the area of triangle $ACD$ is \[\frac{1}{2} \cdot AD \cdot CD = \frac{1}{2} \cdot \frac{64}{17} \cdot \frac{120}{17} = \boxed{\frac{3840}{289}}.\]


Explanation:
Okay, so I need to figure out how many zeros are at the end of 42 factorial. Hmm, let me think. I remember that trailing zeros in a factorial come from factors of 10, right? And since 10 is 2 times 5, each pair of 2 and 5 in the prime factors of 42! will contribute to a trailing zero. But wait, there are usually more 2s than 5s in a factorial, so the number of trailing zeros should depend on the number of 5s. Is that correct?

Let me verify. For example, take 5! which is 120. There's one trailing zero. Let's see how many 5s and 2s there are. 5! = 5×4×3×2×1. The factors of 5 here are just one (from the 5 itself), and factors of 2 are more: 4 contributes two 2s, 2 contributes one, so total of three 2s. So the number of 5s is one, which is less than the number of 2s, so the number of trailing zeros is one. That matches. So yeah, the number of trailing zeros is determined by the number of 5s in the prime factorization of 42!.

So, how do I calculate the number of 5s in 42!? I think the method is to divide 42 by 5, then divide by 25, then 125, etc., until the division results in zero, and then add up all those quotients. Let me recall why that works. Because every multiple of 5 contributes at least one 5, but numbers like 25, 50, etc., contribute an extra 5 because they have two 5s in their prime factors. Similarly, 125 would contribute three 5s, but since 42 is less than 125, we don't have to worry about that here.

So let me try that. First, divide 42 by 5. 42 divided by 5 is 8.4, but we take the floor of that, so 8. Then divide 42 by 25. 25 goes into 42 once, with a remainder of 17, so that's 1. Then divide 42 by 125. 125 is larger than 42, so that's 0. So adding up 8 + 1 + 0 = 9. Therefore, there are 9 trailing zeros in 42!?

Wait, let me check with a smaller number to make sure I didn't mess up the method. Let's take 10! which is 3628800. There are two trailing zeros. Let's apply the same method. 10 divided by 5 is 2, divided by 25 is 0, so total is 2. Correct. Another example: 25! How many zeros? 25/5 = 5, 25/25 = 1, 25/125 = 0. Total is 5 + 1 = 6. Let me check 25! The actual number of trailing zeros is 6, right? Yes, because 25! has factors of 5 from 5, 10, 15, 20, 25. Each of these contributes at least one 5, and 25 contributes an extra one. So total 5 + 1 = 6. So the method works here.

So applying the same logic to 42!:

Number of 5s = floor(42/5) + floor(42/25) + floor(42/125) + ...

Which is 8 + 1 + 0 = 9. Therefore, the number of trailing zeros should be 9. But let me verify with another approach.

Alternatively, maybe I can compute 42! and count the zeros? But that's impractical because 42! is a huge number. However, maybe I can compute the exponent of 5 in 42! by considering how many multiples of 5 are there between 1 and 42, then how many multiples of 25, etc. Which is exactly what I did before. So that seems correct.

Wait, let me think again. For each multiple of 5, we have one 5. But numbers like 25, 50, etc., have two 5s. So for 42, the multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40. That's 8 numbers. Each contributes at least one 5. Then among these, the numbers that are multiples of 25 (which is 5^2) are 25. So 25 contributes an extra 5. So total number of 5s is 8 + 1 = 9. Therefore, the number of 5s is 9, and since there are more than enough 2s, the number of trailing zeros is 9.

Therefore, the answer should be 9. But let me check with another example. Let's take 30! How many zeros? Using the formula: floor(30/5) + floor(30/25) = 6 + 1 = 7. Let me check 30! The actual value is 265252859812191058636308480000000. Wait, that's 30! Let me count the zeros. The number ends with seven zeros. Yes, so that works. So 30! has 7 zeros. So the formula works here.

Therefore, applying the same logic, 42! should have 9 zeros. Wait, but let me check with another way. Let me compute floor(42/5) = 8. Then floor(42/25) = 1. Then floor(42/125) = 0. So total is 8 + 1 = 9. So that's consistent.

Alternatively, if I think about how many times 5 appears in the factors:

Numbers from 1 to 42:

Number of multiples of 5: 42 divided by 5 is 8.4, so 8 numbers. Each contributes at least one 5. Then numbers that are multiples of 25 (which is 5^2) contribute an additional 5. How many multiples of 25 are there in 1-42? 25 and 50, but 50 is beyond 42, so only 25. So that's one more. Then multiples of 125 (5^3) would contribute another 5, but 125 is beyond 42, so no. Therefore, total number of 5s is 8 + 1 = 9.

Therefore, the number of trailing zeros is 9.

But just to be thorough, let me check with another example. Let's take 15!. How many zeros? Using formula: floor(15/5) + floor(15/25) = 3 + 0 = 3. Let me check 15! = 1307674368000. Ends with three zeros. Correct. So formula works here.

Another example: 20! floor(20/5) + floor(20/25) = 4 + 0 = 4. Actual value of 20! is 2432902008176640000, which ends with four zeros. Correct.

So, seems like the formula is working. Therefore, applying it to 42! gives 9.

But just to make sure, let me think if there's any chance that I made a mistake in calculation. For example, floor(42/5) is 8.4, floor is 8. Then floor(42/25) is 1.7, floor is 1. Then floor(42/125) is 0.336, floor is 0. So adding them up, 8 + 1 + 0 = 9.

Alternatively, if I think of 42 divided by 5: 5*8=40, so 42-40=2, so 8. Then 25*1=25, 42-25=17, so 1. Then 125 is too big. So yes, 8 + 1 = 9.

Therefore, I think the answer is 9.

But just to be 100% sure, maybe I can check with another approach. Let me think. The number of trailing zeros is equal to the minimum of the exponents of 2 and 5 in the prime factorization. But since there are more 2s than 5s, it's determined by the number of 5s.

But just for thoroughness, let me check how many 2s are there in 42! So, number of 2s is floor(42/2) + floor(42/4) + floor(42/8) + floor(42/16) + floor(42/32) + ...

Calculating that:

floor(42/2) = 21

floor(42/4) = 10

floor(42/8) = 5

floor(42/16) = 2

floor(42/32) = 1

floor(42/64) = 0

So total number of 2s is 21 + 10 + 5 + 2 + 1 = 39.

So there are 39 factors of 2 and 9 factors of 5. Therefore, the number of trailing zeros is indeed 9, since we can only make 9 pairs of 2 and 5.

Therefore, the answer is 9.

I think that's solid. I don't see any mistakes in my reasoning. So I can confidently say that 42! has 9 trailing zeros.

**Final Answer**
The number of trailing zeros in $42!$ is \boxed{9}.

To determine how many trailing zeros are in the number $ 42! $, we need to understand the nature of trailing zeros in factorials.

---

### Understanding Trailing Zeros

A trailing zero in a number is created when the number is divisible by 10. Since $ 10 = 2 \times 5 $, each pair of factors of 2 and 5 in the prime factorization of a number contributes to a trailing zero.

However, in the prime factorization of a factorial like $ n! $, the number of 2s is always greater than the number of 5s. Therefore, the number of trailing zeros is determined by the number of times 5 appears in the prime factorization of $ n! $.

---

### Counting the Number of 5s in $ 42! $

To count the number of 5s in the prime factorization of $ 42! $, we use the following method:

$$
\text{Number of 5s} = \left\lfloor \frac{42}{5} \right\rfloor + \left\lfloor \frac{42}{25} \right\rfloor + \left\lfloor \frac{42}{125} \right\rfloor + \cdots
$$

This works because:

- Every multiple of 5 contributes at least one 5.
- Every multiple of $ 25 = 5^2 $ contributes an additional 5.
- Every multiple of $ 125 = 5^3 $ contributes yet another 5, and so on.

Let's compute each term:

- $ \left\lfloor \frac{42}{5} \right\rfloor = 8 $
- $ \left\lfloor \frac{42}{25} \right\rfloor = 1 $
- $ \left\lfloor \frac{42}{125} \right\rfloor = 0 $

Adding these up:

$$
8 + 1 + 0 = 9
$$

---

### Final Conclusion

Since there are 9 factors of 5 in $ 42! $, and more than enough factors of 2, the number of trailing zeros in $ 42! $ is:

$$
\boxed{9}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many zeros are at the end of 42 factorial. Hmm, let me think. I remember that trailing zeros in a factorial come from factors of 10, right? And since 10 is 2 times 5, each pair of 2 and 5 in the prime factors of 42! will contribute to a trailing zero. But wait, there are usually more 2s than 5s in a factorial, so the number of trailing zeros should depend on the number of 5s. Is that correct?

Let me verify. For example, take 5! which is 120. There's one trailing zero. Let's see how many 5s and 2s there are. 5! = 5×4×3×2×1. The factors of 5 here are just one (from the 5 itself), and factors of 2 are more: 4 contributes two 2s, 2 contributes one, so total of three 2s. So the number of 5s is one, which is less than the number of 2s, so the number of trailing zeros is one. That matches. So yeah, the number of trailing zeros is determined by the number of 5s in the prime factorization of 42!.

So, how do I calculate the number of 5s in 42!? I think the method is to divide 42 by 5, then divide by 25, then 125, etc., until the division results in zero, and then add up all those quotients. Let me recall why that works. Because every multiple of 5 contributes at least one 5, but numbers like 25, 50, etc., contribute an extra 5 because they have two 5s in their prime factors. Similarly, 125 would contribute three 5s, but since 42 is less than 125, we don't have to worry about that here.

So let me try that. First, divide 42 by 5. 42 divided by 5 is 8.4, but we take the floor of that, so 8. Then divide 42 by 25. 25 goes into 42 once, with a remainder of 17, so that's 1. Then divide 42 by 125. 125 is larger than 42, so that's 0. So adding up 8 + 1 + 0 = 9. Therefore, there are 9 trailing zeros in 42!?

Wait, let me check with a smaller number to make sure I didn't mess up the method. Let's take 10! which is 3628800. There are two trailing zeros. Let's apply the same method. 10 divided by 5 is 2, divided by 25 is 0, so total is 2. Correct. Another example: 25! How many zeros? 25/5 = 5, 25/25 = 1, 25/125 = 0. Total is 5 + 1 = 6. Let me check 25! The actual number of trailing zeros is 6, right? Yes, because 25! has factors of 5 from 5, 10, 15, 20, 25. Each of these contributes at least one 5, and 25 contributes an extra one. So total 5 + 1 = 6. So the method works here.

So applying the same logic to 42!:

Number of 5s = floor(42/5) + floor(42/25) + floor(42/125) + ...

Which is 8 + 1 + 0 = 9. Therefore, the number of trailing zeros should be 9. But let me verify with another approach.

Alternatively, maybe I can compute 42! and count the zeros? But that's impractical because 42! is a huge number. However, maybe I can compute the exponent of 5 in 42! by considering how many multiples of 5 are there between 1 and 42, then how many multiples of 25, etc. Which is exactly what I did before. So that seems correct.

Wait, let me think again. For each multiple of 5, we have one 5. But numbers like 25, 50, etc., have two 5s. So for 42, the multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40. That's 8 numbers. Each contributes at least one 5. Then among these, the numbers that are multiples of 25 (which is 5^2) are 25. So 25 contributes an extra 5. So total number of 5s is 8 + 1 = 9. Therefore, the number of 5s is 9, and since there are more than enough 2s, the number of trailing zeros is 9.

Therefore, the answer should be 9. But let me check with another example. Let's take 30! How many zeros? Using the formula: floor(30/5) + floor(30/25) = 6 + 1 = 7. Let me check 30! The actual value is 265252859812191058636308480000000. Wait, that's 30! Let me count the zeros. The number ends with seven zeros. Yes, so that works. So 30! has 7 zeros. So the formula works here.

Therefore, applying the same logic, 42! should have 9 zeros. Wait, but let me check with another way. Let me compute floor(42/5) = 8. Then floor(42/25) = 1. Then floor(42/125) = 0. So total is 8 + 1 = 9. So that's consistent.

Alternatively, if I think about how many times 5 appears in the factors:

Numbers from 1 to 42:

Number of multiples of 5: 42 divided by 5 is 8.4, so 8 numbers. Each contributes at least one 5. Then numbers that are multiples of 25 (which is 5^2) contribute an additional 5. How many multiples of 25 are there in 1-42? 25 and 50, but 50 is beyond 42, so only 25. So that's one more. Then multiples of 125 (5^3) would contribute another 5, but 125 is beyond 42, so no. Therefore, total number of 5s is 8 + 1 = 9.

Therefore, the number of trailing zeros is 9.

But just to be thorough, let me check with another example. Let's take 15!. How many zeros? Using formula: floor(15/5) + floor(15/25) = 3 + 0 = 3. Let me check 15! = 1307674368000. Ends with three zeros. Correct. So formula works here.

Another example: 20! floor(20/5) + floor(20/25) = 4 + 0 = 4. Actual value of 20! is 2432902008176640000, which ends with four zeros. Correct.

So, seems like the formula is working. Therefore, applying it to 42! gives 9.

But just to make sure, let me think if there's any chance that I made a mistake in calculation. For example, floor(42/5) is 8.4, floor is 8. Then floor(42/25) is 1.7, floor is 1. Then floor(42/125) is 0.336, floor is 0. So adding them up, 8 + 1 + 0 = 9.

Alternatively, if I think of 42 divided by 5: 5*8=40, so 42-40=2, so 8. Then 25*1=25, 42-25=17, so 1. Then 125 is too big. So yes, 8 + 1 = 9.

Therefore, I think the answer is 9.

But just to be 100% sure, maybe I can check with another approach. Let me think. The number of trailing zeros is equal to the minimum of the exponents of 2 and 5 in the prime factorization. But since there are more 2s than 5s, it's determined by the number of 5s.

But just for thoroughness, let me check how many 2s are there in 42! So, number of 2s is floor(42/2) + floor(42/4) + floor(42/8) + floor(42/16) + floor(42/32) + ...

Calculating that:

floor(42/2) = 21

floor(42/4) = 10

floor(42/8) = 5

floor(42/16) = 2

floor(42/32) = 1

floor(42/64) = 0

So total number of 2s is 21 + 10 + 5 + 2 + 1 = 39.

So there are 39 factors of 2 and 9 factors of 5. Therefore, the number of trailing zeros is indeed 9, since we can only make 9 pairs of 2 and 5.

Therefore, the answer is 9.

I think that's solid. I don't see any mistakes in my reasoning. So I can confidently say that 42! has 9 trailing zeros.

**Final Answer**
The number of trailing zeros in $42!$ is \boxed{9}.

To determine how many trailing zeros are in the number $ 42! $, we need to understand the nature of trailing zeros in factorials.

---

### Understanding Trailing Zeros

A trailing zero in a number is created when the number is divisible by 10. Since $ 10 = 2 \times 5 $, each pair of factors of 2 and 5 in the prime factorization of a number contributes to a trailing zero.

However, in the prime factorization of a factorial like $ n! $, the number of 2s is always greater than the number of 5s. Therefore, the number of trailing zeros is determined by the number of times 5 appears in the prime factorization of $ n! $.

---

### Counting the Number of 5s in $ 42! $

To count the number of 5s in the prime factorization of $ 42! $, we use the following method:

$$
\text{Number of 5s} = \left\lfloor \frac{42}{5} \right\rfloor + \left\lfloor \frac{42}{25} \right\rfloor + \left\lfloor \frac{42}{125} \right\rfloor + \cdots
$$

This works because:

- Every multiple of 5 contributes at least one 5.
- Every multiple of $ 25 = 5^2 $ contributes an additional 5.
- Every multiple of $ 125 = 5^3 $ contributes yet another 5, and so on.

Let's compute each term:

- $ \left\lfloor \frac{42}{5} \right\rfloor = 8 $
- $ \left\lfloor \frac{42}{25} \right\rfloor = 1 $
- $ \left\lfloor \frac{42}{125} \right\rfloor = 0 $

Adding these up:

$$
8 + 1 + 0 = 9
$$

---

### Final Conclusion

Since there are 9 factors of 5 in $ 42! $, and more than enough factors of 2, the number of trailing zeros in $ 42! $ is:

$$
\boxed{9}
$$

Solution:
You get a digit $0$ on the end of a number whenever it has a factor of $10$, so the question is really asking, how many $10$s are in the prime factorization of $42!$. Since $10=2\cdot5$, we need to count how many of each there are. We're going to have more $2$s than $5$s, so we actually only need to count how many times $5$ appears in the prime factorization.

Every time a number is a multiple of $5$, it adds a factor of $5$ to the prime factorization. There are $8$ multiples of $5$ between $1$ and $42$. Now look at $25$. It actually has two factors of $5$. We've already counted one of them, so now we need to count one more. This gives a total of $8+1=9$ times the factor $5$ appears, so $42!$ has $\boxed{9}$ zeroes at the end.