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 try to visualize this first.

First, let me sketch triangle ABC. Let me note down the sides: AB is 17 units, AC is 8 units, and BC is 15 units. So, point A, B, and C with those side lengths. D is the foot of the altitude from C to AB, which means CD is perpendicular to AB. So, triangle ACD is a right-angled triangle with right angle at D.

Wait, but to find the area of triangle ACD, I might need the lengths of AD and CD, or maybe some other information. Since CD is the altitude, if I can find the length of CD and AD, then I can compute the area as (AD * CD)/2. Alternatively, maybe I can find coordinates? Maybe coordinate geometry could help here.

Alternatively, since we know all the sides of triangle ABC, maybe I can compute the area of triangle ABC first using Heron's formula, and then since CD is the altitude, maybe I can find CD in terms of that area. Then, if I can find AD, I can compute the area of ACD. Let me try that approach.

First, let me check if triangle ABC is a right-angled triangle. Let me see: 8² + 15² = 64 + 225 = 289. And 17² is 289 as well. Oh! So triangle ABC is a right-angled triangle with right angle at C? Wait, because AB is the hypotenuse, since AB is 17, which is the longest side. Wait, but AC is 8 and BC is 15. So if AC² + BC² = AB², then yes, triangle ABC is right-angled at C. Wait, but hold on: AC is from A to C, BC is from B to C. So if angle C is the right angle, then AB would be the hypotenuse. Yes, so angle C is 90 degrees. Wait, but the problem says D is the foot of the altitude from C to AB. Wait, but if angle C is already a right angle, then the altitude from C to AB would be the same as the leg of the triangle. Wait, but in a right-angled triangle, the altitude from the right angle to the hypotenuse divides the triangle into two similar triangles. Wait, so in that case, maybe CD is the altitude from C to AB. But if angle C is already right, then CD is the altitude. Wait, but in that case, the area of ABC is (AC * BC)/2 = (8 * 15)/2 = 60. Alternatively, using AB as base, the area would be (AB * CD)/2. Therefore, 60 = (17 * CD)/2 => CD = (60 * 2)/17 = 120/17 ≈ 7.0588.

But then, how do I find AD? Since D is the foot of the altitude from C to AB. In a right-angled triangle, there is a formula for the length of the altitude to the hypotenuse, which is (product of the legs)/hypotenuse. So CD = (AC * BC)/AB = (8 * 15)/17 = 120/17. But how do I find AD?

In a right-angled triangle, the length of the segments into which the altitude divides the hypotenuse can be found using geometric mean. Specifically, AD = (AC²)/AB. Wait, is that right? Let me recall.

In a right-angled triangle, the altitude to the hypotenuse creates two smaller similar triangles. So triangle ABC is similar to triangle ACD and CBD. Therefore, ratios of corresponding sides are equal.

So, in particular, AC² = AD * AB. Therefore, AD = AC² / AB. Let me check: AC is 8, AB is 17. Therefore, AD = 64 / 17 ≈ 3.7647. Then, the area of triangle ACD would be (AD * CD)/2.

Let me compute that. AD is 64/17, CD is 120/17. Then area is (64/17 * 120/17)/2 = (7680/289)/2 = 3840/289 ≈ 13.29... But let me check if that is correct. Alternatively, maybe there's a better way to compute the area of ACD. Since triangle ACD is similar to triangle ABC? Wait, no, not necessarily. Wait, but maybe using coordinates?

Alternatively, since we know coordinates, maybe assign coordinates to points A, B, C. Let me try that approach.

Let me place point C at the origin (0, 0) because it's the right angle. Then, since AC is 8 units, and BC is 15 units, let me place point A at (0, 8) and point B at (15, 0). Wait, but then AB would be the hypotenuse. Let me check the distance between A(0,8) and B(15,0). The distance would be sqrt((15-0)^2 + (0 - 8)^2) = sqrt(225 + 64) = sqrt(289) = 17. Yes, that's correct. So coordinates:

C: (0, 0)
A: (0, 8)
B: (15, 0)

Then AB is from (0,8) to (15, 0). Then D is the foot of the altitude from C (which is (0,0)) to AB. Wait, but hold on, if C is already at (0,0), and AB is the hypotenuse. Wait, but in this coordinate system, the foot of the altitude from C to AB. Wait, but since C is already a vertex, but in this case, the altitude from C to AB is CD. But since ABC is right-angled at C, then CD is the same as the altitude. Wait, but in this coordinate system, AB is the hypotenuse, and since triangle ABC is right-angled at C, then CD is the same as the line from C to AB. Wait, but in a right-angled triangle, the altitude from the right angle to the hypotenuse is given by the formula we discussed earlier.

But maybe with coordinates, I can compute D. Let me find the coordinates of D.

The line AB goes from A(0,8) to B(15,0). Let me find the equation of line AB. The slope of AB is (0 - 8)/(15 - 0) = -8/15. Therefore, the equation of AB is y - 8 = (-8/15)(x - 0), so y = (-8/15)x + 8.

Now, the foot of the perpendicular from C(0,0) to AB. The formula for the foot of the perpendicular from a point (x0, y0) to the line ax + by + c = 0 is given by:

D(x, y) = ( (b(bx0 - ay0) - ac ) / (a² + b²), (a(-bx0 + ay0) - bc ) / (a² + b²) )

But maybe it's easier to use parametric equations or vector projections. Alternatively, since we have the equation of AB, let me write AB in standard form.

Starting from y = (-8/15)x + 8, we can rearrange it to (8/15)x + y - 8 = 0. Multiplying both sides by 15 to eliminate fractions: 8x + 15y - 120 = 0. Therefore, standard form is 8x + 15y - 120 = 0.

So, the formula for the foot of perpendicular from point C(0,0) to line AB:

The formula is:

Let me recall that the foot of perpendicular can be found by solving the system of equations: the line AB and the line perpendicular to AB passing through C.

Since AB has slope -8/15, then the perpendicular has slope 15/8. So the equation of the line perpendicular to AB passing through C(0,0) is y = (15/8)x.

Now, find intersection point D between AB and this perpendicular line.

Set y = (15/8)x into AB's equation:

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

Multiply both sides by 15*8 = 120 to eliminate denominators:

15*15x = -8*8x + 8*120

225x = -64x + 960

225x + 64x = 960

289x = 960

x = 960 / 289

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

Therefore, coordinates of D are (960/289, 1800/289).

So now, coordinates:

A is (0, 8), D is (960/289, 1800/289). So to find area of triangle ACD. Since C is (0,0), A is (0,8), D is (960/289, 1800/289).

Alternatively, since we have coordinates, we can use the formula for area of triangle using coordinates. The area can be calculated as:

Area = (1/2)| (x_A(y_C - y_D) + x_C(y_D - y_A) + x_D(y_A - y_C)) |

Plugging in the coordinates:

x_A = 0, y_A = 8

x_C = 0, y_C = 0

x_D = 960/289, y_D = 1800/289

So:

Area = (1/2)| 0*(0 - 1800/289) + 0*(1800/289 - 8) + (960/289)*(8 - 0) |

Simplify:

= (1/2)| 0 + 0 + (960/289)*8 |

= (1/2)*(7680 / 289)

= 3840 / 289

Simplify that fraction? Let me check if 3840 and 289 have a common factor. 289 is 17² = 289. Let me check if 17 divides 3840. 17*225 = 3825. 3840 - 3825 = 15. So 17*225 +15 = 3840, so no, not divisible by 17. So 3840/289 is the simplified fraction.

But let me check if this is correct. Earlier, when I tried using AD and CD, I got the same result: (AD * CD)/2. AD was 64/17 and CD was 120/17. Then, (64/17 * 120/17)/2 = (7680)/ (289 * 2) = 3840 / 289. So same answer. Therefore, this seems consistent.

Alternatively, maybe the problem expects an answer in fraction form, so 3840 over 289. Let me check if that reduces. Since 289 is 17 squared, and 3840 divided by 17 is 225.88... which is not integer, so the fraction is reduced. Therefore, 3840/289 is the answer. Let me check if that's the case.

But wait, let me verify once again, because maybe I made a mistake in coordinates. Let me check coordinates again.

So line AB: points A(0,8) and B(15,0). Equation of AB: y = -8/15 x + 8. Correct. Then the perpendicular from C(0,0) has slope 15/8. Equation y = 15/8 x. Then solving for intersection:

15/8 x = -8/15 x + 8. Multiply both sides by 120: 15*15 x = -8*8 x + 8*120 => 225x = -64x + 960 => 225x + 64x = 960 => 289x = 960 => x=960/289. Then y=15/8 * 960/289 = (15*960)/ (8*289) = (15*120)/289 = 1800/289. Correct.

Then, coordinates of D are correct. Then area of triangle ACD. Since points A(0,8), C(0,0), D(960/289, 1800/289). Alternatively, we can compute the area using base and height. Since AC is vertical from (0,0) to (0,8). The base AC is 8 units. But D is somewhere along AB. Wait, perhaps using coordinates is the safest way.

Alternatively, since we have coordinates of all three points, we can use the formula for area:

Area = ½ |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Plugging in A(0,8), C(0,0), D(960/289, 1800/289):

Area = ½ |0*(0 - 1800/289) + 0*(1800/289 - 8) + 960/289*(8 - 0)|

As before, which gives ½ * 960/289 * 8 = ½ * 7680 / 289 = 3840 / 289. So same result.

Alternatively, maybe we can compute vectors or use determinants.

Alternatively, since we have AD and CD, and triangle ACD is right-angled at D? Wait, is triangle ACD right-angled at D? Wait, D is the foot of altitude from C to AB, so CD is perpendicular to AB. But triangle ACD has a right angle at D only if AD is perpendicular to CD, which is true because CD is perpendicular to AB, and AD is along AB. So yes, triangle ACD is right-angled at D. Therefore, area is (AD * CD)/2.

Earlier, we had AD = AC² / AB = 8² /17 = 64/17, CD = 120/17. Then area is (64/17 * 120/17)/2 = 3840 / 289. So same answer.

Alternatively, since the area of ABC is 60, and area of ACD is a portion of that. Since in a right triangle, the altitude divides the triangle into two smaller triangles, ACD and CBD, which are similar to ABC.

So, area of ACD / area of ABC = (AD / AB)². Wait, because similarity ratio squared.

So area ACD = area ABC * (AD / AB)^2.

AD is 64/17, AB is 17, so AD / AB = 64/17 / 17 = 64 / 289. Then area ACD = 60 * (64 / 289) = (60 * 64)/289 = 3840 / 289. Same result.

So, all methods converge to 3840/289. Therefore, that must be the answer.

But let me check with another method. For instance, using coordinates to find the length AD and then compute area.

Coordinates of A(0,8), D(960/289, 1800/289). So AD is the distance between these two points.

Distance AD = sqrt[(960/289 - 0)^2 + (1800/289 - 8)^2]. Let me compute that.

First, compute 1800/289 - 8. 8 is 2312/289. Therefore, 1800/289 - 2312/289 = (1800 - 2312)/289 = (-512)/289.

Therefore, AD^2 = (960/289)^2 + (-512/289)^2 = [ (960² + 512²) ] / (289²).

Let me compute numerator:

960² = 921600

512² = 262144

So total is 921600 + 262144 = 1,183,744.

Therefore, AD = sqrt(1,183,744)/289. Let me compute sqrt(1,183,744).

But maybe we can factor it. Let me note that 960 and 512 have common factors. 960 = 512 + 448. Wait, maybe not helpful. Alternatively, since we know AD is supposed to be 64/17, so AD squared is (64/17)^2 = 4096 / 289.

But according to above, AD^2 = 1,183,744 / 289². Let me check if 1,183,744 equals 4096 * 289.

Compute 4096 * 289:

Well, 4096 * 289. Let me compute 4096 * 289.

First, note that 289 is 17². Also, 4096 is 2^12.

Alternatively, note that 289 * 4096 = (289 * 4000) + (289 * 96) = 1,156,000 + 27,744 = 1,183,744. Yes! So AD^2 is indeed 1,183,744 / 289² = (4096 * 289)/289² = 4096 / 289. Which is (64²)/17². Therefore, AD is 64/17. So that checks out. Therefore, AD is indeed 64/17. So then area of ACD is (AD * CD)/2 = (64/17 * 120/17)/2 = 3840/289.

Therefore, regardless of the method, I get the same answer. So I think that is correct.

Alternatively, just to make sure, let me check if 3840 divided by 289 is a simplified fraction. As mentioned earlier, 289 is 17². Let me check if 17 divides into 3840. 17*225 = 3825, and 3840 - 3825 = 15. So remainder 15, so no. Therefore, 3840/289 is reduced.

Alternatively, if I write it as a mixed number, but I think as an improper fraction, 3840/289 is acceptable.

Alternatively, maybe the problem expects a decimal? But since it's a competition-style problem, likely expects the fraction.

Therefore, I think the answer is 3840/289.

But just to confirm once more, perhaps check with vectors.

Vector approach: For triangle ACD. We can compute vectors AC and AD, then take half the magnitude of their cross product.

But AC is from A to C, but wait, coordinates:

Wait, point A is (0,8), point C is (0,0), so vector AC is C - A = (0 - 0, 0 - 8) = (0, -8).

Vector AD is D - A = (960/289 - 0, 1800/289 - 8) = (960/289, 1800/289 - 2312/289) = (960/289, -512/289).

The cross product of AC and AD is (0)(-512/289) - (-8)(960/289) = 0 + 7680/289 = 7680/289.

Then the area is half the absolute value of that, which is 3840/289. Same result.

Alright, so that seems to confirm it.

Alternatively, maybe there's a way to do this without coordinates or Heron's formula? Let me think.

Since ABC is right-angled at C, then as I mentioned earlier, the altitude CD divides AB into segments AD and DB. In a right-angled triangle, we have the geometric mean theorem: AD = AC² / AB, DB = BC² / AB, and CD = (AC * BC)/AB.

So since AC = 8, AB = 17, then AD = 8² / 17 = 64/17. Then CD = (8 * 15)/17 = 120/17. Then area of triangle ACD is (AD * CD)/2 = (64/17 * 120/17)/2 = 3840 / 289.

So this method is quicker, but relies on knowing the geometric mean theorem.

So all methods lead to the same answer. Therefore, I think that is correct.

Therefore, 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 $. The key detail is that point $ D $ is the foot of the perpendicular from $ C $ to side $ AB $, meaning that $ CD $ is the altitude from $ C $ to the hypotenuse $ AB $ in triangle $ ABC $. We're to compute the area of triangle $ ACD $.

---

### Step 1: Recognize the type of triangle
We note that triangle $ ABC $ is a **right triangle**. This can be confirmed by applying the Pythagorean Theorem:

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

Hence, triangle $ ABC $ is right-angled at $ C $.

---

### Step 2: Use properties of right triangles
In any right triangle, the **altitude from the right angle to the hypotenuse** divides the triangle into two smaller right triangles that are similar to each other and to the original triangle.

So, the altitude $ CD $ from $ C $ to hypotenuse $ AB $, and the point $ D $ on $ AB $, divides triangle $ ABC $ into triangles $ ACD $ and $ CBD $. These triangles are similar to $ ABC $ and to each other.

Key properties:

- The length of the **altitude $ CD $** from the right angle to the hypotenuse is given by:

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

- The length of the **segment $ AD $** on the hypotenuse is given by:

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

---

### Step 3: Area of triangle $ ACD $
Since triangle $ ACD $ is a right triangle with legs $ AD $ and $ CD $, its area can be found using the right triangle area formula:

$$
\text{Area} = \frac{1}{2} \cdot AD \cdot CD
$$

Substituting the known values:

$$
\text{Area} = \frac{1}{2} \cdot \frac{64}{17} \cdot \frac{120}{17} = \frac{1}{2} \cdot \frac{7680}{289} = \frac{3840}{289}
$$

---

### 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 try to visualize this first.

First, let me sketch triangle ABC. Let me note down the sides: AB is 17 units, AC is 8 units, and BC is 15 units. So, point A, B, and C with those side lengths. D is the foot of the altitude from C to AB, which means CD is perpendicular to AB. So, triangle ACD is a right-angled triangle with right angle at D.

Wait, but to find the area of triangle ACD, I might need the lengths of AD and CD, or maybe some other information. Since CD is the altitude, if I can find the length of CD and AD, then I can compute the area as (AD * CD)/2. Alternatively, maybe I can find coordinates? Maybe coordinate geometry could help here.

Alternatively, since we know all the sides of triangle ABC, maybe I can compute the area of triangle ABC first using Heron's formula, and then since CD is the altitude, maybe I can find CD in terms of that area. Then, if I can find AD, I can compute the area of ACD. Let me try that approach.

First, let me check if triangle ABC is a right-angled triangle. Let me see: 8² + 15² = 64 + 225 = 289. And 17² is 289 as well. Oh! So triangle ABC is a right-angled triangle with right angle at C? Wait, because AB is the hypotenuse, since AB is 17, which is the longest side. Wait, but AC is 8 and BC is 15. So if AC² + BC² = AB², then yes, triangle ABC is right-angled at C. Wait, but hold on: AC is from A to C, BC is from B to C. So if angle C is the right angle, then AB would be the hypotenuse. Yes, so angle C is 90 degrees. Wait, but the problem says D is the foot of the altitude from C to AB. Wait, but if angle C is already a right angle, then the altitude from C to AB would be the same as the leg of the triangle. Wait, but in a right-angled triangle, the altitude from the right angle to the hypotenuse divides the triangle into two similar triangles. Wait, so in that case, maybe CD is the altitude from C to AB. But if angle C is already right, then CD is the altitude. Wait, but in that case, the area of ABC is (AC * BC)/2 = (8 * 15)/2 = 60. Alternatively, using AB as base, the area would be (AB * CD)/2. Therefore, 60 = (17 * CD)/2 => CD = (60 * 2)/17 = 120/17 ≈ 7.0588.

But then, how do I find AD? Since D is the foot of the altitude from C to AB. In a right-angled triangle, there is a formula for the length of the altitude to the hypotenuse, which is (product of the legs)/hypotenuse. So CD = (AC * BC)/AB = (8 * 15)/17 = 120/17. But how do I find AD?

In a right-angled triangle, the length of the segments into which the altitude divides the hypotenuse can be found using geometric mean. Specifically, AD = (AC²)/AB. Wait, is that right? Let me recall.

In a right-angled triangle, the altitude to the hypotenuse creates two smaller similar triangles. So triangle ABC is similar to triangle ACD and CBD. Therefore, ratios of corresponding sides are equal.

So, in particular, AC² = AD * AB. Therefore, AD = AC² / AB. Let me check: AC is 8, AB is 17. Therefore, AD = 64 / 17 ≈ 3.7647. Then, the area of triangle ACD would be (AD * CD)/2.

Let me compute that. AD is 64/17, CD is 120/17. Then area is (64/17 * 120/17)/2 = (7680/289)/2 = 3840/289 ≈ 13.29... But let me check if that is correct. Alternatively, maybe there's a better way to compute the area of ACD. Since triangle ACD is similar to triangle ABC? Wait, no, not necessarily. Wait, but maybe using coordinates?

Alternatively, since we know coordinates, maybe assign coordinates to points A, B, C. Let me try that approach.

Let me place point C at the origin (0, 0) because it's the right angle. Then, since AC is 8 units, and BC is 15 units, let me place point A at (0, 8) and point B at (15, 0). Wait, but then AB would be the hypotenuse. Let me check the distance between A(0,8) and B(15,0). The distance would be sqrt((15-0)^2 + (0 - 8)^2) = sqrt(225 + 64) = sqrt(289) = 17. Yes, that's correct. So coordinates:

C: (0, 0)
A: (0, 8)
B: (15, 0)

Then AB is from (0,8) to (15, 0). Then D is the foot of the altitude from C (which is (0,0)) to AB. Wait, but hold on, if C is already at (0,0), and AB is the hypotenuse. Wait, but in this coordinate system, the foot of the altitude from C to AB. Wait, but since C is already a vertex, but in this case, the altitude from C to AB is CD. But since ABC is right-angled at C, then CD is the same as the altitude. Wait, but in this coordinate system, AB is the hypotenuse, and since triangle ABC is right-angled at C, then CD is the same as the line from C to AB. Wait, but in a right-angled triangle, the altitude from the right angle to the hypotenuse is given by the formula we discussed earlier.

But maybe with coordinates, I can compute D. Let me find the coordinates of D.

The line AB goes from A(0,8) to B(15,0). Let me find the equation of line AB. The slope of AB is (0 - 8)/(15 - 0) = -8/15. Therefore, the equation of AB is y - 8 = (-8/15)(x - 0), so y = (-8/15)x + 8.

Now, the foot of the perpendicular from C(0,0) to AB. The formula for the foot of the perpendicular from a point (x0, y0) to the line ax + by + c = 0 is given by:

D(x, y) = ( (b(bx0 - ay0) - ac ) / (a² + b²), (a(-bx0 + ay0) - bc ) / (a² + b²) )

But maybe it's easier to use parametric equations or vector projections. Alternatively, since we have the equation of AB, let me write AB in standard form.

Starting from y = (-8/15)x + 8, we can rearrange it to (8/15)x + y - 8 = 0. Multiplying both sides by 15 to eliminate fractions: 8x + 15y - 120 = 0. Therefore, standard form is 8x + 15y - 120 = 0.

So, the formula for the foot of perpendicular from point C(0,0) to line AB:

The formula is:

Let me recall that the foot of perpendicular can be found by solving the system of equations: the line AB and the line perpendicular to AB passing through C.

Since AB has slope -8/15, then the perpendicular has slope 15/8. So the equation of the line perpendicular to AB passing through C(0,0) is y = (15/8)x.

Now, find intersection point D between AB and this perpendicular line.

Set y = (15/8)x into AB's equation:

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

Multiply both sides by 15*8 = 120 to eliminate denominators:

15*15x = -8*8x + 8*120

225x = -64x + 960

225x + 64x = 960

289x = 960

x = 960 / 289

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

Therefore, coordinates of D are (960/289, 1800/289).

So now, coordinates:

A is (0, 8), D is (960/289, 1800/289). So to find area of triangle ACD. Since C is (0,0), A is (0,8), D is (960/289, 1800/289).

Alternatively, since we have coordinates, we can use the formula for area of triangle using coordinates. The area can be calculated as:

Area = (1/2)| (x_A(y_C - y_D) + x_C(y_D - y_A) + x_D(y_A - y_C)) |

Plugging in the coordinates:

x_A = 0, y_A = 8

x_C = 0, y_C = 0

x_D = 960/289, y_D = 1800/289

So:

Area = (1/2)| 0*(0 - 1800/289) + 0*(1800/289 - 8) + (960/289)*(8 - 0) |

Simplify:

= (1/2)| 0 + 0 + (960/289)*8 |

= (1/2)*(7680 / 289)

= 3840 / 289

Simplify that fraction? Let me check if 3840 and 289 have a common factor. 289 is 17² = 289. Let me check if 17 divides 3840. 17*225 = 3825. 3840 - 3825 = 15. So 17*225 +15 = 3840, so no, not divisible by 17. So 3840/289 is the simplified fraction.

But let me check if this is correct. Earlier, when I tried using AD and CD, I got the same result: (AD * CD)/2. AD was 64/17 and CD was 120/17. Then, (64/17 * 120/17)/2 = (7680)/ (289 * 2) = 3840 / 289. So same answer. Therefore, this seems consistent.

Alternatively, maybe the problem expects an answer in fraction form, so 3840 over 289. Let me check if that reduces. Since 289 is 17 squared, and 3840 divided by 17 is 225.88... which is not integer, so the fraction is reduced. Therefore, 3840/289 is the answer. Let me check if that's the case.

But wait, let me verify once again, because maybe I made a mistake in coordinates. Let me check coordinates again.

So line AB: points A(0,8) and B(15,0). Equation of AB: y = -8/15 x + 8. Correct. Then the perpendicular from C(0,0) has slope 15/8. Equation y = 15/8 x. Then solving for intersection:

15/8 x = -8/15 x + 8. Multiply both sides by 120: 15*15 x = -8*8 x + 8*120 => 225x = -64x + 960 => 225x + 64x = 960 => 289x = 960 => x=960/289. Then y=15/8 * 960/289 = (15*960)/ (8*289) = (15*120)/289 = 1800/289. Correct.

Then, coordinates of D are correct. Then area of triangle ACD. Since points A(0,8), C(0,0), D(960/289, 1800/289). Alternatively, we can compute the area using base and height. Since AC is vertical from (0,0) to (0,8). The base AC is 8 units. But D is somewhere along AB. Wait, perhaps using coordinates is the safest way.

Alternatively, since we have coordinates of all three points, we can use the formula for area:

Area = ½ |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Plugging in A(0,8), C(0,0), D(960/289, 1800/289):

Area = ½ |0*(0 - 1800/289) + 0*(1800/289 - 8) + 960/289*(8 - 0)|

As before, which gives ½ * 960/289 * 8 = ½ * 7680 / 289 = 3840 / 289. So same result.

Alternatively, maybe we can compute vectors or use determinants.

Alternatively, since we have AD and CD, and triangle ACD is right-angled at D? Wait, is triangle ACD right-angled at D? Wait, D is the foot of altitude from C to AB, so CD is perpendicular to AB. But triangle ACD has a right angle at D only if AD is perpendicular to CD, which is true because CD is perpendicular to AB, and AD is along AB. So yes, triangle ACD is right-angled at D. Therefore, area is (AD * CD)/2.

Earlier, we had AD = AC² / AB = 8² /17 = 64/17, CD = 120/17. Then area is (64/17 * 120/17)/2 = 3840 / 289. So same answer.

Alternatively, since the area of ABC is 60, and area of ACD is a portion of that. Since in a right triangle, the altitude divides the triangle into two smaller triangles, ACD and CBD, which are similar to ABC.

So, area of ACD / area of ABC = (AD / AB)². Wait, because similarity ratio squared.

So area ACD = area ABC * (AD / AB)^2.

AD is 64/17, AB is 17, so AD / AB = 64/17 / 17 = 64 / 289. Then area ACD = 60 * (64 / 289) = (60 * 64)/289 = 3840 / 289. Same result.

So, all methods converge to 3840/289. Therefore, that must be the answer.

But let me check with another method. For instance, using coordinates to find the length AD and then compute area.

Coordinates of A(0,8), D(960/289, 1800/289). So AD is the distance between these two points.

Distance AD = sqrt[(960/289 - 0)^2 + (1800/289 - 8)^2]. Let me compute that.

First, compute 1800/289 - 8. 8 is 2312/289. Therefore, 1800/289 - 2312/289 = (1800 - 2312)/289 = (-512)/289.

Therefore, AD^2 = (960/289)^2 + (-512/289)^2 = [ (960² + 512²) ] / (289²).

Let me compute numerator:

960² = 921600

512² = 262144

So total is 921600 + 262144 = 1,183,744.

Therefore, AD = sqrt(1,183,744)/289. Let me compute sqrt(1,183,744).

But maybe we can factor it. Let me note that 960 and 512 have common factors. 960 = 512 + 448. Wait, maybe not helpful. Alternatively, since we know AD is supposed to be 64/17, so AD squared is (64/17)^2 = 4096 / 289.

But according to above, AD^2 = 1,183,744 / 289². Let me check if 1,183,744 equals 4096 * 289.

Compute 4096 * 289:

Well, 4096 * 289. Let me compute 4096 * 289.

First, note that 289 is 17². Also, 4096 is 2^12.

Alternatively, note that 289 * 4096 = (289 * 4000) + (289 * 96) = 1,156,000 + 27,744 = 1,183,744. Yes! So AD^2 is indeed 1,183,744 / 289² = (4096 * 289)/289² = 4096 / 289. Which is (64²)/17². Therefore, AD is 64/17. So that checks out. Therefore, AD is indeed 64/17. So then area of ACD is (AD * CD)/2 = (64/17 * 120/17)/2 = 3840/289.

Therefore, regardless of the method, I get the same answer. So I think that is correct.

Alternatively, just to make sure, let me check if 3840 divided by 289 is a simplified fraction. As mentioned earlier, 289 is 17². Let me check if 17 divides into 3840. 17*225 = 3825, and 3840 - 3825 = 15. So remainder 15, so no. Therefore, 3840/289 is reduced.

Alternatively, if I write it as a mixed number, but I think as an improper fraction, 3840/289 is acceptable.

Alternatively, maybe the problem expects a decimal? But since it's a competition-style problem, likely expects the fraction.

Therefore, I think the answer is 3840/289.

But just to confirm once more, perhaps check with vectors.

Vector approach: For triangle ACD. We can compute vectors AC and AD, then take half the magnitude of their cross product.

But AC is from A to C, but wait, coordinates:

Wait, point A is (0,8), point C is (0,0), so vector AC is C - A = (0 - 0, 0 - 8) = (0, -8).

Vector AD is D - A = (960/289 - 0, 1800/289 - 8) = (960/289, 1800/289 - 2312/289) = (960/289, -512/289).

The cross product of AC and AD is (0)(-512/289) - (-8)(960/289) = 0 + 7680/289 = 7680/289.

Then the area is half the absolute value of that, which is 3840/289. Same result.

Alright, so that seems to confirm it.

Alternatively, maybe there's a way to do this without coordinates or Heron's formula? Let me think.

Since ABC is right-angled at C, then as I mentioned earlier, the altitude CD divides AB into segments AD and DB. In a right-angled triangle, we have the geometric mean theorem: AD = AC² / AB, DB = BC² / AB, and CD = (AC * BC)/AB.

So since AC = 8, AB = 17, then AD = 8² / 17 = 64/17. Then CD = (8 * 15)/17 = 120/17. Then area of triangle ACD is (AD * CD)/2 = (64/17 * 120/17)/2 = 3840 / 289.

So this method is quicker, but relies on knowing the geometric mean theorem.

So all methods lead to the same answer. Therefore, I think that is correct.

Therefore, 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 $. The key detail is that point $ D $ is the foot of the perpendicular from $ C $ to side $ AB $, meaning that $ CD $ is the altitude from $ C $ to the hypotenuse $ AB $ in triangle $ ABC $. We're to compute the area of triangle $ ACD $.

---

### Step 1: Recognize the type of triangle
We note that triangle $ ABC $ is a **right triangle**. This can be confirmed by applying the Pythagorean Theorem:

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

Hence, triangle $ ABC $ is right-angled at $ C $.

---

### Step 2: Use properties of right triangles
In any right triangle, the **altitude from the right angle to the hypotenuse** divides the triangle into two smaller right triangles that are similar to each other and to the original triangle.

So, the altitude $ CD $ from $ C $ to hypotenuse $ AB $, and the point $ D $ on $ AB $, divides triangle $ ABC $ into triangles $ ACD $ and $ CBD $. These triangles are similar to $ ABC $ and to each other.

Key properties:

- The length of the **altitude $ CD $** from the right angle to the hypotenuse is given by:

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

- The length of the **segment $ AD $** on the hypotenuse is given by:

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

---

### Step 3: Area of triangle $ ACD $
Since triangle $ ACD $ is a right triangle with legs $ AD $ and $ CD $, its area can be found using the right triangle area formula:

$$
\text{Area} = \frac{1}{2} \cdot AD \cdot CD
$$

Substituting the known values:

$$
\text{Area} = \frac{1}{2} \cdot \frac{64}{17} \cdot \frac{120}{17} = \frac{1}{2} \cdot \frac{7680}{289} = \frac{3840}{289}
$$

---

### 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 remember. I think the number of trailing zeros in a factorial depends on how many times the number 10 is a factor in that factorial. Since 10 is 2 times 5, maybe I need to count how many pairs of 2s and 5s there are in the prime factors of 42!.

But wait, I remember that there are usually more 2s than 5s in a factorial, so the number of trailing zeros should be determined by the number of 5s. Is that right? Let me think. For example, in 5! which is 120, there's one 5 and more than one 2, so one zero. Yeah, that seems to check out. So maybe I just need to count the number of 5s in the prime factors of 42!.

Alright, so how do I count the number of times 5 is a factor in the numbers from 1 to 42? I think the formula is to divide 42 by 5, then divide by 25, 125, etc., and add them up. Let me recall. For example, for 100!, it would be 100/5 + 100/25 + 100/125... which is 20 + 4 + 0 = 24. So that gives the number of 5s.

So applying that to 42! Let me compute 42 divided by 5 first. 42 divided by 5 is 8.4, but we take the floor of that, which is 8. Then divide by 25, which is 42 divided by 25. 25 times 1 is 25, 25 times 2 is 50, which is more than 42. So floor(42/25) is 1. Next is 125, which is way bigger than 42, so that's 0. So adding up 8 + 1 + 0 = 9. So does that mean there are 9 trailing zeros in 42!?

Wait, let me verify this with a smaller number. Let's take 5! again. Using this method: 5/5 = 1, 5/25 = 0, so total 1. Which matches, since 5! is 120. How about 10!? 10/5 = 2, 10/25 = 0, so total 2. But 10! is 3628800, which has two zeros. Correct. How about 25!? 25/5 = 5, 25/25 = 1, 25/125 = 0. Total 5 + 1 = 6. Let me check 25!... Well, I don't remember the exact value, but I know that 25! has 6 trailing zeros. That seems right. So the formula works here.

So applying that to 42!:

First division: floor(42/5) = 8. So numbers divisible by 5: 5, 10, 15, 20, 25, 30, 35, 40. That's 8 numbers. Each contributes at least one 5. But some numbers contribute more. For example, 25 is 5 squared, so it contributes two 5s. Similarly, 50 would contribute two, but 50 is beyond 42. So in 42!, the numbers contributing extra 5s are 25. Let me check. 25 is 5*5, so that's one extra 5. So total number of 5s is 8 + 1 = 9.

Therefore, the number of trailing zeros is 9. But wait, let me count the number of 2s just to make sure there are enough. Since there are more 2s than 5s, the number of trailing zeros is determined by the number of 5s. But maybe I should check?

Number of 2s in 42! would be 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

Next term is floor(42/64) = 0. So total number of 2s is 21 + 10 + 5 + 2 + 1 = 39.

Number of 5s is 8 + 1 = 9. So there are plenty of 2s, so the number of trailing zeros is indeed 9.

But let me check with actual computation. Maybe compute 42! and count the zeros? Although that would be tedious. Alternatively, maybe check with another method. Wait, maybe I can use another example. For instance, 20! has floor(20/5) + floor(20/25) = 4 + 0 = 4 trailing zeros. And 20! is 2432902008176640000, which has four zeros at the end. Correct.

So if 20! has 4 zeros, 25! has 6 zeros, then 42! should have 9. Let me think of another way. Let me calculate the number of multiples of 5 in 42 numbers. From 1 to 42, how many multiples of 5? 5*1 up to 5*8=40, so 8 numbers, as before. Then multiples of 25: 25 is 5^2, so there's one multiple (25) which contributes an extra 5. Hence total number of 5s is 8 + 1 = 9. So the answer is 9.

Alternatively, if I think about the formula for trailing zeros in n! is given by the sum_{k=1}^∞ floor(n / 5^k). So for 42, it's floor(42/5) + floor(42/25) + floor(42/125) + ... which is 8 + 1 + 0 + ... = 9. So that's the same result.

Therefore, I think the number of trailing zeros in 42! is 9.

Wait, but just to be thorough, let me check if there are any other numbers beyond 25 that might contribute more 5s? Like 50, but 50 is beyond 42. So no. So only 25 is the multiple of 25 in 1 to 42. Therefore, only one extra 5.

So total number of 5s is 8 + 1 = 9.

Therefore, the answer should be 9.

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

To determine how many **trailing zeros** are at the end of $42!$, we need to understand the structure of factorials and the role of the number **10** in creating trailing zeros.

---

### Understanding Trailing Zeros in Factorials

A factorial $ n! $ is the product of all positive integers from 1 to $ n $. Trailing zeros in a number come from the factor $ 10 $, and since $ 10 = 2 \times 5 $, **each pair of 2 and 5** in the prime factorization of $ n! $ contributes to a trailing zero.

However, in the product of integers from $ 1 $ to $ n $, the number of **2s** in the prime factorization is always **greater** than the number of **5s**. Therefore, the **limiting factor** is the number of **fives**. So, we only need to count how many times **5** divides into the numbers from 1 to $ n $, and this will directly give the number of trailing zeros.

---

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

We use the following method:

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

We'll 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 $ (since 125 > 42)

So total number of 5s is:
$$
8 + 1 + 0 = 9
$$

---

### Confirming the Result

Each multiple of 5 contributes **at least one** factor of 5. Numbers like 25, 50, etc., which are multiples of $ 5^2 $, contribute **an extra 5**. In this case, only 25 is within the range (up to 42), contributing an **additional 5**.

Thus, we have **8 numbers** contributing at least one 5, and **one number (25)** contributing an extra 5. Total 5s = **9**.

Also, because the number of 2s is more abundant than the number of 5s, there are definitely enough 2s to pair with the 5s, so the limiting factor is indeed the number of 5s.

---

### Final Answer

$$
\boxed{9}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many zeros are at the end of 42 factorial. Hmm, let me remember. I think the number of trailing zeros in a factorial depends on how many times the number 10 is a factor in that factorial. Since 10 is 2 times 5, maybe I need to count how many pairs of 2s and 5s there are in the prime factors of 42!.

But wait, I remember that there are usually more 2s than 5s in a factorial, so the number of trailing zeros should be determined by the number of 5s. Is that right? Let me think. For example, in 5! which is 120, there's one 5 and more than one 2, so one zero. Yeah, that seems to check out. So maybe I just need to count the number of 5s in the prime factors of 42!.

Alright, so how do I count the number of times 5 is a factor in the numbers from 1 to 42? I think the formula is to divide 42 by 5, then divide by 25, 125, etc., and add them up. Let me recall. For example, for 100!, it would be 100/5 + 100/25 + 100/125... which is 20 + 4 + 0 = 24. So that gives the number of 5s.

So applying that to 42! Let me compute 42 divided by 5 first. 42 divided by 5 is 8.4, but we take the floor of that, which is 8. Then divide by 25, which is 42 divided by 25. 25 times 1 is 25, 25 times 2 is 50, which is more than 42. So floor(42/25) is 1. Next is 125, which is way bigger than 42, so that's 0. So adding up 8 + 1 + 0 = 9. So does that mean there are 9 trailing zeros in 42!?

Wait, let me verify this with a smaller number. Let's take 5! again. Using this method: 5/5 = 1, 5/25 = 0, so total 1. Which matches, since 5! is 120. How about 10!? 10/5 = 2, 10/25 = 0, so total 2. But 10! is 3628800, which has two zeros. Correct. How about 25!? 25/5 = 5, 25/25 = 1, 25/125 = 0. Total 5 + 1 = 6. Let me check 25!... Well, I don't remember the exact value, but I know that 25! has 6 trailing zeros. That seems right. So the formula works here.

So applying that to 42!:

First division: floor(42/5) = 8. So numbers divisible by 5: 5, 10, 15, 20, 25, 30, 35, 40. That's 8 numbers. Each contributes at least one 5. But some numbers contribute more. For example, 25 is 5 squared, so it contributes two 5s. Similarly, 50 would contribute two, but 50 is beyond 42. So in 42!, the numbers contributing extra 5s are 25. Let me check. 25 is 5*5, so that's one extra 5. So total number of 5s is 8 + 1 = 9.

Therefore, the number of trailing zeros is 9. But wait, let me count the number of 2s just to make sure there are enough. Since there are more 2s than 5s, the number of trailing zeros is determined by the number of 5s. But maybe I should check?

Number of 2s in 42! would be 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

Next term is floor(42/64) = 0. So total number of 2s is 21 + 10 + 5 + 2 + 1 = 39.

Number of 5s is 8 + 1 = 9. So there are plenty of 2s, so the number of trailing zeros is indeed 9.

But let me check with actual computation. Maybe compute 42! and count the zeros? Although that would be tedious. Alternatively, maybe check with another method. Wait, maybe I can use another example. For instance, 20! has floor(20/5) + floor(20/25) = 4 + 0 = 4 trailing zeros. And 20! is 2432902008176640000, which has four zeros at the end. Correct.

So if 20! has 4 zeros, 25! has 6 zeros, then 42! should have 9. Let me think of another way. Let me calculate the number of multiples of 5 in 42 numbers. From 1 to 42, how many multiples of 5? 5*1 up to 5*8=40, so 8 numbers, as before. Then multiples of 25: 25 is 5^2, so there's one multiple (25) which contributes an extra 5. Hence total number of 5s is 8 + 1 = 9. So the answer is 9.

Alternatively, if I think about the formula for trailing zeros in n! is given by the sum_{k=1}^∞ floor(n / 5^k). So for 42, it's floor(42/5) + floor(42/25) + floor(42/125) + ... which is 8 + 1 + 0 + ... = 9. So that's the same result.

Therefore, I think the number of trailing zeros in 42! is 9.

Wait, but just to be thorough, let me check if there are any other numbers beyond 25 that might contribute more 5s? Like 50, but 50 is beyond 42. So no. So only 25 is the multiple of 25 in 1 to 42. Therefore, only one extra 5.

So total number of 5s is 8 + 1 = 9.

Therefore, the answer should be 9.

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

To determine how many **trailing zeros** are at the end of $42!$, we need to understand the structure of factorials and the role of the number **10** in creating trailing zeros.

---

### Understanding Trailing Zeros in Factorials

A factorial $ n! $ is the product of all positive integers from 1 to $ n $. Trailing zeros in a number come from the factor $ 10 $, and since $ 10 = 2 \times 5 $, **each pair of 2 and 5** in the prime factorization of $ n! $ contributes to a trailing zero.

However, in the product of integers from $ 1 $ to $ n $, the number of **2s** in the prime factorization is always **greater** than the number of **5s**. Therefore, the **limiting factor** is the number of **fives**. So, we only need to count how many times **5** divides into the numbers from 1 to $ n $, and this will directly give the number of trailing zeros.

---

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

We use the following method:

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

We'll 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 $ (since 125 > 42)

So total number of 5s is:
$$
8 + 1 + 0 = 9
$$

---

### Confirming the Result

Each multiple of 5 contributes **at least one** factor of 5. Numbers like 25, 50, etc., which are multiples of $ 5^2 $, contribute **an extra 5**. In this case, only 25 is within the range (up to 42), contributing an **additional 5**.

Thus, we have **8 numbers** contributing at least one 5, and **one number (25)** contributing an extra 5. Total 5s = **9**.

Also, because the number of 2s is more abundant than the number of 5s, there are definitely enough 2s to pair with the 5s, so the limiting factor is indeed the number of 5s.

---

### Final Answer

$$
\boxed{9}
$$

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.