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 ADC, which is right-angled at D, we have AC² = AD² + CD². Similarly, in triangle BDC, 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 ADC: AC² = AD² + CD² => 8² = x² + CD² => 64 = x² + CD². From triangle BDC: 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 = Simplify numerator and denominator by dividing by 2: 64/17 ≈ 3.7647...

So AD is 64/17. 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 = (64²)/(17²) = 4096/289. Then CD² = 64 - 4096/289. Let me convert 64 to 289 denominator:

64 = 64 * 289 / 289 = Let me compute 64 * 289. Wait, that might be tedious. Alternatively, 64 is 64/1, so to subtract 4096/289 from 64, I can write 64 as (64 * 289)/289. Let me compute 64 * 289. Hmm, 64 * 289. Let me note that 289 is 17², which is 289. 64 * 289. Let me compute 64 * 200 = 12,800; 64 * 89 = ?

Wait, maybe there's a better way. Alternatively, since CD² = 64 - (64/17)². Let me compute that:

CD² = 64 - (4096/289) = (64 * 289 - 4096)/289. Let me compute 64 * 289. Let me note that 64 * 289 = 64 * (200 + 89) = 64*200 + 64*89 = 12,800 + 5,696 = 18,496. Wait, 64*89: 60*89=5,340 and 4*89=356, so total 5,340 + 356 = 5,696. Then 12,800 + 5,696 = 18,496. Then subtract 4,096: 18,496 - 4,096 = 14,400. Therefore CD² = 14,400 / 289. Therefore CD = sqrt(14,400 / 289) = 120 / 17. Because sqrt(14,400) is 120 and sqrt(289) is 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 = 7,680. Then divide by 17 * 17 = 289. Then divide by 2: 7,680 / 289 / 2 = 7,680 / 578. Let me simplify that fraction.

Divide numerator and denominator by 2: 3,840 / 289. Let me check if 289 divides into 3,840. 289 * 13 = 3,757. 3,840 - 3,757 = 83. So 3,840 / 289 = 13 + 83/289. Wait, but maybe I made a mistake here? Let me check my calculations again.

Wait, area of ACD is (AD * CD)/2. AD is 64/17, CD is 120/17. So (64/17 * 120/17)/2 = (64*120)/(17*17*2) = (7680)/578. Let me divide numerator and denominator by 2: 3840 / 289. Let me check if this reduces further. 289 is 17². Let me check if 17 divides into 3840. 17*225 = 3825. 3840 - 3825 = 15. So 3840 = 17*225 + 15. Therefore, 3840/289 is 225/17 + 15/289. But maybe I can leave it as 3840/289? Wait, but maybe there's a mistake earlier?

Alternatively, maybe I can compute the area of ABC first and then find the ratio? Let me try that approach to check.

Area of ABC can be found using Heron's formula. The sides are AB=17, AC=8, BC=15. Let me compute the semi-perimeter: s = (17 + 8 + 15)/2 = (40)/2 = 20. Then area = sqrt[s(s - AB)(s - AC)(s - BC)] = sqrt[20*(20 - 17)*(20 - 8)*(20 - 15)] = sqrt[20*3*12*5]. Let me compute that: 20*3=60, 12*5=60, so sqrt[60*60] = sqrt[3600] = 60. So area of ABC is 60.

Alternatively, since CD is the altitude from C to AB, then area of ABC is also (AB * CD)/2 = (17 * CD)/2. Therefore, 60 = (17 * CD)/2 => CD = (60 * 2)/17 = 120/17. Which matches the earlier result. So CD is indeed 120/17.

So then, if area of ABC is 60, and area of ACD is (AD / AB) * area of ABC? Wait, because if D divides AB into AD and DB, then the area of ACD would be (AD / AB) * area of ABC? Wait, is that correct? Because the height is the same for both triangles ACD and ABC, right? Since both have the same altitude CD. Wait, no. Wait, ABC has base AB and height CD. ACD has base AD and same height CD. Therefore, area of ACD is (AD / AB) * area of ABC.

So if that's the case, then area of ACD is (AD / 17) * 60. Since AD is 64/17, then area is (64/17 / 17) * 60 = (64/289)*60 = 3840/289. Which is the same as before. So that's consistent.

But 3840 divided by 289... Let me check if that reduces. Since 289 is 17², and 3840 divided by 17 is 225.88... Wait, earlier I saw that 17*225 = 3825, so 3840 - 3825 = 15. So 3840/289 = 225/17 + 15/289. But maybe the answer is supposed to be a fraction? Let me check if 3840 and 289 have common factors. 289 is 17². Let me check if 17 divides into 3840. 17*200 = 3400, 3840 - 3400 = 440. 17*25 = 425, 440 - 425 = 15. So no, 17 doesn't divide into 3840. Therefore, 3840/289 is the simplified fraction.

But let me check if my first approach was correct. Because 3840 divided by 289... Let me compute decimal value to check. 289*13 = 3757, as before. 3840 - 3757 = 83. So 3840/289 ≈ 13 + 83/289 ≈ 13.287... But maybe there's a mistake here? Because if area of ABC is 60, and AD is 64/17 ≈ 3.7647, then area of ACD is (AD * CD)/2 ≈ (3.7647 * 7.0588)/2. Let me compute that. 3.7647 * 7.0588 ≈ Let me approximate:

First, 3.7647 * 7 = 26.3529, and 3.7647 * 0.0588 ≈ approx 0.221. So total approx 26.57. Divide by 2: approx 13.285. Which matches the decimal value of 3840/289 ≈ 13.287. So that seems okay.

But maybe there's a simpler way? Let me think again. Alternatively, since we know coordinates? Maybe assign coordinates to the triangle. Let me try that. Let me place point A at (0, 0), point B at (17, 0). Then point C is somewhere in the plane. We know AC = 8, BC = 15. Let me find coordinates of C.

Let me denote coordinates: A(0, 0), B(17, 0), C(x, y). Then AC = 8: sqrt(x² + y²) = 8 => x² + y² = 64. BC = 15: sqrt((x - 17)² + y²) = 15 => (x - 17)² + y² = 225. Subtract 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 is the same as AD earlier. So x-coordinate of C is 64/17. Then, from AC equation, x² + y² = 64. So y² = 64 - (64/17)². Which is same as before. Which gives y = CD = 120/17. Therefore, coordinates of C are (64/17, 120/17).

Then, D is the foot of 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, if AB is on the x-axis, then the altitude from C to AB is vertical? Wait, no. Wait, the altitude from C to AB is perpendicular to AB. Since AB is horizontal, its slope is 0, so the altitude is vertical, yes. 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, so D has same x-coordinate as C? Wait, no. Wait, suppose AB is horizontal, so the line AB is y=0. The altitude from C(x, y) to AB is vertical? Wait, no. Wait, the altitude is perpendicular to AB. Since AB is horizontal, the altitude is vertical. Therefore, yes, the altitude is vertical, so it goes from C(x, y) straight down to AB at (x, 0). Therefore, D is (x, 0). Therefore, since we found x-coordinate of C is 64/17, then D is (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 A(0,0), D(64/17, 0), C(64/17, 120/17). Therefore, triangle ACD has coordinates A(0,0), C(64/17, 120/17), D(64/17, 0). So this 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.

But maybe the problem expects a simpler answer? Let me check if 3840/289 reduces or if I made a mistake in calculation. Wait, 289 is 17², and 3840 divided by 17 is 225.88... Wait, but maybe there's another way to compute the area?

Alternatively, since we know the area of ABC is 60, and AD is 64/17, then area of ACD is (AD / AB) * area ABC = (64/17 / 17) * 60 = 64*60 / 289 = 3840 / 289. Which is same as before.

Alternatively, maybe I can check with vectors or coordinate geometry. Let me compute vectors. Coordinates of A(0,0), D(64/17, 0), C(64/17, 120/17). So vector AD is (64/17, 0), vector AC is (64/17, 120/17). The area of triangle ACD can be calculated via determinant formula: ½ | (x_A(y_D - y_C) + x_D(y_C - y_A) + x_C(y_A - y_D)) |. But since A is (0,0), D is (64/17, 0), C is (64/17, 120/17). Plugging into formula:

½ | 0*(0 - 120/17) + (64/17)*(120/17 - 0) + (64/17)*(0 - 0) | = ½ | 0 + (64/17)*(120/17) + 0 | = ½ * (7680 / 289) = 3840 / 289. Same result.

So seems like 3840/289 is the answer. Let me check if that fraction can be simplified. Since 289 is 17², and 3840 divided by 17 is 225.88... which is not integer. 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, remainder 3840 - 3757 = 83.

So GCD(289, 83).

Divide 289 by 83: 83*3 = 249, remainder 289 - 249 = 40.

GCD(83, 40).

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

GCD(40, 3).

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

GCD(3, 1).

Which is 1. Therefore, GCD is 1. Therefore, 3840/289 is reduced completely.

But maybe the problem expects a decimal? But in math competitions, usually fractions are preferred. However, let me check if there's a miscalculation. Wait, area of ABC is 60, and area of ACD is 3840/289 ≈ 13.287. Let me check with another approach.

Alternatively, since we know coordinates of all points, we can compute vectors or use base-height.

Alternatively, since AD = 64/17, and CD = 120/17, then area is ½ * 64/17 * 120/17 = 3840/289.

Alternatively, maybe there's a mistake in assuming that area of ACD is (AD / AB) * area ABC? Wait, let me think. If two triangles share the same height, then their areas are in the ratio of their bases. In this case, triangle ACD and triangle ABC share the same height from C to AB. Therefore, their areas are in the ratio of AD / AB. Therefore, area ACD = (AD / AB) * area ABC. Which is (64/17 / 17) * 60 = 64*60 / 289 = 3840 / 289. So that's correct.

Alternatively, maybe the problem expects an answer as a reduced fraction or mixed number? But 3840/289 is already reduced. Let me check if 289 divides into 3840. 289*13 = 3757, 289*13.287 ≈ 3840. So no, it's an improper fraction.

Alternatively, maybe I made a mistake in Heron's formula? Let me check that again.

Sides: AB=17, AC=8, BC=15. Semi-perimeter s=(17+8+15)/2=40/2=20. Area = sqrt[s(s - AB)(s - AC)(s - BC)] = sqrt[20*(20 - 17)*(20 - 8)*(20 - 15)] = sqrt[20*3*12*5].

Compute 20*3=60, 12*5=60, so sqrt[60*60] = 60. Correct. So area of ABC is indeed 60.

Therefore, seems like 3840/289 is correct. Let me check if that's equal to something else? Let me compute 3840 divided by 289. Let me do the division:

289*13 = 3757, as before. 3840 - 3757 = 83. So 3840/289 = 13 + 83/289.

But maybe the problem expects the answer as a fraction? So 3840/289. Let me check if that can be simplified. Since GCD is 1, as we saw, so that's the answer.

Alternatively, maybe I made a mistake in calculation of AD? Let me check again.

From earlier steps, we had:

From triangle ADC: 8² = x² + CD²

From triangle BDC: 15² = (17 - x)² + CD²

Subtracting gives 225 - 64 = (17 - x)² - x² => 161 = 289 - 34x => -128 = -34x => x = 128/34 = 64/17. Correct.

So AD is 64/17. Then CD² = 64 - (64/17)². Which we calculated as 14,400 / 289. Which is (120/17)^2. Correct.

Therefore, area of ACD is ½ * 64/17 * 120/17 = 3840/289.

Alternatively, maybe the problem expects the answer as a decimal? But in Olympiad problems, fractions are preferred. So 3840/289. Let me check if that reduces. Wait, 289 is 17², and 3840 divided by 17 is 225.88... So no.

Alternatively, maybe I made a mistake in the coordinate approach? Let me check coordinates again.

Coordinates of C: (64/17, 120/17). Coordinates of D: (64/17, 0). Coordinates of A: (0, 0). So triangle ACD is a right triangle with legs AD = 64/17 and CD = 120/17. Therefore, area is ½ * 64/17 * 120/17 = same as before.

Alternatively, maybe the problem is expecting an integer? But 3840 divided by 289 is approximately 13.287, which is not integer. Wait, but area of ABC is 60. Maybe there's a miscalculation?

Wait, let me check Heron's formula again. Maybe I miscalculated.

s = 20.

s - AB = 20 - 17 = 3.

s - AC = 20 - 8 = 12.

s - BC = 20 - 15 = 5.

So area is sqrt(20 * 3 * 12 * 5).

Compute 20 * 3 = 60, 12 * 5 = 60. Then sqrt(60 * 60) = 60. Correct.

So area of ABC is 60.

Alternatively, maybe the answer is 60*(AD / AB) = 60*(64/17)/17 = 60*64/289 = 3840/289.

Alternatively, maybe there's a different approach. Let me think.

Alternatively, since CD is the altitude, and we found CD = 120/17, and AD = 64/17, then area of ACD is ½ * AD * CD = ½ * (64/17) * (120/17) = same as before.

Alternatively, maybe I can use similar triangles?

But I don't see an immediate similar triangle here.

Alternatively, maybe using coordinate geometry, find the equation of the altitude and find D?

But we already did that.

Alternatively, maybe using vectors?

But again, same result.

Alternatively, maybe I can check with another formula for area.

Alternatively, since we know coordinates of all points, compute vectors AC and AD and take cross product?

Vector AC is (64/17, 120/17). Vector AD is (64/17, 0). The area is ½ |AC × AD|.

Cross product in 2D is scalar magnitude: (64/17)(0) - (120/17)(64/17) = 0 - (7680/289) = -7680/289. Absolute value divided by 2: 7680/289 / 2 = 3840/289. Same result.

Therefore, I think this is correct.

But maybe the problem expects the answer as a mixed number or something? But in the context of a competition problem, it's more likely to be a reduced fraction. So 3840/289. Let me check if that can be simplified.

Wait, 3840 divided by 17 is 225.88... So no.

Alternatively, maybe I made a mistake in the problem statement? Let me check again.

"In triangle ABC, AB = 17, AC = 8, and BC = 15. Let D be the foot of the altitude from C to AB. Find the area of triangle ACD."

Yes, that's what was given.

Alternatively, maybe there's a calculation mistake in Heron's formula? Wait, if AB=17, AC=8, BC=15, then triangle inequality holds? AB + AC = 25 > 15, AB + BC = 32 > 8, AC + BC = 23 > 17. Yes, so triangle exists.

Alternatively, maybe I can compute the area of ACD using coordinates. Coordinates of A(0,0), C(64/17, 120/17), D(64/17, 0). Using the formula for area of triangle with coordinates:

Area = ½ |x_A(y_C - y_D) + x_C(y_D - y_A) + x_D(y_A - y_C)|

Plugging in:

½ |0*(120/17 - 0) + (64/17)*(0 - 0) + (64/17)*(0 - 120/17)|

= ½ |0 + 0 + (64/17)*(-120/17)|

= ½ | -7680 / 289 | = ½ * 7680 / 289 = 3840 / 289. Same result.

Therefore, I think this is correct.

Alternatively, maybe the answer is 120/17 * 64/17 / 2 = 3840/289.

Alternatively, maybe the problem expects the answer as a decimal? But in math competitions, fractions are standard.

Alternatively, maybe I can write it as a mixed number? 3840 divided by 289. Let me compute 289*13 = 3757, 3840 - 3757 = 83. So 13 and 83/289. But that's not simpler.

Alternatively, maybe there's a calculation mistake in the beginning? Let me check once again.

From the two equations:

From triangle ADC: 64 = x² + CD²

From triangle BDC: 225 = (17 - x)² + CD²

Subtracting gives 225 - 64 = (17 - x)² - x² => 161 = 289 - 34x => -128 = -34x => x = 128/34 = 64/17. Correct.

Then CD² = 64 - (64/17)². Let me compute 64 as 64/1, so:

CD² = 64 - (4096/289) = (64*289 - 4096)/289.

Compute 64*289:

Well, 64*289. Let me compute 64*(200 + 80 + 9) = 64*200 + 64*80 + 64*9 = 12,800 + 5,120 + 576 = 12,800 + 5,120 = 17,920 + 576 = 18,496. Then subtract 4,096: 18,496 - 4,096 = 14,400. Therefore CD² = 14,400 / 289. Therefore CD = 120/17. Correct.

Therefore, area of ACD is ½ * 64/17 * 120/17 = 3840/289.

Therefore, I think this is correct.

Alternatively, maybe the problem is from an Olympiad where answers are integers? But maybe not. Let me check with another approach.

Alternatively, since area of ABC is 60, and area of ACD is ?

Alternatively, using coordinates, coordinates of C(64/17, 120/17). Coordinates of D(64/17, 0). Coordinates of A(0,0). So the triangle ACD is a right triangle with legs AD = 64/17 and CD = 120/17. Therefore, area is ½ * 64/17 * 120/17 = 3840/289.

Alternatively, maybe the problem expects the answer as a decimal? If so, 3840 divided by 289. Let me compute that.

Well, 289*13 = 3757, as before. 3840 - 3757 = 83. So 83/289 ≈ 0.287. So total ≈ 13.287. But maybe the exact fraction is needed.

Alternatively, maybe I can check with another method. Let me think.

Alternatively, using trigonometry. Let me find angle at A or something.

But maybe not necessary.

Alternatively, since we know coordinates of all points, maybe compute the vectors and use the formula.

But again, same result.

Alternatively, maybe I can check with mass point or something else? Not sure.

Alternatively, maybe I can check with the area ratios.

Since area of ABC is 60, and area of ACD is ?

If I can find the ratio of AD to AB, which is 64/17 divided by 17, which is 64/289. Therefore, area of ACD is 64/289 * 60 = 3840/289.

Alternatively, maybe the answer is 120/17 * 64/17 / 2 = same thing.

Therefore, I think I have to conclude that the area is 3840/289.

But let me check if that fraction can be simplified. Since 3840 and 289. 289 is 17². 17 divides into 3840? 17*200=3400, 3840-3400=440. 17*25=425, 440-425=15. So no. Therefore, 3840/289 is simplest form.

Alternatively, maybe the problem expects the answer as a mixed number? But in most cases, improper fraction is acceptable.

Alternatively, maybe I made a mistake in the problem statement? Let me check again.

"In triangle ABC, AB = 17, AC = 8, and BC = 15. Let D be the foot of the altitude from C to AB. Find the area of triangle ACD."

Yes, that's correct.

Alternatively, maybe I can check with another formula for area. For example, using coordinates, the area of ACD can be calculated as the area of ABC minus area of BCD.

Area of ABC is 60. Area of BCD can be calculated as (BD * CD)/2. BD = AB - AD = 17 - 64/17 = (289 - 64)/17 = 225/17. CD is 120/17. Therefore, area of BCD is (225/17 * 120/17)/2 = (27000)/578 = 13500/289. Then area of ACD = area ABC - area BCD = 60 - 13500/289. Let me compute that.

Convert 60 to 289 denominator: 60 = 60 * 289 / 289 = 17340 / 289. Then subtract 13500 / 289: 17340 - 13500 = 3840. Therefore, area of ACD is 3840 / 289. Same result.

Therefore, this confirms the answer.

Therefore, I think the answer is 3840/289.

But let me check if that reduces. Since 3840 divided by 17 is 225.88... and 289 is 17². So no.

Alternatively, maybe the problem expects the answer as a decimal? But in the context of a math competition, fractions are preferred.

Alternatively, maybe I can write it as a mixed number? 13 and 83/289. But again, not sure.

Alternatively, maybe there's a calculation mistake? Let me check once again.

If area of ABC is 60, and area of ACD is 3840/289 ≈ 13.287, and area of BCD is 60 - 13.287 ≈ 46.713. Let me check area of BCD with BD = 225/17 ≈ 13.235, CD = 120/17 ≈ 7.058. Then area is ½ * 13.235 * 7.058 ≈ ½ * 93.4 ≈ 46.7, which matches. So that seems correct.

Therefore, I think the answer is 3840/289.

But let me check if that fraction can be simplified. Since 3840 and 289. 289 is 17². 3840 divided by 17 is 225.88... So no.

Alternatively, maybe the problem had a typo? But assuming the problem is correct, then 3840/289 is the answer.

Alternatively, maybe I can write it as \boxed{\dfrac{3840}{289}}.

But let me check if that's the simplest form.

Alternatively, maybe I can divide numerator and denominator by something? Let me check if 3840 and 289 have any common factors.

As we did before, GCD(3840, 289) = 1. Therefore, yes, it's reduced.

Therefore, I think that is the answer.

**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: Use Heron’s Formula to Find the Area of Triangle $ ABC $

We first compute the area of triangle $ ABC $ using **Heron’s formula**.

- Semi-perimeter:
  $$
  s = \frac{AB + AC + BC}{2} = \frac{17 + 8 + 15}{2} = 20
  $$

- Area of triangle $ ABC $:
  $$
  \text{Area} = \sqrt{s(s - AB)(s - AC)(s - BC)} = \sqrt{20(20 - 17)(20 - 8)(20 - 15)} = \sqrt{20 \cdot 3 \cdot 12 \cdot 5}
  $$
  $$
  = \sqrt{3600} = 60
  $$

So, the area of triangle $ ABC $ is $ 60 $.

---

### Step 2: Use Coordinates to Find the Area of Triangle $ ACD $

Let’s place triangle $ ABC $ on the coordinate plane for clarity:

- Let $ A = (0, 0) $
- Let $ B = (17, 0) $
- Let $ C = (x, y) $

We are given:
- $ AC = 8 \Rightarrow x^2 + y^2 = 64 $
- $ BC = 15 \Rightarrow (x - 17)^2 + y^2 = 225 $

Subtracting the first equation from the second:
$$
(x - 17)^2 + y^2 - (x^2 + y^2) = 225 - 64 \Rightarrow x^2 - 34x + 289 - x^2 = 161
$$
$$
-34x + 289 = 161 \Rightarrow -34x = -128 \Rightarrow x = \frac{64}{17}
$$

Now substitute $ x = \frac{64}{17} $ into $ x^2 + y^2 = 64 $ to find $ y $:
$$
\left(\frac{64}{17}\right)^2 + y^2 = 64 \Rightarrow \frac{4096}{289} + y^2 = 64
$$
$$
y^2 = 64 - \frac{4096}{289} = \frac{18496 - 4096}{289} = \frac{14400}{289} \Rightarrow y = \frac{120}{17}
$$

So, point $ C = \left( \frac{64}{17}, \frac{120}{17} \right) $

Since $ D $ is the foot of the perpendicular from $ C $ to $ AB $, and $ AB $ lies on the x-axis, the x-coordinate of $ D $ is the same as that of $ C $, and the y-coordinate is 0. So:

- $ D = \left( \frac{64}{17}, 0 \right) $

---

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

Triangle $ ACD $ is a right triangle with:

- Base $ AD = \frac{64}{17} $
- Height $ CD = \frac{120}{17} $

So, the area is:
$$
\text{Area} = \frac{1}{2} \cdot AD \cdot CD = \frac{1}{2} \cdot \frac{64}{17} \cdot \frac{120}{17} = \frac{1}{2} \cdot \frac{7680}{289} = \frac{3840}{289}
$$

---

### 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 ADC, which is right-angled at D, we have AC² = AD² + CD². Similarly, in triangle BDC, 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 ADC: AC² = AD² + CD² => 8² = x² + CD² => 64 = x² + CD². From triangle BDC: 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 = Simplify numerator and denominator by dividing by 2: 64/17 ≈ 3.7647...

So AD is 64/17. 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 = (64²)/(17²) = 4096/289. Then CD² = 64 - 4096/289. Let me convert 64 to 289 denominator:

64 = 64 * 289 / 289 = Let me compute 64 * 289. Wait, that might be tedious. Alternatively, 64 is 64/1, so to subtract 4096/289 from 64, I can write 64 as (64 * 289)/289. Let me compute 64 * 289. Hmm, 64 * 289. Let me note that 289 is 17², which is 289. 64 * 289. Let me compute 64 * 200 = 12,800; 64 * 89 = ?

Wait, maybe there's a better way. Alternatively, since CD² = 64 - (64/17)². Let me compute that:

CD² = 64 - (4096/289) = (64 * 289 - 4096)/289. Let me compute 64 * 289. Let me note that 64 * 289 = 64 * (200 + 89) = 64*200 + 64*89 = 12,800 + 5,696 = 18,496. Wait, 64*89: 60*89=5,340 and 4*89=356, so total 5,340 + 356 = 5,696. Then 12,800 + 5,696 = 18,496. Then subtract 4,096: 18,496 - 4,096 = 14,400. Therefore CD² = 14,400 / 289. Therefore CD = sqrt(14,400 / 289) = 120 / 17. Because sqrt(14,400) is 120 and sqrt(289) is 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 = 7,680. Then divide by 17 * 17 = 289. Then divide by 2: 7,680 / 289 / 2 = 7,680 / 578. Let me simplify that fraction.

Divide numerator and denominator by 2: 3,840 / 289. Let me check if 289 divides into 3,840. 289 * 13 = 3,757. 3,840 - 3,757 = 83. So 3,840 / 289 = 13 + 83/289. Wait, but maybe I made a mistake here? Let me check my calculations again.

Wait, area of ACD is (AD * CD)/2. AD is 64/17, CD is 120/17. So (64/17 * 120/17)/2 = (64*120)/(17*17*2) = (7680)/578. Let me divide numerator and denominator by 2: 3840 / 289. Let me check if this reduces further. 289 is 17². Let me check if 17 divides into 3840. 17*225 = 3825. 3840 - 3825 = 15. So 3840 = 17*225 + 15. Therefore, 3840/289 is 225/17 + 15/289. But maybe I can leave it as 3840/289? Wait, but maybe there's a mistake earlier?

Alternatively, maybe I can compute the area of ABC first and then find the ratio? Let me try that approach to check.

Area of ABC can be found using Heron's formula. The sides are AB=17, AC=8, BC=15. Let me compute the semi-perimeter: s = (17 + 8 + 15)/2 = (40)/2 = 20. Then area = sqrt[s(s - AB)(s - AC)(s - BC)] = sqrt[20*(20 - 17)*(20 - 8)*(20 - 15)] = sqrt[20*3*12*5]. Let me compute that: 20*3=60, 12*5=60, so sqrt[60*60] = sqrt[3600] = 60. So area of ABC is 60.

Alternatively, since CD is the altitude from C to AB, then area of ABC is also (AB * CD)/2 = (17 * CD)/2. Therefore, 60 = (17 * CD)/2 => CD = (60 * 2)/17 = 120/17. Which matches the earlier result. So CD is indeed 120/17.

So then, if area of ABC is 60, and area of ACD is (AD / AB) * area of ABC? Wait, because if D divides AB into AD and DB, then the area of ACD would be (AD / AB) * area of ABC? Wait, is that correct? Because the height is the same for both triangles ACD and ABC, right? Since both have the same altitude CD. Wait, no. Wait, ABC has base AB and height CD. ACD has base AD and same height CD. Therefore, area of ACD is (AD / AB) * area of ABC.

So if that's the case, then area of ACD is (AD / 17) * 60. Since AD is 64/17, then area is (64/17 / 17) * 60 = (64/289)*60 = 3840/289. Which is the same as before. So that's consistent.

But 3840 divided by 289... Let me check if that reduces. Since 289 is 17², and 3840 divided by 17 is 225.88... Wait, earlier I saw that 17*225 = 3825, so 3840 - 3825 = 15. So 3840/289 = 225/17 + 15/289. But maybe the answer is supposed to be a fraction? Let me check if 3840 and 289 have common factors. 289 is 17². Let me check if 17 divides into 3840. 17*200 = 3400, 3840 - 3400 = 440. 17*25 = 425, 440 - 425 = 15. So no, 17 doesn't divide into 3840. Therefore, 3840/289 is the simplified fraction.

But let me check if my first approach was correct. Because 3840 divided by 289... Let me compute decimal value to check. 289*13 = 3757, as before. 3840 - 3757 = 83. So 3840/289 ≈ 13 + 83/289 ≈ 13.287... But maybe there's a mistake here? Because if area of ABC is 60, and AD is 64/17 ≈ 3.7647, then area of ACD is (AD * CD)/2 ≈ (3.7647 * 7.0588)/2. Let me compute that. 3.7647 * 7.0588 ≈ Let me approximate:

First, 3.7647 * 7 = 26.3529, and 3.7647 * 0.0588 ≈ approx 0.221. So total approx 26.57. Divide by 2: approx 13.285. Which matches the decimal value of 3840/289 ≈ 13.287. So that seems okay.

But maybe there's a simpler way? Let me think again. Alternatively, since we know coordinates? Maybe assign coordinates to the triangle. Let me try that. Let me place point A at (0, 0), point B at (17, 0). Then point C is somewhere in the plane. We know AC = 8, BC = 15. Let me find coordinates of C.

Let me denote coordinates: A(0, 0), B(17, 0), C(x, y). Then AC = 8: sqrt(x² + y²) = 8 => x² + y² = 64. BC = 15: sqrt((x - 17)² + y²) = 15 => (x - 17)² + y² = 225. Subtract 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 is the same as AD earlier. So x-coordinate of C is 64/17. Then, from AC equation, x² + y² = 64. So y² = 64 - (64/17)². Which is same as before. Which gives y = CD = 120/17. Therefore, coordinates of C are (64/17, 120/17).

Then, D is the foot of 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, if AB is on the x-axis, then the altitude from C to AB is vertical? Wait, no. Wait, the altitude from C to AB is perpendicular to AB. Since AB is horizontal, its slope is 0, so the altitude is vertical, yes. 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, so D has same x-coordinate as C? Wait, no. Wait, suppose AB is horizontal, so the line AB is y=0. The altitude from C(x, y) to AB is vertical? Wait, no. Wait, the altitude is perpendicular to AB. Since AB is horizontal, the altitude is vertical. Therefore, yes, the altitude is vertical, so it goes from C(x, y) straight down to AB at (x, 0). Therefore, D is (x, 0). Therefore, since we found x-coordinate of C is 64/17, then D is (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 A(0,0), D(64/17, 0), C(64/17, 120/17). Therefore, triangle ACD has coordinates A(0,0), C(64/17, 120/17), D(64/17, 0). So this 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.

But maybe the problem expects a simpler answer? Let me check if 3840/289 reduces or if I made a mistake in calculation. Wait, 289 is 17², and 3840 divided by 17 is 225.88... Wait, but maybe there's another way to compute the area?

Alternatively, since we know the area of ABC is 60, and AD is 64/17, then area of ACD is (AD / AB) * area ABC = (64/17 / 17) * 60 = 64*60 / 289 = 3840 / 289. Which is same as before.

Alternatively, maybe I can check with vectors or coordinate geometry. Let me compute vectors. Coordinates of A(0,0), D(64/17, 0), C(64/17, 120/17). So vector AD is (64/17, 0), vector AC is (64/17, 120/17). The area of triangle ACD can be calculated via determinant formula: ½ | (x_A(y_D - y_C) + x_D(y_C - y_A) + x_C(y_A - y_D)) |. But since A is (0,0), D is (64/17, 0), C is (64/17, 120/17). Plugging into formula:

½ | 0*(0 - 120/17) + (64/17)*(120/17 - 0) + (64/17)*(0 - 0) | = ½ | 0 + (64/17)*(120/17) + 0 | = ½ * (7680 / 289) = 3840 / 289. Same result.

So seems like 3840/289 is the answer. Let me check if that fraction can be simplified. Since 289 is 17², and 3840 divided by 17 is 225.88... which is not integer. 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, remainder 3840 - 3757 = 83.

So GCD(289, 83).

Divide 289 by 83: 83*3 = 249, remainder 289 - 249 = 40.

GCD(83, 40).

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

GCD(40, 3).

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

GCD(3, 1).

Which is 1. Therefore, GCD is 1. Therefore, 3840/289 is reduced completely.

But maybe the problem expects a decimal? But in math competitions, usually fractions are preferred. However, let me check if there's a miscalculation. Wait, area of ABC is 60, and area of ACD is 3840/289 ≈ 13.287. Let me check with another approach.

Alternatively, since we know coordinates of all points, we can compute vectors or use base-height.

Alternatively, since AD = 64/17, and CD = 120/17, then area is ½ * 64/17 * 120/17 = 3840/289.

Alternatively, maybe there's a mistake in assuming that area of ACD is (AD / AB) * area ABC? Wait, let me think. If two triangles share the same height, then their areas are in the ratio of their bases. In this case, triangle ACD and triangle ABC share the same height from C to AB. Therefore, their areas are in the ratio of AD / AB. Therefore, area ACD = (AD / AB) * area ABC. Which is (64/17 / 17) * 60 = 64*60 / 289 = 3840 / 289. So that's correct.

Alternatively, maybe the problem expects an answer as a reduced fraction or mixed number? But 3840/289 is already reduced. Let me check if 289 divides into 3840. 289*13 = 3757, 289*13.287 ≈ 3840. So no, it's an improper fraction.

Alternatively, maybe I made a mistake in Heron's formula? Let me check that again.

Sides: AB=17, AC=8, BC=15. Semi-perimeter s=(17+8+15)/2=40/2=20. Area = sqrt[s(s - AB)(s - AC)(s - BC)] = sqrt[20*(20 - 17)*(20 - 8)*(20 - 15)] = sqrt[20*3*12*5].

Compute 20*3=60, 12*5=60, so sqrt[60*60] = 60. Correct. So area of ABC is indeed 60.

Therefore, seems like 3840/289 is correct. Let me check if that's equal to something else? Let me compute 3840 divided by 289. Let me do the division:

289*13 = 3757, as before. 3840 - 3757 = 83. So 3840/289 = 13 + 83/289.

But maybe the problem expects the answer as a fraction? So 3840/289. Let me check if that can be simplified. Since GCD is 1, as we saw, so that's the answer.

Alternatively, maybe I made a mistake in calculation of AD? Let me check again.

From earlier steps, we had:

From triangle ADC: 8² = x² + CD²

From triangle BDC: 15² = (17 - x)² + CD²

Subtracting gives 225 - 64 = (17 - x)² - x² => 161 = 289 - 34x => -128 = -34x => x = 128/34 = 64/17. Correct.

So AD is 64/17. Then CD² = 64 - (64/17)². Which we calculated as 14,400 / 289. Which is (120/17)^2. Correct.

Therefore, area of ACD is ½ * 64/17 * 120/17 = 3840/289.

Alternatively, maybe the problem expects the answer as a decimal? But in Olympiad problems, fractions are preferred. So 3840/289. Let me check if that reduces. Wait, 289 is 17², and 3840 divided by 17 is 225.88... So no.

Alternatively, maybe I made a mistake in the coordinate approach? Let me check coordinates again.

Coordinates of C: (64/17, 120/17). Coordinates of D: (64/17, 0). Coordinates of A: (0, 0). So triangle ACD is a right triangle with legs AD = 64/17 and CD = 120/17. Therefore, area is ½ * 64/17 * 120/17 = same as before.

Alternatively, maybe the problem is expecting an integer? But 3840 divided by 289 is approximately 13.287, which is not integer. Wait, but area of ABC is 60. Maybe there's a miscalculation?

Wait, let me check Heron's formula again. Maybe I miscalculated.

s = 20.

s - AB = 20 - 17 = 3.

s - AC = 20 - 8 = 12.

s - BC = 20 - 15 = 5.

So area is sqrt(20 * 3 * 12 * 5).

Compute 20 * 3 = 60, 12 * 5 = 60. Then sqrt(60 * 60) = 60. Correct.

So area of ABC is 60.

Alternatively, maybe the answer is 60*(AD / AB) = 60*(64/17)/17 = 60*64/289 = 3840/289.

Alternatively, maybe there's a different approach. Let me think.

Alternatively, since CD is the altitude, and we found CD = 120/17, and AD = 64/17, then area of ACD is ½ * AD * CD = ½ * (64/17) * (120/17) = same as before.

Alternatively, maybe I can use similar triangles?

But I don't see an immediate similar triangle here.

Alternatively, maybe using coordinate geometry, find the equation of the altitude and find D?

But we already did that.

Alternatively, maybe using vectors?

But again, same result.

Alternatively, maybe I can check with another formula for area.

Alternatively, since we know coordinates of all points, compute vectors AC and AD and take cross product?

Vector AC is (64/17, 120/17). Vector AD is (64/17, 0). The area is ½ |AC × AD|.

Cross product in 2D is scalar magnitude: (64/17)(0) - (120/17)(64/17) = 0 - (7680/289) = -7680/289. Absolute value divided by 2: 7680/289 / 2 = 3840/289. Same result.

Therefore, I think this is correct.

But maybe the problem expects the answer as a mixed number or something? But in the context of a competition problem, it's more likely to be a reduced fraction. So 3840/289. Let me check if that can be simplified.

Wait, 3840 divided by 17 is 225.88... So no.

Alternatively, maybe I made a mistake in the problem statement? Let me check again.

"In triangle ABC, AB = 17, AC = 8, and BC = 15. Let D be the foot of the altitude from C to AB. Find the area of triangle ACD."

Yes, that's what was given.

Alternatively, maybe there's a calculation mistake in Heron's formula? Wait, if AB=17, AC=8, BC=15, then triangle inequality holds? AB + AC = 25 > 15, AB + BC = 32 > 8, AC + BC = 23 > 17. Yes, so triangle exists.

Alternatively, maybe I can compute the area of ACD using coordinates. Coordinates of A(0,0), C(64/17, 120/17), D(64/17, 0). Using the formula for area of triangle with coordinates:

Area = ½ |x_A(y_C - y_D) + x_C(y_D - y_A) + x_D(y_A - y_C)|

Plugging in:

½ |0*(120/17 - 0) + (64/17)*(0 - 0) + (64/17)*(0 - 120/17)|

= ½ |0 + 0 + (64/17)*(-120/17)|

= ½ | -7680 / 289 | = ½ * 7680 / 289 = 3840 / 289. Same result.

Therefore, I think this is correct.

Alternatively, maybe the answer is 120/17 * 64/17 / 2 = 3840/289.

Alternatively, maybe the problem expects the answer as a decimal? But in math competitions, fractions are standard.

Alternatively, maybe I can write it as a mixed number? 3840 divided by 289. Let me compute 289*13 = 3757, 3840 - 3757 = 83. So 13 and 83/289. But that's not simpler.

Alternatively, maybe there's a calculation mistake in the beginning? Let me check once again.

From the two equations:

From triangle ADC: 64 = x² + CD²

From triangle BDC: 225 = (17 - x)² + CD²

Subtracting gives 225 - 64 = (17 - x)² - x² => 161 = 289 - 34x => -128 = -34x => x = 128/34 = 64/17. Correct.

Then CD² = 64 - (64/17)². Let me compute 64 as 64/1, so:

CD² = 64 - (4096/289) = (64*289 - 4096)/289.

Compute 64*289:

Well, 64*289. Let me compute 64*(200 + 80 + 9) = 64*200 + 64*80 + 64*9 = 12,800 + 5,120 + 576 = 12,800 + 5,120 = 17,920 + 576 = 18,496. Then subtract 4,096: 18,496 - 4,096 = 14,400. Therefore CD² = 14,400 / 289. Therefore CD = 120/17. Correct.

Therefore, area of ACD is ½ * 64/17 * 120/17 = 3840/289.

Therefore, I think this is correct.

Alternatively, maybe the problem is from an Olympiad where answers are integers? But maybe not. Let me check with another approach.

Alternatively, since area of ABC is 60, and area of ACD is ?

Alternatively, using coordinates, coordinates of C(64/17, 120/17). Coordinates of D(64/17, 0). Coordinates of A(0,0). So the triangle ACD is a right triangle with legs AD = 64/17 and CD = 120/17. Therefore, area is ½ * 64/17 * 120/17 = 3840/289.

Alternatively, maybe the problem expects the answer as a decimal? If so, 3840 divided by 289. Let me compute that.

Well, 289*13 = 3757, as before. 3840 - 3757 = 83. So 83/289 ≈ 0.287. So total ≈ 13.287. But maybe the exact fraction is needed.

Alternatively, maybe I can check with another method. Let me think.

Alternatively, using trigonometry. Let me find angle at A or something.

But maybe not necessary.

Alternatively, since we know coordinates of all points, maybe compute the vectors and use the formula.

But again, same result.

Alternatively, maybe I can check with mass point or something else? Not sure.

Alternatively, maybe I can check with the area ratios.

Since area of ABC is 60, and area of ACD is ?

If I can find the ratio of AD to AB, which is 64/17 divided by 17, which is 64/289. Therefore, area of ACD is 64/289 * 60 = 3840/289.

Alternatively, maybe the answer is 120/17 * 64/17 / 2 = same thing.

Therefore, I think I have to conclude that the area is 3840/289.

But let me check if that fraction can be simplified. Since 3840 and 289. 289 is 17². 17 divides into 3840? 17*200=3400, 3840-3400=440. 17*25=425, 440-425=15. So no. Therefore, 3840/289 is simplest form.

Alternatively, maybe the problem expects the answer as a mixed number? But in most cases, improper fraction is acceptable.

Alternatively, maybe I made a mistake in the problem statement? Let me check again.

"In triangle ABC, AB = 17, AC = 8, and BC = 15. Let D be the foot of the altitude from C to AB. Find the area of triangle ACD."

Yes, that's correct.

Alternatively, maybe I can check with another formula for area. For example, using coordinates, the area of ACD can be calculated as the area of ABC minus area of BCD.

Area of ABC is 60. Area of BCD can be calculated as (BD * CD)/2. BD = AB - AD = 17 - 64/17 = (289 - 64)/17 = 225/17. CD is 120/17. Therefore, area of BCD is (225/17 * 120/17)/2 = (27000)/578 = 13500/289. Then area of ACD = area ABC - area BCD = 60 - 13500/289. Let me compute that.

Convert 60 to 289 denominator: 60 = 60 * 289 / 289 = 17340 / 289. Then subtract 13500 / 289: 17340 - 13500 = 3840. Therefore, area of ACD is 3840 / 289. Same result.

Therefore, this confirms the answer.

Therefore, I think the answer is 3840/289.

But let me check if that reduces. Since 3840 divided by 17 is 225.88... and 289 is 17². So no.

Alternatively, maybe the problem expects the answer as a decimal? But in the context of a math competition, fractions are preferred.

Alternatively, maybe I can write it as a mixed number? 13 and 83/289. But again, not sure.

Alternatively, maybe there's a calculation mistake? Let me check once again.

If area of ABC is 60, and area of ACD is 3840/289 ≈ 13.287, and area of BCD is 60 - 13.287 ≈ 46.713. Let me check area of BCD with BD = 225/17 ≈ 13.235, CD = 120/17 ≈ 7.058. Then area is ½ * 13.235 * 7.058 ≈ ½ * 93.4 ≈ 46.7, which matches. So that seems correct.

Therefore, I think the answer is 3840/289.

But let me check if that fraction can be simplified. Since 3840 and 289. 289 is 17². 3840 divided by 17 is 225.88... So no.

Alternatively, maybe the problem had a typo? But assuming the problem is correct, then 3840/289 is the answer.

Alternatively, maybe I can write it as \boxed{\dfrac{3840}{289}}.

But let me check if that's the simplest form.

Alternatively, maybe I can divide numerator and denominator by something? Let me check if 3840 and 289 have any common factors.

As we did before, GCD(3840, 289) = 1. Therefore, yes, it's reduced.

Therefore, I think that is the answer.

**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: Use Heron’s Formula to Find the Area of Triangle $ ABC $

We first compute the area of triangle $ ABC $ using **Heron’s formula**.

- Semi-perimeter:
  $$
  s = \frac{AB + AC + BC}{2} = \frac{17 + 8 + 15}{2} = 20
  $$

- Area of triangle $ ABC $:
  $$
  \text{Area} = \sqrt{s(s - AB)(s - AC)(s - BC)} = \sqrt{20(20 - 17)(20 - 8)(20 - 15)} = \sqrt{20 \cdot 3 \cdot 12 \cdot 5}
  $$
  $$
  = \sqrt{3600} = 60
  $$

So, the area of triangle $ ABC $ is $ 60 $.

---

### Step 2: Use Coordinates to Find the Area of Triangle $ ACD $

Let’s place triangle $ ABC $ on the coordinate plane for clarity:

- Let $ A = (0, 0) $
- Let $ B = (17, 0) $
- Let $ C = (x, y) $

We are given:
- $ AC = 8 \Rightarrow x^2 + y^2 = 64 $
- $ BC = 15 \Rightarrow (x - 17)^2 + y^2 = 225 $

Subtracting the first equation from the second:
$$
(x - 17)^2 + y^2 - (x^2 + y^2) = 225 - 64 \Rightarrow x^2 - 34x + 289 - x^2 = 161
$$
$$
-34x + 289 = 161 \Rightarrow -34x = -128 \Rightarrow x = \frac{64}{17}
$$

Now substitute $ x = \frac{64}{17} $ into $ x^2 + y^2 = 64 $ to find $ y $:
$$
\left(\frac{64}{17}\right)^2 + y^2 = 64 \Rightarrow \frac{4096}{289} + y^2 = 64
$$
$$
y^2 = 64 - \frac{4096}{289} = \frac{18496 - 4096}{289} = \frac{14400}{289} \Rightarrow y = \frac{120}{17}
$$

So, point $ C = \left( \frac{64}{17}, \frac{120}{17} \right) $

Since $ D $ is the foot of the perpendicular from $ C $ to $ AB $, and $ AB $ lies on the x-axis, the x-coordinate of $ D $ is the same as that of $ C $, and the y-coordinate is 0. So:

- $ D = \left( \frac{64}{17}, 0 \right) $

---

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

Triangle $ ACD $ is a right triangle with:

- Base $ AD = \frac{64}{17} $
- Height $ CD = \frac{120}{17} $

So, the area is:
$$
\text{Area} = \frac{1}{2} \cdot AD \cdot CD = \frac{1}{2} \cdot \frac{64}{17} \cdot \frac{120}{17} = \frac{1}{2} \cdot \frac{7680}{289} = \frac{3840}{289}
$$

---

### 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 2, 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!. 10! is 3628800, which has two trailing zeros. Let's apply the formula. 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!... Well, I don't remember the exact value, but I know that 10! has 2, 15! would have 3 (from 15/5=3, 15/25=0), 20! would have 4 (20/5=4, 20/25=0), 25! would have 5 + 1 = 6. That seems right. So the method works.

Therefore, applying it to 42!:

First step: 42 divided by 5 is 8.4, floor to 8.

Second step: 42 divided by 25 is 1.68, floor to 1.

Third step: 42 divided by 125 is 0.336, floor to 0.

So total number of 5s is 8 + 1 = 9. Therefore, there are 9 trailing zeros. But let me check with another approach to be safe.

Alternatively, I can list all the multiples of 5 in 42! and count how many 5s they contribute. The multiples of 5 between 1 and 42 are 5, 10, 15, 20, 25, 30, 35, 40. That's 8 numbers. Each of these contributes at least one 5. However, some numbers contribute more than one. Specifically, numbers that are multiples of 25 contribute an extra 5. In the list above, 25 is the only multiple of 25 (since 25*2=50 is beyond 42). So 25 contributes an extra 5. Therefore, total number of 5s is 8 + 1 = 9. Same result. So that seems to confirm it.

But just to be thorough, let me think if there's another way someone might approach this problem and maybe get confused. For example, someone might think that since 42 divided by 5 is 8.4, so 8 zeros? But that would be wrong because they forget about the numbers that contribute more than one 5. So in this case, 25 is such a number. So adding that extra 1 gives the correct count.

Alternatively, someone might think that since 5*8=40, and then 42-40=2, so maybe 8? But again, missing the extra 5 from 25. So yeah, the formula is definitely the way to go.

Another check: Let me compute 42! and see how many zeros it has. Wait, but 42! is a huge number. I can't compute it directly, but maybe I can compute the number of zeros using another method? Or maybe check with a known value?

Alternatively, think about 40! which is known to have 9 trailing zeros? Wait, let me check. For 40!:

Number of 5s: 40/5 = 8, 40/25 = 1, 40/125 = 0. Total 8 + 1 = 9. So 40! has 9 trailing zeros. Then 41! is 41*40!, so multiplying by 41, which doesn't add any factors of 5, so 41! also has 9 trailing zeros. Similarly, 42! is 42*41!, so multiplying by 42, which is 6*7, so no factors of 5. Therefore, 42! also has 9 trailing zeros. Therefore, the answer is 9. That seems consistent.

Wait, but let me check 45! just to see. For 45! it would be 45/5=9, 45/25=1, 45/125=0, total 10. So 45! has 10 zeros. So between 40! and 45!, the number of zeros increases by 1 when we reach 45. So 42! is between 40! and 45!, so it should still have 9 zeros. Therefore, that seems correct.

Alternatively, if I think about the formula again, the number of trailing zeros in n! is given by the sum floor(n/5) + floor(n/25) + floor(n/125) + ... until the division yields zero. So for n=42:

floor(42/5) = 8

floor(42/25) = 1

floor(42/125) = 0

So total is 8 + 1 = 9. Yep. So that's the answer.

But just to make sure I didn't make any calculation errors. Let me recalculate:

First term: 42 divided by 5. 5*8=40, 42-40=2, so yes, floor is 8.

Second term: 42 divided by 25. 25*1=25, 42-25=17, so floor is 1.

Third term: 42 divided by 125. 125 is bigger than 42, so floor is 0. So adding them up, 8+1=9. Correct.

Therefore, I think the answer is 9. But just to be 100% sure, let me think if there's any other factors that could contribute to trailing zeros? For example, factors of 2 and 5. But since we already accounted for all the 5s, and there are more 2s, so the number of 2s is not a limiting factor here. So the number of 5s is the limiting factor, so the number of trailing zeros is equal to the number of 5s. Therefore, 9 is correct.

Alternatively, if I wanted to check with another example, say 25! which we said had 6 zeros. Let me compute floor(25/5) + floor(25/25) = 5 + 1 = 6. Correct. And 25! is 15511210043330985984000000, which indeed has 6 trailing zeros. So that works.

Another example: 100! has floor(100/5) + floor(100/25) + floor(100/125) = 20 + 4 + 0 = 24 trailing zeros. Which I think is correct. So the formula works.

Therefore, applying the same formula to 42! gives 9. Therefore, I think the answer is 9.

**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 factorial $ n! $, the number of 2s in the prime factorization is always greater than the number of 5s. Therefore, the number of trailing zeros in $ n! $ 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:

$$
\left\lfloor \frac{42}{5} \right\rfloor + \left\lfloor \frac{42}{25} \right\rfloor + \left\lfloor \frac{42}{125} \right\rfloor + \cdots
$$

This works because:

- Every multiple of 5 contributes at least one 5.
- Every multiple of 25 contributes an additional 5 (since $ 25 = 5^2 $).
- Every multiple of 125 contributes yet another 5 (since $ 125 = 5^3 $), 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 together:

$$
8 + 1 + 0 = 9
$$

---

### Final Answer

Thus, 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 2, 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!. 10! is 3628800, which has two trailing zeros. Let's apply the formula. 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!... Well, I don't remember the exact value, but I know that 10! has 2, 15! would have 3 (from 15/5=3, 15/25=0), 20! would have 4 (20/5=4, 20/25=0), 25! would have 5 + 1 = 6. That seems right. So the method works.

Therefore, applying it to 42!:

First step: 42 divided by 5 is 8.4, floor to 8.

Second step: 42 divided by 25 is 1.68, floor to 1.

Third step: 42 divided by 125 is 0.336, floor to 0.

So total number of 5s is 8 + 1 = 9. Therefore, there are 9 trailing zeros. But let me check with another approach to be safe.

Alternatively, I can list all the multiples of 5 in 42! and count how many 5s they contribute. The multiples of 5 between 1 and 42 are 5, 10, 15, 20, 25, 30, 35, 40. That's 8 numbers. Each of these contributes at least one 5. However, some numbers contribute more than one. Specifically, numbers that are multiples of 25 contribute an extra 5. In the list above, 25 is the only multiple of 25 (since 25*2=50 is beyond 42). So 25 contributes an extra 5. Therefore, total number of 5s is 8 + 1 = 9. Same result. So that seems to confirm it.

But just to be thorough, let me think if there's another way someone might approach this problem and maybe get confused. For example, someone might think that since 42 divided by 5 is 8.4, so 8 zeros? But that would be wrong because they forget about the numbers that contribute more than one 5. So in this case, 25 is such a number. So adding that extra 1 gives the correct count.

Alternatively, someone might think that since 5*8=40, and then 42-40=2, so maybe 8? But again, missing the extra 5 from 25. So yeah, the formula is definitely the way to go.

Another check: Let me compute 42! and see how many zeros it has. Wait, but 42! is a huge number. I can't compute it directly, but maybe I can compute the number of zeros using another method? Or maybe check with a known value?

Alternatively, think about 40! which is known to have 9 trailing zeros? Wait, let me check. For 40!:

Number of 5s: 40/5 = 8, 40/25 = 1, 40/125 = 0. Total 8 + 1 = 9. So 40! has 9 trailing zeros. Then 41! is 41*40!, so multiplying by 41, which doesn't add any factors of 5, so 41! also has 9 trailing zeros. Similarly, 42! is 42*41!, so multiplying by 42, which is 6*7, so no factors of 5. Therefore, 42! also has 9 trailing zeros. Therefore, the answer is 9. That seems consistent.

Wait, but let me check 45! just to see. For 45! it would be 45/5=9, 45/25=1, 45/125=0, total 10. So 45! has 10 zeros. So between 40! and 45!, the number of zeros increases by 1 when we reach 45. So 42! is between 40! and 45!, so it should still have 9 zeros. Therefore, that seems correct.

Alternatively, if I think about the formula again, the number of trailing zeros in n! is given by the sum floor(n/5) + floor(n/25) + floor(n/125) + ... until the division yields zero. So for n=42:

floor(42/5) = 8

floor(42/25) = 1

floor(42/125) = 0

So total is 8 + 1 = 9. Yep. So that's the answer.

But just to make sure I didn't make any calculation errors. Let me recalculate:

First term: 42 divided by 5. 5*8=40, 42-40=2, so yes, floor is 8.

Second term: 42 divided by 25. 25*1=25, 42-25=17, so floor is 1.

Third term: 42 divided by 125. 125 is bigger than 42, so floor is 0. So adding them up, 8+1=9. Correct.

Therefore, I think the answer is 9. But just to be 100% sure, let me think if there's any other factors that could contribute to trailing zeros? For example, factors of 2 and 5. But since we already accounted for all the 5s, and there are more 2s, so the number of 2s is not a limiting factor here. So the number of 5s is the limiting factor, so the number of trailing zeros is equal to the number of 5s. Therefore, 9 is correct.

Alternatively, if I wanted to check with another example, say 25! which we said had 6 zeros. Let me compute floor(25/5) + floor(25/25) = 5 + 1 = 6. Correct. And 25! is 15511210043330985984000000, which indeed has 6 trailing zeros. So that works.

Another example: 100! has floor(100/5) + floor(100/25) + floor(100/125) = 20 + 4 + 0 = 24 trailing zeros. Which I think is correct. So the formula works.

Therefore, applying the same formula to 42! gives 9. Therefore, I think the answer is 9.

**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 factorial $ n! $, the number of 2s in the prime factorization is always greater than the number of 5s. Therefore, the number of trailing zeros in $ n! $ 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:

$$
\left\lfloor \frac{42}{5} \right\rfloor + \left\lfloor \frac{42}{25} \right\rfloor + \left\lfloor \frac{42}{125} \right\rfloor + \cdots
$$

This works because:

- Every multiple of 5 contributes at least one 5.
- Every multiple of 25 contributes an additional 5 (since $ 25 = 5^2 $).
- Every multiple of 125 contributes yet another 5 (since $ 125 = 5^3 $), 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 together:

$$
8 + 1 + 0 = 9
$$

---

### Final Answer

Thus, 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.