In a right triangle \\(ABC\\), where \\( \angle C \\) is right angled, if the ratio of the lengths of \\( AC \\) and \\( BC \\) is \\( 5:12 \\) and the length of the hypotenuse \\( AB \\) is \\( 13 \\) cm, then what will be the lengths of \\( AC \\) and \\( BC \\)? Also, calculate the area of ​​the triangle and state whether it is more than \\( 40 \\) sq. cm. Also, if \\( D \\) is a point on \\( AC \\) and the length of \\( AD \\) is \\( 7 \\) cm, then what will be the length of \\( DC\\)?

### Step 1: Determine the lengths of \( AC \) and \( BC \)

We are given that the ratio of the lengths of \( AC \) and \( BC \) is \( 5:12 \), and the hypotenuse \( AB \) is \( 13 \) cm. Since \( \triangle ABC \) is a right triangle, we can apply the Pythagorean theorem:

\[
AB^2 = AC^2 + BC^2
\]

Let the lengths of \( AC \) and \( BC \) be \( 5x \) and \( 12x \), respectively, where \( x \) is a scaling factor. Then, according to the Pythagorean theorem:

\[
(5x)^2 + (12x)^2 = 13^2
\]

Simplifying:

\[
25x^2 + 144x^2 = 169
\]
\[
169x^2 = 169
\]
\[
x^2 = 1
\]
\[
x = 1
\]

Therefore:
- \( AC = 5x = 5 \) cm
- \( BC = 12x = 12 \) cm

### Step 2: Calculate the area of the triangle

The area of a right triangle is given by:

\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]

Here, we can consider \( AC \) as the base and \( BC \) as the height, so:

\[
\text{Area} = \frac{1}{2} \times 5 \times 12 = \frac{1}{2} \times 60 = 30 \, \text{sq. cm}
\]

Since the area is \( 30 \, \text{sq. cm} \), it is **less than** \( 40 \, \text{sq. cm} \).

### Step 3: Determine the length of \( DC \)

If \( D \) is a point on \( AC \) such that the length of \( AD \) is \( 7 \) cm, we can find \( DC \) by subtracting the length of \( AD \) from \( AC \):

\[
DC = AC - AD = 5 - 7 = -2 \, \text{cm}
\]

However, a negative length doesn't make sense geometrically. This means \( AD \) cannot be greater than \( AC \), so there seems to be an issue with the value of \( AD = 7 \, \text{cm} \), as it exceeds the total length of \( AC = 5 \, \text{cm} \).

### Summary:
- \( AC = 5 \, \text{cm} \)
- \( BC = 12 \, \text{cm} \)
- The area of the triangle is \( 30 \, \text{sq. cm} \), which is **less than** \( 40 \, \text{sq. cm} \).
- The value of \( AD = 7 \, \text{cm} \) leads to a contradiction since \( AD > AC \), so this condition is impossible as stated.

Would you like further details on any of these calculations?

### Related Questions:
1. What happens if \( AD \) is less than \( AC \)?
2. How can we find the missing length using similar triangles in this case?
3. How would the area change if the ratio of \( AC \) to \( BC \) were different?
4. What happens if the hypotenuse were shorter or longer than 13 cm?
5. How does the Pythagorean theorem generalize to non-right triangles?

**Tip:** Always ensure that all lengths in a triangle adhere to geometric constraints, such as the total length of a segment being greater than or equal to any of its sub-segments.

In a right triangle ABC, where \( \angle C \) is right-angled, if the ratio of the lengths of AC and BC is 5:12 and the length of the hypotenuse AB is 13 cm, then what will be the lengths of AC and BC? Also, calculate the area of ​​the triangle and state whether it is more than 40 sq. cm. Also, if D is a point on AC and the length of AD is 7 cm, then what will be the length of DC?

Solve a right triangle problem where the ratio of AC to BC is 5:12 and AB is 13 cm. Learn how to find the lengths of AC and BC, calculate the area, and check if it's more than 40 sq. cm. Apply the Pythagorean theorem and basic geometry concepts. Suitable for students in Grades 7-9.

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