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To solve this problem, let’s analyze the given information and find the missing values.

### Given:
- Triangle \( ABC \) has a right altitude \( CD = 5 \) (where \( D \) is the foot of the altitude from \( C \) to \( AB \)).
- \( \angle A = 30^\circ \) and \( \angle B = 45^\circ \).

### Objective:
1. **Find the lengths of sides \( AC \) and \( BC \).**
2. **Calculate the area of triangle \( ABC \).**
3. **Calculate the perimeter of triangle \( ABC \).**

### Solution Steps:

1. **Determine \( \angle C \):**
   Since the angles in a triangle sum up to \( 180^\circ \),
   \[
   \angle C = 180^\circ - 30^\circ - 45^\circ = 105^\circ.
   \]

2. **Using right triangle properties in \( \triangle ACD \) and \( \triangle BCD \):**

   - In \( \triangle ACD \), we have \( \angle ACD = 90^\circ \) and \( \angle A = 30^\circ \).
     - Since \( CD = 5 \), we can use the properties of a 30-60-90 triangle:
       - In a 30-60-90 triangle, the ratio of the sides opposite the 30°, 60°, and 90° angles is \( 1 : \sqrt{3} : 2 \).
       - Here, \( CD \) (opposite the 60°) is \( 5 \), so:
         \[
         AD = \frac{CD}{\sqrt{3}} = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3}.
         \]
         \[
         AC = 2 \times AD = 2 \times \frac{5\sqrt{3}}{3} = \frac{10\sqrt{3}}{3}.
         \]

   - In \( \triangle BCD \), we have \( \angle BCD = 90^\circ \) and \( \angle B = 45^\circ \).
     - Since it is a 45-45-90 triangle, the sides are in the ratio \( 1 : 1 : \sqrt{2} \).
       - Here, \( CD = 5 \), so:
         \[
         BD = CD = 5.
         \]
         \[
         BC = BD \times \sqrt{2} = 5\sqrt{2}.
         \]

3. **Calculate the area of \( \triangle ABC \):**
   The area \( A \) of a triangle can be found using:
   \[
   A = \frac{1}{2} \times \text{base} \times \text{height}.
   \]
   Here, the base \( AB = AD + BD = \frac{5\sqrt{3}}{3} + 5 \) and the height is \( CD = 5 \).

   So,
   \[
   A = \frac{1}{2} \times \left(\frac{5\sqrt{3}}{3} + 5\right) \times 5 = \frac{5}{2} \times \left(\frac{5\sqrt{3}}{3} + 5\right).
   \]

4. **Calculate the perimeter of \( \triangle ABC \):**
   \[
   \text{Perimeter} = AB + AC + BC = \left(\frac{5\sqrt{3}}{3} + 5\right) + \frac{10\sqrt{3}}{3} + 5\sqrt{2}.
   \]

Would you like the calculations to be simplified further, or do you have any specific questions about any of these steps?

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### Additional Questions to Explore:
1. How would the calculations change if \( CD \) was a different length?
2. Can you prove why a 30-60-90 triangle has the side ratios \( 1 : \sqrt{3} : 2 \)?
3. How does trigonometry help in solving non-right triangles?
4. What would happen to the area if the triangle was scalene and did not have any specific angle measures?
5. How could the Law of Sines or Cosines be used to verify the perimeter?

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### Tip:
When working with right triangles, always check if they follow special angle patterns (like 30-60-90 or 45-45-90), as this can simplify calculations significantly.

Find the area and perimeter of the triangle with given angles (30° and 45°) and altitude CD = 5, as shown in the figure.

This solution shows how to find the area and perimeter of a triangle with angles of 30° and 45° and a given altitude of 5. It uses properties of special triangles (30-60-90 and 45-45-90) and trigonometry to determine side lengths. Suitable for high school students.

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