To solve the problem:

Given Data:

1. Triangle $ABC$ has an area of 24 cm².
2. $\angle BAC = \angle ACD$.
3. $AC: BC = 4:3$.
4. $\angle ACD$ is a right angle.

Goal:

Find the area of triangle $ACD$.

Step 1: Relationship Between Triangles

Since $\angle BAC = \angle ACD$ and triangle $ACD$ is a right triangle, triangles $ABC$ and $ACD$ are similar (by AA similarity).

The ratio of their corresponding sides will be the same as the square root of their area ratio.

Step 2: Determine Dimensions of Triangle $ABC$

The ratio $AC: BC = 4:3$ implies: Let:

• $AC = 4x$,
• $BC = 3x$.

Using the area formula for triangle $ABC$: $\text{Area of } \triangle ABC = \frac{1}{2} \times AC \times BC \times \sin(\angle BAC).$ Substitute $AC = 4x$ and $BC = 3x$: $24 = \frac{1}{2} \times (4x) \times (3x) \times \sin(\angle BAC).$ $24 = 6x^2 \sin(\angle BAC).$ $\sin(\angle BAC) = \frac{24}{6x^2} = \frac{4}{x^2}.$

Step 3: Area of Triangle $ACD$

Using similarity: $\frac{\text{Area of } \triangle ACD}{\text{Area of } \triangle ABC} = \left(\frac{AD}{AB}\right)^2.$ From similarity, $\frac{AD}{AB} = \frac{AC}{AC + BC} = \frac{4x}{4x + 3x} = \frac{4}{7}$.

Thus: $\frac{\text{Area of } \triangle ACD}{24} = \left(\frac{4}{7}\right)^2.$ $\text{Area of } \triangle ACD = 24 \times \frac{16}{49}.$ $\text{Area of } \triangle ACD = \frac{384}{49} \, \text{cm}^2 \, \approx \, 7.84 \, \text{cm}^2.$

Final Answer:

The area of triangle $ACD$ is approximately 7.84 cm².

Let me know if you'd like further clarification or detailed steps!

Related Questions:

1. What is the length of side $AC$ in triangle $ABC$?
2. How does the area ratio relate to side ratios in similar triangles?
3. Can you calculate $\angle BAC$ if the sine is known?
4. What is the perimeter of triangle $ACD$?
5. How do properties of right triangles simplify calculations?

Tip:

In geometry, always check for similarity or congruence when dealing with related triangles—it simplifies computations!