https://file.aiquickdraw.com/mathbot/file/1729524249799nj5wg7pg.jpg

Let's solve the problems based on the figure provided.

### Given Information:
- \( BC = 10 \, \text{cm} \)
- \( \angle ABC = 90^\circ \)
- \( \angle ADC = 90^\circ \)
- \( \angle ACB = 45^\circ \)
- \( \angle CAD = 60^\circ \)

### a) **Measure of \( \angle BAC \):**

In triangle \( ABC \), we already know:
- \( \angle ABC = 90^\circ \)
- \( \angle ACB = 45^\circ \)

Since the sum of angles in any triangle is \( 180^\circ \):
\[
\angle BAC = 180^\circ - \angle ABC - \angle ACB = 180^\circ - 90^\circ - 45^\circ = 45^\circ
\]

Thus, \( \angle BAC = 45^\circ \).

---

### b) **Length of \( AB \):**

Since \( \triangle ABC \) is a right triangle with \( \angle ACB = 45^\circ \), this makes \( \triangle ABC \) a 45°-45°-90° triangle.

In a 45°-45°-90° triangle, the sides follow the ratio:
\[
1:1:\sqrt{2}
\]
This means:
\[
AB = BC \times \sqrt{2} = 10 \times \sqrt{2} \approx 10 \times 1.414 = 14.14 \, \text{cm}
\]

---

### c) **Perimeter of Triangle \( ABC \):**

We know:
- \( BC = 10 \, \text{cm} \)
- \( AB = 14.14 \, \text{cm} \)
- Since \( \triangle ABC \) is a 45°-45°-90° triangle, \( AC = AB = 14.14 \, \text{cm} \).

Thus, the perimeter is:
\[
\text{Perimeter of } \triangle ABC = AB + BC + AC = 14.14 + 10 + 14.14 = 38.28 \, \text{cm}
\]

---

### d) **Area of Quadrilateral \( ABCD \):**

The quadrilateral \( ABCD \) consists of two right triangles \( ABC \) and \( ACD \).

1. **Area of Triangle \( ABC \):**
\[
\text{Area of } \triangle ABC = \frac{1}{2} \times BC \times AB = \frac{1}{2} \times 10 \times 10 = 50 \, \text{cm}^2
\]

2. **Area of Triangle \( ACD \):**
Since \( \triangle ACD \) is also a right triangle with \( \angle ADC = 90^\circ \), we can use the known side \( AC = 14.14 \, \text{cm} \). To calculate \( CD \), we can apply the Law of Sines in \( \triangle ACD \).

Using the Law of Sines:
\[
\frac{CD}{\sin 45^\circ} = \frac{AC}{\sin 60^\circ} \Rightarrow CD = \frac{14.14 \times \sin 45^\circ}{\sin 60^\circ}
\]
\[
CD = \frac{14.14 \times 0.707}{0.866} \approx 11.55 \, \text{cm}
\]

Now, the area of \( \triangle ACD \) is:
\[
\text{Area of } \triangle ACD = \frac{1}{2} \times AC \times CD = \frac{1}{2} \times 14.14 \times 11.55 \approx 81.65 \, \text{cm}^2
\]

Thus, the total area of quadrilateral \( ABCD \) is:
\[
\text{Area of } ABCD = \text{Area of } \triangle ABC + \text{Area of } \triangle ACD = 50 + 81.65 = 131.65 \, \text{cm}^2
\]

---

Let me know if you'd like further clarifications, or if you want details for any part of the solution! 

---

### Related Questions:
1. How would the perimeter of quadrilateral \( ABCD \) be calculated?
2. Can you calculate the length of \( AD \) using the Pythagorean theorem?
3. What are the other types of special right triangles besides 45°-45°-90°?
4. How does the Law of Sines work in a general triangle?
5. How would the solution change if \( \angle ACB \) were different?

### Tip:
For right triangles, identifying if they follow special angle ratios like 45°-45°-90° or 30°-60°-90° can simplify the calculation of side lengths.

In the figure BC = 10 cm, ∠ABC = ∠ADC = 90°, ∠ACB = 45°, ∠CAD = 60°. Solve for the following: (a) What is the measure of ∠BAC? (b) What is the length of AB? (c) What is the perimeter of the triangle ABC? (d) What is the area of the quadrilateral ABCD?

Learn how to solve for angles, side lengths, and area in a geometric problem involving a 45°-45°-90° triangle and a quadrilateral. This problem includes step-by-step solutions using the angle sum property, the 45°-45°-90° triangle ratio, the Law of Sines, and more. Suitable for Grades 9-11.

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