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Let's break down and solve the problem based on the image you provided. Here's a summary of the given information and the questions:

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

**Questions:**
1. What is the measure of \( \angle BAC \)?
2. What is the length of \( AB \)?
3. What is the perimeter of triangle \( ABC \)?
4. What is the area of quadrilateral \( ABCD \)?

### Solution Steps:

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

Using the fact that the sum of angles in a triangle is \(180^\circ\), for triangle \(ABC\):
\[
\angle BAC = 180^\circ - \angle ABC - \angle ACB
\]
Substitute the given values:
\[
\angle BAC = 180^\circ - 90^\circ - 45^\circ = 45^\circ
\]

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

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

We know that triangle \( ABC \) is a right triangle with \( \angle ACB = 45^\circ \). In a right triangle where one of the angles is \( 45^\circ \), the two legs are equal in length.

Since \( BC = 10 \, \text{cm} \), we can use the following relationship for a \(45^\circ-45^\circ-90^\circ\) triangle:
\[
AB = BC = 10 \, \text{cm}
\]

Thus, the length of \( AB \) is also \( 10 \, \text{cm} \).

#### (c) Perimeter of triangle \( ABC \)

The perimeter of a triangle is the sum of the lengths of all its sides. In triangle \(ABC\), the sides are \( AB = 10 \, \text{cm} \), \( BC = 10 \, \text{cm} \), and the hypotenuse \( AC \).

To find \( AC \), use the Pythagorean theorem:
\[
AC^2 = AB^2 + BC^2 = 10^2 + 10^2 = 100 + 100 = 200
\]
\[
AC = \sqrt{200} = 10\sqrt{2} \, \text{cm} \approx 14.14 \, \text{cm}
\]

Now, the perimeter is:
\[
\text{Perimeter} = AB + BC + AC = 10 + 10 + 14.14 = 34.14 \, \text{cm}
\]

#### (d) Area of quadrilateral \( ABCD \)

The quadrilateral \( ABCD \) is composed of two triangles: \( ABC \) and \( ACD \).

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

2. **Area of triangle \( ACD \):**
   To find the area of triangle \( ACD \), we need the height and base. The base \( AD \) can be found using trigonometry. From \( \angle CAD = 60^\circ \) and \( \angle ADC = 90^\circ \), we can find the height \( CD \).

   Since \( \sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} \), we can find \( CD \) as:
   \[
   \sin(60^\circ) = \frac{CD}{AC} \quad \Rightarrow \quad CD = AC \times \sin(60^\circ)
   \]
   \[
   CD = 10\sqrt{2} \times \frac{\sqrt{3}}{2} = 10\sqrt{6} \, \text{cm}
   \]

   For now, we'd calculate the area using approximation of the triangle formula. Shall I compute or clarify this?

Would you like more detailed steps for \( \text{Area of Quadrilateral ABCD} \)?  

### Next Steps:
1. Would you like to calculate other quantities for quadrilateral exactly?
2. Shall I clarify question or

In the figure, BC = 10 cm, ∠ABC = ∠ADC = 90°, ∠ACB = 45°, and ∠CAD = 60°. What is the measure of ∠BAC? What is the length of AB? What is the perimeter of triangle ABC? What is the area of quadrilateral ABCD?

Solve a geometry problem involving a right triangle ABC with BC = 10 cm, ∠ABC = 90°, ∠ACB = 45°, and a quadrilateral ABCD. Learn how to find angles, side lengths, perimeter, and area step by step using trigonometry and geometry formulas.

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