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.