AB = AC but it's not equal to 10cm
Only the radius of a circle around it equals 10cm

Thank you for the clarification. So, in this problem, \(AB = AC\) but they are not necessarily equal to 10 cm. Instead, 10 cm is the radius of the circumcircle of the triangle. Let's revise the solution accordingly.

### Revised Problem Information:

- Triangle \( ABC \) is isosceles, with \( AB = AC \).
- The circumcircle of the triangle has a radius of 10 cm.
- Let \( \alpha = \angle BAC \).

### Revised Solution Approach:

#### Part (a): Expressing the Area of \( \triangle ABC \) as a Function of \( \alpha \)

Since \( AB = AC \), and these are the legs of the isosceles triangle, the center of the circumcircle is the midpoint of \( BC \) and lies on the perpendicular bisector of \( BC \). This setup allows us to use the radius of the circle to express the sides of the triangle.

- The radius of the circumcircle \( R \) is given as 10 cm.
- For an isosceles triangle inscribed in a circle (circumscribed), the formula relating the radius \( R \), side \( s \), and angle \( \alpha \) at the apex (vertex angle \( BAC \)) is:

  \[
  R = \frac{s}{2 \sin\left(\frac{\alpha}{2}\right)}
  \]

  Plugging \( R = 10 \) cm:

  \[
  10 = \frac{s}{2 \sin\left(\frac{\alpha}{2}\right)}
  \]

  \[
  s = 20 \sin\left(\frac{\alpha}{2}\right)
  \]

Thus, the length of each side \( AB = AC = s \) can be expressed in terms of \( \alpha \).

- To find the area \( A \) of \( \triangle ABC \), we use the general formula for the area of a triangle given two sides and the included angle:

  \[
  \text{Area} = \frac{1}{2} \times AB \times AC \times \sin(\alpha)
  \]

  Substituting \( AB = AC = s \):

  \[
  \text{Area} = \frac{1}{2} \times s \times s \times \sin(\alpha)
  \]

  \[
  \text{Area} = \frac{1}{2} \times \left(20 \sin\left(\frac{\alpha}{2}\right)\right)^2 \times \sin(\alpha)
  \]

  \[
  \text{Area} = 200 \sin^2\left(\frac{\alpha}{2}\right) \sin(\alpha)
  \]

#### Part (b): Calculating the Area When \( \alpha = 90^\circ \)

- Substitute \( \alpha = 90^\circ \) into the area expression:

  \[
  \text{Area} = 200 \sin^2\left(\frac{90^\circ}{2}\right) \sin(90^\circ)
  \]

  \[
  \text{Area} = 200 \sin^2(45^\circ) \times 1
  \]

  \[
  \text{Area} = 200 \left(\frac{\sqrt{2}}{2}\right)^2
  \]

  \[
  \text{Area} = 200 \times \frac{1}{2}
  \]

  \[
  \text{Area} = 100 \, \text{cm}^2
  \]

### Summary:

1. The area of \( \triangle ABC \) as a function of \( \alpha \) is \( 200 \sin^2\left(\frac{\alpha}{2}\right) \sin(\alpha) \).
2. The area of \( \triangle ABC \) is \( 100 \, \text{cm}^2 \) when \( \alpha = 90^\circ \).

Would you like more details or have any questions?

### Related Questions:

1. How does the length of \( BC \) change as \( \alpha \) varies?
2. What would the area of \( \triangle ABC \) be if \( \alpha = 60^\circ \)?
3. How can the height of \( \triangle ABC \) be calculated using \( R \) and \( \alpha \)?
4. How do the properties of an isosceles triangle inscribed in a circle differ from those of a general triangle?
5. How can you calculate the perimeter of \( \triangle ABC \) as a function of \( \alpha \)?

### Tip:

For problems involving circles and triangles, drawing auxiliary lines such as altitudes, medians, or angle bisectors can often reveal relationships that make it easier to solve for unknowns.

Understand how to solve a geometry problem involving an isosceles triangle with sides AB = AC and a circumcircle radius of 10 cm. Explore the relationship between triangle sides and the circumcircle radius using trigonometric functions. Suitable for high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation