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.