To solve the problem and find the area of the isosceles triangle $\triangle ABC$, I will extract the coordinates from the graph.

The vertices of the triangle appear to be approximately:

• $A(0, 0)$
• $B(4, 0)$
• $C(0, 3)$

Steps:

1. Base $AB$: The length of $AB$ is 4 units.
2. Height $AC$: The length of $AC$ is 3 units.

Area of $\triangle ABC$:

$\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 4 \times 3 = 6 \text{ square units}$

The area of $\triangle ABC$ is $6$ square units.

Would you like any further details or have any questions?

Here are 5 related questions to consider:

1. How do you determine the lengths of the sides of a triangle given the coordinates?
2. What are the characteristics of an isosceles triangle?
3. How can you calculate the area of a triangle using different methods (e.g., Heron's formula)?
4. How would the area change if the triangle were not isosceles?
5. Can you find the perimeter of this triangle?

Tip: When dealing with geometric figures on a coordinate plane, always identify and confirm the exact coordinates of the vertices before proceeding with calculations.