To find the largest area of an isosceles triangle with two equal sides of length 3, we need to maximize the area of the triangle as a function of the angle between the two equal sides. \[\] Formula for the area of the triangle The area \(A\) of an isosceles triangle with two sides of length 3 and an included angle \(\theta\) is given by the formula: \[ A = \frac{1}{2} \times 3 \times 3 \times \sin(\theta) \] \[ A = \frac{9}{2} \sin(\theta) \] \[\]Maximizing the area The function \(\sin(\theta)\) reaches its maximum value of 1 when \(\theta = 90^\circ\). Therefore, the maximum area occurs when the angle between the two equal sides is \(90^\circ\). \[\] Calculate the largest area Substituting \(\sin(90^\circ) = 1\) into the area formula: \[ A_{\text{max}} = \frac{9}{2} \times 1 = \frac{9}{2} \] So, the largest area of the isosceles triangle is: \[ \frac{9}{2} \]