KM \( =6 \mathrm{~cm} \) \( A Y=3 \mathrm{~cm} \) Area of \( \triangle K M J=34 \mathrm{~cm}^{2} \) the area of \( \triangle A Y X \). Your answer may be exact or rounded to the nearest hundredth.

To find the area of triangle A Y X given the context, we can start by relating the two triangles using their bases and heights. Assuming that triangles KMJ and AYX are similar (which is common in problems involving proportional relationships and similar shapes), we can use the bases KM and AY to set up a ratio for determining the area of AYX. Since the length of KM is 6 cm and AY is 3 cm, the ratio of their bases is 3:6, or 1:2. The area of triangle KMJ is given as 34 cm². When calculating the area of triangle AYX using the ratio of the squares of the lengths (since area is proportional to the square of the side lengths), it will be: \[ \text{Area}_{AYX} = \text{Area}_{KMJ} \times \left(\frac{\text{base } AY}{\text{base } KM}\right)^2 = 34 \times \left(\frac{3}{6}\right)^2 = 34 \times \left(\frac{1}{2}\right)^2 = 34 \times \frac{1}{4} = 8.5 \text{ cm}^2. \] So, the area of triangle A Y X is 8.5 cm².