It is given that the triangles are similar, that is,
![\triangle ABC\sim\triangle XYZ](https://img.qammunity.org/2023/formulas/mathematics/high-school/pdt1eqwmoo4dkc7dyuhulejzdgxknuff9q.png)
It is required to find the area of the larger triangle.
Recall that the scale factor, k of similar figures is the ratio of their corresponding sides:
![k=(AC)/(XZ)](https://img.qammunity.org/2023/formulas/mathematics/high-school/eezfxkts2t4j87pn31evplix2biuij8lq2.png)
Substitute AC=5 and XZ=15 into the equation:
![k=(5)/(15)=(1)/(3)](https://img.qammunity.org/2023/formulas/mathematics/high-school/7ytcxdnlsxfgm24qzgc5udmfn3alh7fsqn.png)
Hence, the scale factor is 1/3.
Recall that as per the Area of Similar Figures, the ratio of areas for two similar figures with a scale factor, k is:
![k^2](https://img.qammunity.org/2023/formulas/mathematics/college/bj6boyph4x07eql6xoflecw156836rua0v.png)
This implies that:
![\frac{\text{area of }\triangle ABC}{\text{area of }\triangle XYZ}=k^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/7ijtcm23g6uig475h02tak4xpkbhegyg97.png)
Substitute the values of the area of triangle ABC and the scale factor into the proportion:
![\Rightarrow\frac{48}{\text{area of }\triangle XYZ}=((1)/(3))^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/youtf16oc4r4vqazzjsnvmmvtjvzhiyfbk.png)
Let the area of ΔXYZ be A, and solve for A in the equation:
![\begin{gathered} (48)/(A)=(1)/(9) \\ \Rightarrow A=9*48=432ft^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/puxnmfsbq41lb0pjlks7qw8ybhbzplfm2s.png)
The required answer is 432 square ft.
The last choice is the answer.