Identify the structure of compound with C₉H₁₀ formula using IR and NMR data
The molecular formula of the compound is $\ce{C9H10}$. The number of unsaturations in the compound of $\ce{C_nH_m}$ is $\frac{2n+1-n}{2}$. Thus, the number of unsaturations in $\ce{C9H10}$ is $\frac{2\times 9+1-10}{2}=5$. You already found out that compound is aromatic by FT-IR, and hence $\ce{C6}$ of the formula counted out with 4 unsaturaations (3 double bond and a ring for aromatic nucleus).The remaining number of unsaturarion is $5-4=1$, and remaining number of $\ce{C}$ in the formula is 3. The remaining unsaturation could be a double bond or a ring.
When look at $\ce{^1H}$-$\mathrm{NMR}$ spectrum, chemical shifts of all but aromatic peaks are less than $\pu{3.00 ppm}$. Therefore, it is safe to conclude that the compound is a aromatic bicyclic rather than aromatic alkene. Therefore, the compound is indane:

The protons on 1-$\ce{C}$ and 3-$\ce{C}$ are chemically and magnetically equivalent, and benzylic, hence the their relevant signal would be down-field to that of protons on 2-$\ce{C}$, which is usual acyclic aliphatic carbon. The benzylic protons have two vicinal protons, thus the corresponding peak is a triplet. Meanwhile, protons on 2-$\ce{C}$ have four vicinal protons (benzylic protons), thus the corresponding peak is a quintet (5 peaks).
Note: The integration given is a ratio, not correct number of protons.