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Given datai-Areec of froting\begin{array}{l}A=\frac{\text { Totel arial load }}{\text { allowable 5oil pressume }} \\=\frac{Q_{1}+Q_{2}}{9}=\frac{(500+750) 10^{3}}{120 \times 10^{3}} \\=10.41 \mathrm{~m}^{2} \\\end{array}Takemoment about A\begin{array}{l} \sum M_{A}=0 \\-Q_{2} \times L_{3}+\left[\left(Q+2_{2}\right) \times x\right]=0 \\-(750 \times 45)+[(500+750) x]=0 \\1250 x=3375 \\x=2.7 \mathrm{~m}\end{array}Now from tree body diagrume.\begin{aligned} \frac{i}{2} & =x+L_{2} \quad \text { Therefore, the lengtn of } \\ L & =2[2.7+1] \text { footing is } 7.4 \mathrm{~m} \\ \Rightarrow[L & =7.4 \mathrm{~m}]\end{aligned}\begin{array}{l}\text { Now, Area }=L \times W \\W=\frac{A}{L}=\frac{10.41}{7.4} \\\Rightarrow w=1.406 \mathrm{~m} \\\end{array}\rightarrow Therefore, The width of footing is 1.406 \mathrm{~m} Pleuse like my Answer. Thmk you ...