Triangle EFG has vertices E(–3, 4), F(–5, –1), and G(1, 1). The triangle is translated so that the coordinates of the image are
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First off, let's convert the decimal to a fraction, notice, we have two decimals, so we'll use in the denominator, a 1 with two zeros then, two decimals, two zeros, thus 
now, we know then the ratio dimensions for the new photograph,
![\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\ -------------------------------\\\\ \cfrac{7}{4}\implies \cfrac{4+3}{4}\implies \cfrac{4}{4}+\cfrac{3}{4}\implies 1+\boxed{\cfrac{3}{4}}\impliedby \textit{perimeter is }\frac{3}{4}\textit{ larger} \\\\\\ \stackrel{areas'~ratio}{\cfrac{s^2}{s^2}}\implies \cfrac{3^2}{4^2}\implies \cfrac{9}{16}\impliedby \textit{area is }\frac{9}{16}\textit{ larger than original}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Ctextit%7Bsimilar%20shape%7D%7D%7B%5Ctextit%7Bsimilar%20shape%7D%7D%5Cqquad%20%5Ccfrac%7Bs%7D%7Bs%7D%3D%5Ccfrac%7B%5Csqrt%7Bs%5E2%7D%7D%7B%5Csqrt%7Bs%5E2%7D%7D%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Ccfrac%7B7%7D%7B4%7D%5Cimplies%20%5Ccfrac%7B4%2B3%7D%7B4%7D%5Cimplies%20%5Ccfrac%7B4%7D%7B4%7D%2B%5Ccfrac%7B3%7D%7B4%7D%5Cimplies%201%2B%5Cboxed%7B%5Ccfrac%7B3%7D%7B4%7D%7D%5Cimpliedby%20%5Ctextit%7Bperimeter%20is%20%7D%5Cfrac%7B3%7D%7B4%7D%5Ctextit%7B%20larger%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bareas%27~ratio%7D%7B%5Ccfrac%7Bs%5E2%7D%7Bs%5E2%7D%7D%5Cimplies%20%5Ccfrac%7B3%5E2%7D%7B4%5E2%7D%5Cimplies%20%5Ccfrac%7B9%7D%7B16%7D%5Cimpliedby%20%5Ctextit%7Barea%20is%20%7D%5Cfrac%7B9%7D%7B16%7D%5Ctextit%7B%20larger%20than%20original%7D)
now, we know then the ratio dimensions for the new photograph,
