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# Q.5 Find the perimeter of (i) triangle ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?

(i) Perimeter of triangleABE = AB+BE+AE = $$\frac{5}{2}+2\frac{3}{4}+3\frac{3}{5} =\frac{5}{2}+\frac{11}{4}+\frac{18}{5} =\frac{50+55+72}{20}=\frac{177}{20}=8\frac{17}{20}$$
= $$2 × [2\frac{3}{4}+\frac{7}{6}] = 2 × [ \frac{11}{4}+ \frac{7}{6}] = 2× \frac{33+14}{12} = 2×\frac{47}{12}= \frac{47}{6} = 7\frac{5}{6}$$
Since $$8\frac{17}{20} > 7\frac{5}{6}$$