Find the perimeters of (i) ∆ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?

(i) According to the given figure, the perimeter of $∆\mathrm{ABE}$ is given by:

$\begin{array}{rcl} & =& \mathrm{AB}+\mathrm{BE}+\mathrm{AE}\\ & =& \frac{5}{2}+2\frac{3}{4}+3\frac{3}{5}\\ & =& \frac{5}{2}+\frac{11}{4}+\frac{18}{5}\\ & =& \frac{5\times 10+11\times 5+18\times 4}{20}\\ & =& \frac{50+55+72}{20}\\ & =& \frac{177}{20}=8\frac{17}{20}\mathrm{cm}\end{array}$

(ii) The perimeter of rectangle BCDE is given as:

$\begin{array}{rcl} & =& 2\times \left(\mathrm{BE}+\mathrm{DE}\right)\\ & =& 2\times \left(2\frac{3}{4}+\frac{7}{6}\right)\\ & =& 2\times \left(\frac{11}{4}+\frac{7}{6}\right)\\ & =& 2\times \left(\frac{11\times 3+7\times 2}{12}\right)\\ & =& 2\times \left(\frac{33+14}{12}\right)\\ & =& 2\times \left(\frac{47}{12}\right)\\ & =& \frac{47}{6}=7\frac{5}{6}\mathrm{cm}\end{array}$

Now, to find the greater perimeter, the LCM of $20, 6$ is $60$. Therefore,

$\frac{177}{20}\times \frac{3}{3}=\frac{531}{60}; \frac{47}{6}\times \frac{10}{10}=\frac{470}{60}$

Comparing the above fractions: 

$$\begin{gathered} \frac{531}{60}>\frac{470}{60} \\ 8\frac{17}{20}>7\frac{5}{6} \end{gathered}$$

Hence, perimeter of triangle ABE is greater than perimeter of rectangle BCDE.