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

Solution

(i) Perimeter of ΔABE = AB + BE + EA

$= (\frac{5}{2} + 2 \frac{3}{4} + 3 \frac{3}{5}) = (\frac{5}{2} + \frac{11}{4} + \frac{18}{5}) = (\frac{5 \times 10}{2 \times 10} + \frac{11 \times 5}{4 \times 5} + \frac{18 \times 4}{5 \times 4}) = (\frac{50 + 55 + 72}{20}) = \frac{177}{20} = 18 \frac{17}{20} c m$

(ii)

Perimeter of rectangle = 2 (Length + Breadth)

$Perimeter of rectangle = 2 [\frac{11}{4} + \frac{7}{6}] = 2 [\frac{11 \times 3}{4 \times 3} + \frac{7 \times 2}{6 \times 2}] = 2 [\frac{33 + 14}{12}] = 2 \times \frac{47}{12} = \frac{47}{6} = 7 \frac{5}{6} cm$

Perimeter of ΔABE = $\frac{177}{20} c m$

Changing them to like fractions, we obtain

$\frac{177}{20} = \frac{177 \times 3}{20 \times 3} = \frac{531}{60} \frac{43}{6} = \frac{43 \times 10}{6 \times 10} = \frac{430}{60}$

As 531 > 430,

$\frac{177}{20} > \frac{43}{6}$

Perimeter (ΔABE) > Perimeter (BCDE)

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