Find the perimeters of (i) $\triangle ABE$ (ii) the rectangle $BCDE$ in this figure. Whose perimeter is greater?
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Given:

$ABCDE$ is the given figure.

To do:

We have to find the perimeters of (i) $\triangle ABE$ and (ii) the rectangle $BCDE$.

Solution:

Perimeter of a triangle of sides of lengths $a$ units, $b$ units and $c$ units is $(a+b+c) $ units.

Perimeter of a rectangle of length $l$ units and breadth $b$ units is $2(l+b)$ units.

(i) The lengths of the sides of the triangle $ABE$ are $AB=\frac{5}{2}7 cm, AE=3\frac{3}{5}\ cm$ and $BE=2\frac{3}{4}\ cm$.

The perimeter of the triangle $ABE= (\frac{5}{2} +3\frac{3}{5} +2\frac{3}{4})\ cm$

$=(\frac{5}{2} +\frac{3\times5+3}{5} +\frac{2\times4+3}{4})\ cm$

$=(\frac{5}{2} +\frac{15+3}{5} +\frac{8+3}{4})\ cm$

$=(\frac{5}{2} +\frac{18}{5} +\frac{11}{4})\ cm$

$=(\frac{5\times 10+18\times 4+11\times5}{20})\ cm$

$=\frac{50+72+55}{20}\ cm$

$=\frac{177}{20}\ cm$

$=8\frac{17}{20}\ cm$

(ii) The lengths of the sides of the rectangle are $2\frac{3}{4}\ cm$ and $\frac{7}{6}\ cm$.

The perimeter of the given rectangle $= 2( 2\frac{3}{4} +\frac{7}{6})\ cm$

$=2(\frac{2\times4+3}{4} +\frac{7}{6})\ cm$

$=2(\frac{8+3}{4}+\frac{7}{6})\ cm$

$=2(\frac{11}{4}+\frac{7}{6})\ cm$

$=2(\frac{11\times3+7\times2}{12})\ cm$           (LCM of 4 and 6 is 12)

$=2(\frac{33+14}{12})\ cm$

$=2\times \frac{47}{12}\ cm$

$=\frac{47}{6}\ cm$

$=7\frac{5}{6}\ cm$

The perimeter of triangle $ABE$ is $8\frac{17}{20}\ cm$ and the perimeter of rectangle  $BCDE$ is $7\frac{5}{6}\ cm$.

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Updated on: 10-Oct-2022

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