Question

Find the perimeters of

(i) $△ABE,$

(ii) the rectangle $BCDE$ in this figure.

Which figure has a greater perimeter and by how much?

Answer 1

Perimeter of $\Delta$ABE $=$ $AB+BE+AE$

$=$ $3\frac{1}{3}$+$2\frac{1}{5}$+$4\frac{2}{3}$

$=$$\frac{10}{3}$ +$\frac{11}{5}$+$\frac{14}{3}$

$=$$\frac{50+33+70}{15}$

$=$$\frac{153}{15}$

$=$$10\frac{3}{15}$ cm

Perimeter of Rectangle $BCDE$ $=$ $2\left(l+b\right)$

$=2\left(2\frac{1}{5}+1\frac{2}{3}\right)$

$=2\left(\frac{11}{5}+\frac{5}{3}\right)$

$=2\left(\frac{33+25}{15}\right)$

$=2\left(\frac{58}{15}\right)$

$=\frac{116}{15}$

$=7\frac{11}{15}$ cm

So,$\Delta$ABE has greater perimeter

Difference of perimeter $=$$\frac{153}{15}-\frac{116}{15}$

$=\frac{37}{15}$

$=2\frac{7}{15}$ cm

Hence $\Delta$ABE has greater perimeter by $=2\frac{7}{15}$ cm