Question

A) A rectangular sheet of paper is $12\frac{1}{2}cm$ long and $10\frac{2}{3}cm$ wide. Find its perimeter.

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

Answer 1

A) Perimeter of a rectangle $=2(l+b)$

$=2(12\frac{1}{2}+10\frac{2}{3})$

$=2(\frac{25}{2}+\frac{32}{3})$

$=2\times \left(\frac{75+64}{6}\right)$
$=2\times \frac{139}{6}$

$=\frac{139}{3}cm$
$=46\frac{1}{3}cm$

B) (i) Perimeter of $\Delta ABC=A+B+C$

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

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

$=\frac{177}{20}cm$
$=8\frac{17}{20}cm$

(ii) Perimeter of rectangle $BCDE=2(l+b)$

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

$=\frac{11}{2}+\frac{7}{3}$

$=\frac{33+14}{6}$
$=\frac{47}{6}cm$
$=7\frac{5}{6}cm$

As $8\frac{17}{20}>8$ & $7\frac{5}{6}<8$

So, the perimeter of the triangle is greater.