Question

Find the perimeters of
$\left(i\right)$ $△ABE$
$\left(ii\right)$ the rectangle $BCED$ in this figure, Whose perimeter is greater?

Answer 1

Given $AB=\frac{5}{2}cm,BE=\frac{11}{4}cm,AE=\frac{18}{5}cm,DE=\frac{7}{6}cm$

(i) perimeter of $\Delta ABE=AB+BE+AE$

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

$LCM(2,4,5)=20=\frac{5\times 10+11\times 5+18}{20}$

$=\frac{50+55+18}{20}=\frac{123}{20}$

= 6.15 cm

(ii) Perimeter of rectangle BCED

= 2(BC+CD)

=2(BC+CD)

= BC+CD+DE+EB

= DE+BE+DE+BE

Since opposite sides are equal

= 2(DE+BE)

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

$=\frac{2}{24}(28+66)$

$=\frac{1}{12}\times 94=\frac{94}{12}=7.833$

= 7.833 cm