Question

In the figure to the right, AE + DC = 1 1/5 cm, AB = 1 3/4 cm, DE = 1 1/4 cm, and BC = 1 3/10 cm. Find the perimeter of the figure.

Answer 1

Answer:

$5\dfrac{1}{2}\ cm$

Step-by-step explanation:

The perimeter of the shown figure is the sum of the lengths of all sides. So, the perimeter is

$AE+ED+DC+CB+AB$

Using commutative and reflexive properties, rewrite it as follows:

$(AE+DC)+DE+BC+AB$

Therefore, the perimeter is

$P\\ \\=1\dfrac{1}{5}+1 \dfrac{1}{4}+1\dfrac{3}{10}+1\dfrac{3}{4}\\ \\=(1+1+1+1)+\left(\dfrac{1}{5}+\dfrac{3}{10}\right)+\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\\ \\=4+\left(\dfrac{2}{10}+\dfrac{3}{10}\right)+1\\ \\=5+\dfrac{5}{10}\\ \\=5+\dfrac{1}{2}\\ \\=5\dfrac{1}{2}\ cm$