Question

What is the perimeter of the figure? A.68 1/2 yd B.68 5/12 YD C.67 1/2 YD D.67 5/14 YD
Also sorry for reposting. I forgot to mark the answers

Answer 1

I will give a medal to anyone that answers

Answer 2

Add the lengths of the 4 sides.

Answer 3

\(10 \dfrac{1}{3} + 24 \dfrac{1}{4} + 15 \dfrac{2}{3} + 18 \dfrac{1}{4} \)

First, remember that a mixed numeral can be thought of as the sum of whole number and a fraction:

\(=10 + \dfrac{1}{3} + 24 + \dfrac{1}{4} + 15 + \dfrac{2}{3} + 18 + \dfrac{1}{4} \)

Now we use the commutative and associative properties of addition to rearrange the numbers in the sum above:

\(=10 + 24 + 15 + 18 + \dfrac{1}{3} + \dfrac{2}{3}+ \dfrac{1}{4} + \dfrac{1}{4} \)

Now we add the whole numbers together, and we add the fractions with the same denominators together:

\(=10 + 24 + 15 + 18 + \dfrac{1}{3} + \dfrac{2}{3}+ \dfrac{1}{4} + \dfrac{1}{4} \)

\(=67 + \dfrac{3}{3}+ \dfrac{2}{4} \)

Now we notice that 3/3 is simply 1, and 2/4 is the same as 1/2:

\(=67 + 1+ \dfrac{1} {2 } \)

\(=68 \dfrac{1} {2 } \)

Since all units are in yards, the answer is:

\( 68 \dfrac{1} {2 } \sf~yd\)