MATH SOLVE

9 months ago

Q:
# What is the perimeter of the figure in the diagram (round to three decimal places if necessary)?A) 44 B) 36.472 C) 36 D) 28.472

Accepted Solution

A:

Perimeter (p) = length around a shape.

So any horizontal sides we can add based upon their length in x-values and any vertical sides we can add based upon their length in y-values. The diagonal sides we can find by the distance formula:

[tex]d = \sqrt{( {(x2 - x1)}^{2} + {(y2 - y1)}^{2} } [/tex]

So the bottom base goes from x=1 to x=8, then up 1, right 1, and up from y=3 to y=9, then left from x=9 to x=6, down from y=9 to y=6, left from x=6 to x=3.

So far that is: (8-1)+1+1+(9-3)+(9-6)+(9-6)+(6-3)

= 7+2+6+3+3+3 = 24

Now for the diagonal:

[tex]d = \sqrt{( {(x2 - x1)}^{2} + {(y2 - y1)}^{2}) } \\ = \sqrt{( {(3- 1)}^{2} + {(6 - 2)}^{2} )} \\ = \sqrt{( {2}^{2} + {4}^{2} )} = \sqrt{(4 + 16)} \\ = \sqrt{20} = 2 \sqrt{5} = 4.47[/tex]

P = 24 + 4.47 = 28.47

D) 28.472

So any horizontal sides we can add based upon their length in x-values and any vertical sides we can add based upon their length in y-values. The diagonal sides we can find by the distance formula:

[tex]d = \sqrt{( {(x2 - x1)}^{2} + {(y2 - y1)}^{2} } [/tex]

So the bottom base goes from x=1 to x=8, then up 1, right 1, and up from y=3 to y=9, then left from x=9 to x=6, down from y=9 to y=6, left from x=6 to x=3.

So far that is: (8-1)+1+1+(9-3)+(9-6)+(9-6)+(6-3)

= 7+2+6+3+3+3 = 24

Now for the diagonal:

[tex]d = \sqrt{( {(x2 - x1)}^{2} + {(y2 - y1)}^{2}) } \\ = \sqrt{( {(3- 1)}^{2} + {(6 - 2)}^{2} )} \\ = \sqrt{( {2}^{2} + {4}^{2} )} = \sqrt{(4 + 16)} \\ = \sqrt{20} = 2 \sqrt{5} = 4.47[/tex]

P = 24 + 4.47 = 28.47

D) 28.472