Question 1037844
quadrilateral STUV, S(-5,3) T(7,-2) U(7,-6) V(-5,-6) 
<pre><b>
{{{drawing(400,2400/7,-6,8,-7,5, grid(1),

red(line(-5,-6,7,-6),line(7,-6,7,-2), line(7,-2,-5,3),line(-5,3,-5,-6)),
locate(-4.9,3.5,S), locate(-4.9,-6,V),locate(7.1,-2,T),locate(7.1,-6,U)  )}}}

The perimeter is just how many units it is all the way around
the quadrilateral.  You can get 3 of the sides just by counting
the blocks on the graph paper.

From S to V is 9 units straight down.
From V to U is 12 units left to right.
From U to T is 4 units straight up.

The only line that we have to calculate is the slanted one from S to T.

We do that by drawing a horizontal line from T over to the line SV, like
below.  We will label the point W(-5,-2) where it intersects line SV:

{{{drawing(400,2400/7,-6,8,-7,5, grid(1),

red(line(-5,-6,7,-6),line(7,-6,7,-2), line(7,-2,-5,3),line(-5,3,-5,-6),line(-5,-2,7,-2)),
locate(-4.9,3.5,S), locate(-4.9,-6,V),locate(7.1,-2,T),locate(7.1,-6,U),
locate(-5.4,-2,W)
  )}}}

To calculate to upper side ST of the quadrilateral, we use the
Pythagorean theorem on right triangle SWT, which states that

{{{ST^2}}}{{{""=""}}}{{{SW^2+WT^2}}}

We count blocks and get that WT = 12, the same as VU. And we
see that SW is 5 unit long, by counting blocks.

So we substitute 12 for SW and 5 for WT

{{{ST^2}}}{{{""=""}}}{{{5^2+12^2}}}

{{{ST^2}}}{{{""=""}}}{{{25+144}}}

{{{ST^2}}}{{{""=""}}}{{{169}}}

{{{ST}}}{{{""=""}}}{{{sqrt(169)}}}

From the calculator we can get that square
root to be 13.

Then the perimeter is the sum of all four
sides:

SV + VU + UT + ST = 9 + 12 + 4 + 13 = 38

So the perimeter is 38.

Edwin</pre></b>