desperate for help!!!?

Question

can somebody help me find the perimeter of the triangle PQR with vertices P (-6,-3), Q(1,-1), and R(1,-5)? and also of the pentagon ABCDE with vertices A(6,-2), B(-4,7), C(0,4), D(0,0), and E(-4,-3)? thank you!

Answer 1

Use the Pythagorean distance formula for each segment of the figure:

D = sqrt( [x2 - x1]² + [y2 - y1]² ),

where sqrt() is the square root function.

A slightly more condensed way of writing this is:

D = sqrt ( Δx² + Δy² ),

where Δx means "the difference in x," or "the change in x."

For example, between points P and R, the difference between the the two x values is

Δx = { 1 - (-6) } = 7,

while the difference in y is

Δy = { (-3) - (-5) } = -2.

Therefore the distance between P and R is, (remembering that exponents always come before signs in order of operations....)

D(PQ) = sqrt { 7² + (-2)² }

= sqrt ( 49 + 4 )

= sqrt ( 53 )

≈ 7.28.

In order to find the perimeter of PQR, you need to add the length of each individual segment, or mathematically;

Perimeter = D(PQ) + D(QR) + D(RP)....

Hope that makes sense,
~W.O.M.B.A.T.

Answer 2

that's easy...just use the distance formula so that you can get the measures of the sides of the triangle..after that you use the formula for the perimeter of the triangle....

do the same with the pentagon....

i hope this helps...

Answer 3

drink pepsi