The vertices of xyz are x(-3,-6), Y(21,-6), and z(21,4). What is the perimeter of the triangle

Note that this a right-angled triangle with the right angle at Y.

Use coordinates of the point to find XY and YZ

XY = 21 - (-3)

= 24

YZ = 4 - (-6)

= 10

Use Pythagoras theorem to XZ

XZ = sqrt[24^2 + 10^2]

= 26

Perimeter = 26 + 24 + 10

= 60

Answer Link

aachen aachen · Answer 2 · 2019-10-15T06:23:20+02:00

Answer:

60 units

Step-by-step explanation:

Given: The vertices of XYZ are [tex]X(-3,-6)[/tex], [tex]Y(21,-6)[/tex], and [tex]Z(21,4)[/tex]

To find: Perimeter of the triangle.

Solution:

We know that the distance between two points [tex]A(x_{1},y_{1})[/tex] and [tex]B(x_{2},y_{2})[/tex] is [tex]\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }[/tex]

We have, the vertices of triangle XYZ as [tex]X(-3,-6)[/tex], [tex]Y(21,-6)[/tex], and [tex]Z(21,4)[/tex]

So,

[tex]XY=\sqrt{(21+3)^{2}+(-6+6)^{2}}=24[/tex]

[tex]YZ=\sqrt{(21-21)^{2}+(4+6)^{2}}=10[/tex]

[tex]ZX=\sqrt{(21+3)^{2}+(4+6)^{2}[/tex]

[tex]ZX=\sqrt{(24)^{2}+(10)^{2}[/tex]

[tex]ZX=\sqrt{576+100}[/tex]

[tex]ZX=\sqrt{676}=26[/tex]

Now, we have [tex]XY=24[/tex], [tex]YZ=10[/tex], and [tex]ZX=26[/tex]

Perimeter of Δ[tex]XYZ=24+10+26=60[/tex]

Hence, perimeter of ΔXYZ is 60 units.