Question

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

Answer 1

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 2

Answer:

60 units

Step-by-step explanation:

Given: The vertices of XYZ are $X(-3,-6)$, $Y(21,-6)$, and $Z(21,4)$

To find: Perimeter of the triangle.

Solution:

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

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

So,

$XY=\sqrt{(21+3)^{2}+(-6+6)^{2}}=24$

$YZ=\sqrt{(21-21)^{2}+(4+6)^{2}}=10$

$ZX=\sqrt{(21+3)^{2}+(4+6)^{2}$

$ZX=\sqrt{(24)^{2}+(10)^{2}$

$ZX=\sqrt{576+100}$

$ZX=\sqrt{676}=26$

Now, we have $XY=24$, $YZ=10$, and $ZX=26$

Perimeter of Δ$XYZ=24+10+26=60$

Hence, perimeter of ΔXYZ is 60 units.