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.
