Question

consider △pqr and △stu below.
note that △pqr has vertices p(-6, -3), q(-3, -2), and r(-5, -6) and △stu has vertices s(3, 3), t(4, 6), and u(6, 2). find the side lengths below, then determine whether △pqr ≅ △stu. give exact answers (not decimal approximations).
part 1 of 2
pq =
qr =
pr =
st =
tu =
su =

Explanation:

Step1: Find PQ length

Use distance formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). For \(P(-6, -3)\) and \(Q(-3, -2)\):
\(PQ = \sqrt{(-3 - (-6))^2 + (-2 - (-3))^2} = \sqrt{(3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10}\)

Step2: Find QR length

For \(Q(-3, -2)\) and \(R(-5, -6)\):
\(QR = \sqrt{(-5 - (-3))^2 + (-6 - (-2))^2} = \sqrt{(-2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}\)

Step3: Find PR length

For \(P(-6, -3)\) and \(R(-5, -6)\):
\(PR = \sqrt{(-5 - (-6))^2 + (-6 - (-3))^2} = \sqrt{(1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}\)

Step4: Find ST length

For \(S(3, 3)\) and \(T(4, 6)\):
\(ST = \sqrt{(4 - 3)^2 + (6 - 3)^2} = \sqrt{(1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10}\)

Step5: Find TU length

For \(T(4, 6)\) and \(U(6, 2)\):
\(TU = \sqrt{(6 - 4)^2 + (2 - 6)^2} = \sqrt{(2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}\)

Step6: Find SU length

For \(S(3, 3)\) and \(U(6, 2)\):
\(SU = \sqrt{(6 - 3)^2 + (2 - 3)^2} = \sqrt{(3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}\)

Answer:

\(PQ = \sqrt{10}\), \(QR = 2\sqrt{5}\), \(PR = \sqrt{10}\), \(ST = \sqrt{10}\), \(TU = 2\sqrt{5}\), \(SU = \sqrt{10}\)