Question

kuta software - infinite geometry
midsegment of a triangle
name
date
period
in each triangle, m, n, and p are the midpoints of the sides. name a segment parallel to the one given.
1)
2)
find the missing length indicated.

1. find cd
2. find ac
3. find kj
4. find ik
5. find df
6. find pq

Explanation:

Step1: Use the midsegment theorem

The midsegment theorem states that the midsegment of a triangle is parallel to the third side and half its length.

Step2: Solve for each problem

1. In the first triangle, since \(M\), \(N\), and \(P\) are mid - points, \(\overline{CD}\parallel\overline{PN}\)
2. In the second triangle, \(\overline{MN}\parallel\overline{QS}\)
3. For \(CD\):

• Given \(XZ = 16\), and \(CD\) is a midsegment. By the midsegment theorem \(CD=\frac{1}{2}XZ\)
• \(CD = 8\)

1. For \(AC\):

• Given \(TS = 4\), and \(TS\) is a midsegment. Then \(AC=2\times TS\)
• \(AC = 8\)

1. For \(KJ\):

• Given \(AC = 18\), and \(KJ\) is a midsegment. So \(KJ=\frac{1}{2}AC\)
• \(KJ = 9\)

1. For \(IK\):

• Given \(SR = 5\), and \(SR\) is a midsegment. Then \(IK = 2\times SR\)
• \(IK=10\)

1. For \(DF\):

• Given \(VU = 9\), and \(VU\) is a midsegment. So \(DF=2\times VU\)
• \(DF = 18\)

1. For \(PQ\):

• First, find \(TS\): \(TS=30 - 7=23\)
• Since \(PQ\) is a midsegment of \(\triangle TUS\), \(PQ=\frac{1}{2}TS\)
• \(PQ = 11.5\)

Answer:

1. \(\overline{PN}\)
2. \(\overline{MN}\)
3. \(8\)
4. \(8\)
5. \(9\)
6. \(10\)
7. \(18\)
8. \(11.5\)