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7. Congruence & Triangles
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United States of AmericaTennessee
Integrated Math 2

7.06 Medians and altitudes in triangles

Lesson
Practice
Lesson

Concept summary

For triangles, we define the following two specific line segments:

Median

A line segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side.

A triangle with a dashed segment drawn from one of the vertices to the opposite side, dividing the segment into two congruent segments.
Altitude

A line segment from the vertex of a triangle that is perpendicular to the opposite side.

A triangle with a dashed segment drawn from one of the vertices perpendicular to the opposite side.

Note that a median is always internal as it joins a vertex to the opposite midpoint, but an altitude can be external to the triangle.

It is possible for a line segment to be both a median and an altitude of a triangle - this occurs when the triangle is iscoseles, and the median connects to the midpoint of the base.

Worked examples

Example 1

Determine if \overline{AX} is a median, altitude, both, or neither.

Obtuse Triangle A B C is drawn such that angle A C B is an obtuse angle and side B C is a horizontal segment. From A, a dashed segment A X exterior to the triangle is drawn perpendicular to dashed segment C X. B, C, and X are collinear.

Solution

We can see \overline{AX} passes through the vertex A and meets the extension of the side BC at a right angle. So \overline{AX} is an altitude.

We can also see that X cannot be the midpoint of \overline{BC}. So \overline{AX} cannot be a median.

Reflection

An altitude will not always be able to meet the opposite side of the triangle perpendicularly. In those cases, the altitude instead meets the extension of the opposite side, as seen in this example.

Example 2

In the following diagram, \overline{BD} is a median. If all measurements are in inches, determine the length of leg \overline{CB}.

From Segment A C with midpoint D, two connected triangles are drawn. Triangle C D B and triangle D A B share a common side D B. Segment C D measures 7 x + 2, segment D A measures 3 x + 18, B D is 3x and C B is 5 x + 6.

Approach

Since \overline{BD} is a median, we know that AD = DC. We can use this to form an equation in terms of x, then solve for x and use this to calculate CB.

Solution

\displaystyle AD\displaystyle =\displaystyle DC\overline{BD} is a median of \overline{AC}
\displaystyle 3x + 18\displaystyle =\displaystyle 7x + 2Substitute known values
\displaystyle 16\displaystyle =\displaystyle 4xSubtract 3x and 2 from both sides
\displaystyle 4\displaystyle =\displaystyle xDivide both sides by 4

We can now substitute x = 4 into the expression for CB.

\displaystyle CB\displaystyle =\displaystyle 5x + 6Given
\displaystyle =\displaystyle 5 \cdot 4 + 6Substitute x = 4
\displaystyle =\displaystyle 26Simplify

So we have that CB = 26 inches.

Reflection

We can confirm that x = 4 is correct by substituting into the expressions for AD and DC and checking that they are equal.

\displaystyle AD\displaystyle =\displaystyle 3x + 18
\displaystyle =\displaystyle 3 \cdot 4 + 18
\displaystyle =\displaystyle 30

and

\displaystyle DC\displaystyle =\displaystyle 7x + 2
\displaystyle =\displaystyle 7 \cdot 4 + 2
\displaystyle =\displaystyle 30

This confirms that x = 4 is the value that makes D the midpoint of \overline{AC}.

Outcomes

M2.G.CO.C.8

Use definitions and theorems about triangles to solve problems and to justify relationships in geometric figures.

M2.MP1

Make sense of problems and persevere in solving them.

M2.MP3

Construct viable arguments and critique the reasoning of others.

M2.MP4

Model with mathematics.

M2.MP5

Use appropriate tools strategically.

M2.MP6

Attend to precision.

M2.MP7

Look for and make use of structure.

M2.MP8

Look for and express regularity in repeated reasoning.

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