Median of a Triangle

Median of a Triangle

Median of a Triangle

A line segment from a vertex of a triangle to the midpoint of its opposite side is called the median of the triangle. There are three medians of a triangle. Medians of a triangle are concurrent i.e. they passes through a single point which is called the centroid of the triangle.
Three medians of the triangle ABC
In the figure AP, BQ and CR are three medians of the triangle ABC. Point O is the centroid of the triangle.

Properties of medians of a triangle

1.    The median of a triangle divides it into two equal parts. (Median bisects the triangle.)
2.    Three medians of a triangle are concurrent i.e. they passes through a single point which is called centroid of the triangle.
3.   Centroid of a triangle is at a distance two-third of median from the vertex.

Workout Examples

Example 1: From the given figure, find the area of ΔAMC.
Example 1: Triangle

Solution: From the figure,
                      Area of ΔABC = ½ BC × AD ---------> area of Δ = ½ base × height
                                               = ½ × 6 × 8
                                               = 24 cm2
                      
                      Area of ΔAMC = ½ of ΔABC ----------> median bisects the triangle
                                               = ½ × 24 cm2
                                               = 12 cm2


Example 2: In the adjoining figure, D is the mid point of AC. If 3DE = BC = 12cm. Find the area of ΔABC.
Example 2: Triangle

Solution: Here, 3DE = BC = 12cm
                       BC = 12cm, and 3DE = 12cm or, DE = 12/3 = 4cm
                BD joined.
                      Area of ΔBCD = ½ BC × DE ---------> area of Δ = ½ base × height
                                               = ½ × 12 × 4
                                               = 24 cm2

                     Area of ΔABD = Area of ΔBCD ---------> median divides the triangle into two equal parts
                                              = 24 cm2
                                
                     Area of ΔABC = Area of ΔBCD + ΔABD
                                              = 24 + 24
                                              = 48 cm2


Example 3: In the given figure, D, E, F and H are the mid points of BC, AD, CE and BF respectively. Prove that: ΔEFH = 1/8 ΔABC.
Example 3: Triangle

Solution: From the figure,
                     ΔEFH = ½ of ΔEBF ----------> median EH bisects the ΔEBF
                                = ½ × ½ of ΔEBC ---------> median BF bisects the ΔEBC
                                = ½ × ½ × 2ΔEDC ---------> median ED also bisects the ΔEBC
                                = ½ × ½ × 2 × ½ of ΔADC ---------> median CE bisects the ΔADC
                                = ½ × ½ × 2 × ½ × ½ of ΔABC ---------> median AD bisects the ΔABC
                                = 1/8 ΔABC.   Proved.

               
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