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.

In the figure AP, BQ and CR are three medians of the triangle ABC. Point O is the centroid of the triangle.

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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.

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.

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.

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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