EXAMPLE 1 Use the centroid of a triangle

SOLUTION

SQ = 23 SW Concurrency of Medians of a

Triangle Theorem

8 = 23 SW Substitute 8 for SQ.

12 = SW Multiply each side by the reciprocal, .23

Then QW = SW – SQ =12 – 8 =4.

So, QW = 4 and SW = 12.

In RST, Q is the centroid and SQ = 8. Find QW and SW.

EXAMPLE 2 Standardized Test Practice

SOLUTION

Sketch FGH. Then use the Midpoint Formula to find the midpoint K of FH and sketch median GK .

K( ) =2 + 6 , 5 + 12 2 K(4, 3)

The centroid is two thirds of the distance from each vertex to the midpoint of the opposite side.

EXAMPLE 2 Standardized Test Practice

The distance from vertex G(4, 9) to K(4, 3) is

9 – 3 = 6 units. So, the centroid is (6) = 4 units

down from G on GK .

23

The coordinates of the centroid P are (4, 9 – 4), or (4, 5).

The correct answer is B.

GUIDED PRACTICE for Examples 1 and 2

There are three paths through a triangular park. Each path goes from the midpoint of one edge to the opposite corner. The paths meet at point P.

1. If SC = 2100 feet, find PS and PC.

SC = 32 PC

2100 = 32 PC Substitute 2100 for SC.

1400 = PC Multiply each side by the reciprocal, .32

Then PS = SC – PC =2100 – 1400 = 700.

So, PS = 700 and PC = 1400.

Concurrency of Medians of a Triangle Theorem

SOLUTION

GUIDED PRACTICE for Examples 1 and 2

2. If BT = 1000 feet, find TC and BC.

Substitute 1000 for BT.

So, BC = 2000 and TC = 1000.

BT = TC

1000 = TC

T is midpoint of side BC.

BC = 2 TC

BC = 2000

SOLUTION

GUIDED PRACTICE for Examples 1 and 2

3. If PT = 800 feet, find PA and TA.

PT = 12 PA

800 = 12 PA Substitute 800 for PT.

1600 = PA Multiply each side by the reciprocal, .12

Then TA = PT + PA =1600 + 800 =2400.

So, TA = 2400 and PA = 1600.

Concurrency of Medians of

a Triangle Theorem

SOLUTION