EXAMPLE 1 Use the centroid of a triangle SOLUTION SQ = 2 3 SW Concurrency of Medians of a Triangle Theorem 8 =8 = 2 3 SW Substitute 8 for SQ. 12 = SW Multiply.

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Presentation transcript:

EXAMPLE 1 Use the centroid of a triangle SOLUTION SQ = 2 3 SW Concurrency of Medians of a Triangle Theorem 8 =8 = 2 3 SW Substitute 8 for SQ. 12 = SW Multiply each side by the reciprocal,. 2 3 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, 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. 2 3 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 = 3 2 PC 2100 = 3 2 PC Substitute 2100 for SC = PC Multiply each side by the reciprocal,. 3 2 Then PS = SC – PC = 2100 – 1400 = 700. So, PS = 700 and PC = 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 = 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 = 1 2 PA 800 = 1 2 PA Substitute 800 for PT = PA Multiply each side by the reciprocal,. 1 2 Then TA = PT + PA = = So, TA = 2400 and PA = Concurrency of Medians of a Triangle Theorem SOLUTION