body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

Pearson Geometry Common Core, 2011

PG

Pearson Geometry Common Core, 2011 View details

Mid-Chapter Quiz

Continue to next subchapter

Exercise 10 Page 316

The centroid is the point of concurrency of the medians of a triangle. Use the Concurrency of Medians Theorem.

$Q P = 3$ and $Q B = 9$

Practice makes perfect

The centroid is the point of concurrency of the medians of a triangle. In the given diagram, $P$ is the centroid.

Using the Concurrency of Medians Theorem, we can write an equation that we can use to find one of the desired lengths,

Q B .

P B = \frac{2}{3} Q B

We are given that

P B = 6 .

We can substitute this into the above equation to find the length of segment

Q B .

P B = \frac{2}{3} Q B

$P B = 6$

6 = \frac{2}{3} Q B

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

6 = \frac{2 Q B}{3}

$LHS \cdot 3 = RHS \cdot 3$

18 = 2 Q B

$LHS / 2 = RHS / 2$

9 = Q B

Rearrange equation

Q B = 9

With the Segment Addition Postulate, we can rewrite

Q B

as the sum of the two smaller segments.

Q B = Q P + P B

Since we know that

Q B = 9

and

P B = 6,

we can solve the equation above for

Q P .

9 = Q P + 6 \Leftrightarrow Q P = 3