Question

use medians and altitudes
★ i can use medians of triangles to solve problems. ★
in △pqr, nq = 6, rk = 3, and pk = 4. find each length.

1. km
2. kq
3. lk
4. lr
5. nk
6. pm

in △abc, cp = 30, ep = 18, and bf = 39. find each length.

1. pd
2. fp
3. bp
4. cd
5. pa
6. ea

★ i can use the median point of concurrency, the centroid, to find segment lengths. ★
in △str, h is the centroid, eh = 6, dh = 4, and sm = 24. find each length.

1. sh
2. hm
3. th
4. hr
5. td
6. er

Explanation:

Step1: Recall Centroid Properties

The centroid of a triangle divides each median into a ratio of \(2:1\), with the longer segment being closer to the vertex.

Step2: Solve for KM (Problem 1)

In \(\triangle PQR\), \(K\) is the centroid (intersection of medians). \(PK = 4\), so \(KM=\frac{1}{2}PK\) (since centroid divides median \(PM\) into \(PK:KM = 2:1\)).
\(KM=\frac{4}{2}=2\).

Step3: Solve for KQ (Problem 2)

\(NQ = 6\), and \(K\) is centroid, so \(KQ=\frac{2}{3}NQ\) (centroid divides median \(NQ\) into \(KQ:KN = 2:1\)).
\(KQ=\frac{2}{3}\times6 = 4\).

Step4: Solve for LK (Problem 3)

\(RK = 3\), centroid divides median \(LR\) into \(LK:KR = 2:1\), so \(LK = 2\times KR=2\times3 = 6\).

Step5: Solve for LR (Problem 4)

\(LR = LK + KR\), \(LK = 6\), \(KR = 3\), so \(LR = 6 + 3 = 9\).

Step6: Solve for NK (Problem 5)

\(NQ = 6\), centroid divides \(NQ\) into \(KQ:KN = 2:1\), so \(KN=\frac{1}{3}NQ=\frac{6}{3}=2\) (Note: \(NK = KN\)).

Step7: Solve for PM (Problem 6)

\(PK = 4\), \(KM = 2\), so \(PM = PK + KM = 4 + 2 = 6\).

Step8: Solve for PD (Problem 7, \(\triangle ABC\))

\(CP = 30\), centroid \(P\) divides median \(CD\) into \(CP:PD = 2:1\), so \(PD=\frac{1}{2}CP=\frac{30}{2}=15\).

Step9: Solve for FP (Problem 8)

\(BF = 39\), centroid \(P\) divides median \(BF\) into \(FP:PB = 2:1\), so \(FP=\frac{2}{3}BF=\frac{2}{3}\times39 = 26\).

Step10: Solve for BP (Problem 9)

\(BF = 39\), \(BP=\frac{1}{3}BF=\frac{39}{3}=13\) (since \(BP:PF = 1:2\), so \(BP=\frac{1}{3}BF\)).

Step11: Solve for CD (Problem 10)

\(CP = 30\), \(PD = 15\), so \(CD = CP + PD = 30 + 15 = 45\).

Step12: Solve for PA (Problem 11)

\(EP = 18\), centroid \(P\) divides median \(EA\) into \(EP:PA = 2:1\), so \(PA=\frac{1}{2}EP=\frac{18}{2}=9\).

Step13: Solve for EA (Problem 12)

\(EA = EP + PA = 18 + 9 = 27\).

Step14: Solve for SH (Problem 13, \(\triangle STR\))

\(SM = 24\), centroid \(H\) divides median \(SM\) into \(SH:HM = 2:1\), so \(SH=\frac{2}{3}SM=\frac{2}{3}\times24 = 16\).

Step15: Solve for HM (Problem 14)

\(HM=\frac{1}{3}SM=\frac{24}{3}=8\).

Step16: Solve for TH (Problem 15)

\(DH = 4\), centroid \(H\) divides median \(TD\) into \(TH:HD = 2:1\), so \(TH = 2\times DH = 2\times4 = 8\).

Step17: Solve for HR (Problem 16)

\(EH = 6\), centroid divides median \(ER\) into \(EH:HR = 2:1\), so \(HR=\frac{1}{2}EH=\frac{6}{2}=3\).

Step18: Solve for TD (Problem 17)

\(TH = 8\), \(HD = 4\), so \(TD = TH + HD = 8 + 4 = 12\).

Step19: Solve for ER (Problem 18)

\(EH = 6\), \(HR = 3\), so \(ER = EH + HR = 6 + 3 = 9\).

Answer:

s (Selected Problems Shown):

1. \(KM = \boldsymbol{2}\)
2. \(KQ = \boldsymbol{4}\)
3. \(LK = \boldsymbol{6}\)
4. \(LR = \boldsymbol{9}\)
5. \(NK = \boldsymbol{2}\)
6. \(PM = \boldsymbol{6}\)
7. \(PD = \boldsymbol{15}\)
8. \(FP = \boldsymbol{26}\)
9. \(BP = \boldsymbol{13}\)
10. \(CD = \boldsymbol{45}\)
11. \(PA = \boldsymbol{9}\)
12. \(EA = \boldsymbol{27}\)
13. \(SH = \boldsymbol{16}\)
14. \(HM = \boldsymbol{8}\)
15. \(TH = \boldsymbol{8}\)
16. \(HR = \boldsymbol{3}\)
17. \(TD = \boldsymbol{12}\)
18. \(ER = \boldsymbol{9}\)