QUESTION IMAGE
Question
b. incenter
c. centroid
d. orthocenter
- classify point c.
a. circumcenter
b. incenter
c. centroid
d. orthocenter
c. centroid
d. orthocent
- classify point c
a. circumcenter
b. incenter
c. centroid
d. orthocenter
use the diagram below for questions 13-15 in which h is the circumcenter of $\triangle bcd$, $bc = 18$ and $hd = 14$.
- find $be$.
- find $ch$.
- find $eh$.
use the diagram below for questions 16-18 in which p is the incenter of $\triangle jkl$, $pn = 21$ and $ml = 27$.
- find $po$.
- find $lo$.
- find $pl$.
Step1: Solve for BE (Q13)
H is circumcenter, so BE is half BC.
$BE = \frac{BC}{2} = \frac{18}{2} = 9$
Step2: Solve for CH (Q14)
Circumcenter is equidistant from vertices, so $CH = HD = 14$
Step3: Solve for EH (Q15)
Use Pythagorean theorem in $\triangle BEH$: $EH = \sqrt{BH^2 - BE^2}$. $BH=HD=14$, so
$EH = \sqrt{14^2 - 9^2} = \sqrt{196 - 81} = \sqrt{115}$
Step4: Solve for PO (Q16)
P is incenter, so $PO = PN = 21$
Step5: Solve for LO (Q17)
Incenter creates congruent tangent segments, so $LO = ML = 27$
Step6: Solve for PL (Q18)
Use Pythagorean theorem in $\triangle PLO$: $PL = \sqrt{LO^2 + PO^2}$
$PL = \sqrt{27^2 + 21^2} = \sqrt{729 + 441} = \sqrt{1170} = 3\sqrt{130}$
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- $9$
- $14$
- $\sqrt{115}$
- $21$
- $27$
- $3\sqrt{130}$