use the diagram below for questions 16-18 in which p is the incenter of δjkl, pn = 21 and ml = 27.
19. find po.
po =
20. find lo.
lo =
21. find pl.
pl =
use the diagram below for questions 21-24 in which g is the centroid of δabc, fc = 35, ag = 42, bf = 57 and dg = 14.
32. find ac.
ac =
33. find bg.
bg =
34. find gc.
gc =
35. find ae.
ae =

Question

Accepted Answer

# Explanation:
## Step1: Recall Incenter Properties
The incenter $ P $ of a triangle is equidistant from all sides. So, $ PO = PN $ (since $ PN \perp JK $ and $ PO \perp KL $, both are distances from incenter to sides). Given $ PN = 21 $, thus $ PO = 21 $.
## Step2: Recall Angle Bisector and Segment
For $ LO $, since $ ML = 27 $ and $ M $ and $ O $ are points of tangency from $ L $ to the incircle, $ ML = LO $ (tangents from a point to a circle are equal). So $ LO = 27 $.
## Step3: Use Pythagorean Theorem for $ PL $
In right triangle $ PLO $, $ PL $ is the hypotenuse, $ PO = 21 $, $ LO = 27 $. By Pythagoras: $ PL=\sqrt{PO^{2}+LO^{2}}=\sqrt{21^{2}+27^{2}}=\sqrt{441 + 729}=\sqrt{1170}=\sqrt{9\times130}=3\sqrt{130}\approx34.2 $. Wait, but maybe using tangents: Wait, no, maybe I misread. Wait, the inradius is $ PN = 21 $, and $ LO = ML = 27 $. Then $ PL $ is the hypotenuse of right triangle with legs 21 and 27. So $ PL=\sqrt{21^{2}+27^{2}}=\sqrt{441 + 729}=\sqrt{1170}=3\sqrt{130}\approx34.2 $. But maybe the problem expects using the inradius and tangent length. Wait, maybe the diagram has $ ML = LO = 27 $, and $ PO = 21 $, so $ PL=\sqrt{21^{2}+27^{2}} $. Alternatively, maybe the triangle is such that $ PL $ is calculated as $ \sqrt{21^2 + 27^2} $.

For Centroid (Questions 32 - 35):
## Step32: Centroid and Median
Centroid $ G $ divides median $ AF $ (where $ F $ is midpoint of $ AC $)? Wait, no, $ FC = 35 $, since $ F $ is midpoint (centroid is on median, so $ AF $ is median, so $ AF = 2\times FC $? Wait, no, centroid divides median in ratio $ 2:1 $, but $ FC $: if $ F $ is midpoint of $ AC $, then $ AC = 2\times FC $. Given $ FC = 35 $, so $ AC = 2\times35 = 70 $.
## Step33: Centroid and Median $ BE $
Centroid $ G $ divides median $ BE $ in ratio $ 2:1 $. $ DG = 14 $, but $ D $ is midpoint of $ AB $, so $ AG = 42 $, and centroid divides $ AG $ (median $ AD $) in ratio $ 2:1 $, so $ AG = \frac{2}{3}AD $, but for $ BG $: Wait, $ BF = 57 $, but maybe $ BE $ is median. Wait, maybe $ DG = 14 $, so $ AG = 42 $, so $ AD = AG + GD = 42 + 14 = 56 $, so median $ AD = 56 $, so $ BD = AD = 56 $? No, $ D $ is midpoint, so $ AD = BD $. Wait, maybe for $ BG $: Centroid divides median $ BE $ into $ BG:GE = 2:1 $. But we need length of $ BE $. Wait, maybe $ DG = 14 $, $ AG = 42 $, so $ AD = 56 $, so the median $ AD = 56 $, so the length from $ B $ to $ G $: Wait, maybe another approach. Wait, the centroid $ G $ has $ AG = 42 $, so $ GD = \frac{1}{2}AG = 21 $? Wait, no, centroid divides median in ratio $ 2:1 $, so $ AG:GD = 2:1 $, so $ GD = \frac{1}{2}AG = 21 $, but the problem says $ DG = 14 $. Maybe the diagram is different. Wait, maybe the first part (19 - 21) is about incenter, 32 - 35 about centroid.

Let's re-express:

### Question 19: $ PO $
Incenter $ P $, $ PN \perp JK $, $ PO \perp KL $, so $ PO = PN = 21 $ (incenter is equidistant from sides). So $ PO = 21 $.

### Question 20: $ LO $
Tangents from $ L $ to incircle: $ ML = LO $ (tangents from a point to a circle are equal), $ ML = 27 $, so $ LO = 27 $.

### Question 21: $ PL $
In right triangle $ PLO $, $ PO = 21 $, $ LO = 27 $, so $ PL = \sqrt{21^2 + 27^2} = \sqrt{441 + 729} = \sqrt{1170} = 3\sqrt{130} \approx 34.2 $. But maybe the problem expects integer, so maybe I made a mistake. Wait, maybe $ ML = LO = 27 $, $ PO = 21 $, so $ PL = \sqrt{21^2 + 27^2} = \sqrt{1170} \approx 34.2 $, but maybe the diagram has $ ML = 27 $, $ PO = 21 $, so $ PL = \sqrt{21^2 + 27^2} $.

### Question 32: $ AC $
Centroid $ G $, $ F $ is midpoint of $ AC $ (since $ AF $ is median), so $ AC = 2 \times FC $. Given $ FC = 35 $, so $ AC = 70 $.

### Question 33: $ BG $
Centroid divides median in ratio $ 2:1 $. $ DG = 14 $, $ AG = 42 $, so $ AD = AG + GD = 42 + 14 = 56 $ (median $ AD $). Then, median $ BE $: Wait, maybe $ BF = 57 $, but $ F $ is midpoint of $ AC $, so $ BF $ is median? No, $ B $ to $ F $ (midpoint of $ AC $) is median. So $ BF = 57 $, centroid $ G $ divides $ BF $ into $ BG:GF = 2:1 $, so $ BG = \frac{2}{3}BF = \frac{2}{3} \times 57 = 38 $.

### Question 34: $ GC $
Wait, $ GC $: Maybe $ G $ is centroid, $ DG = 14 $, $ AG = 42 $, so $ AD = 56 $, median $ AD $. Then $ GC $: Maybe from centroid to vertex? No, $ GC $ is part of median? Wait, maybe $ GC $ is related to $ DG $. Wait, no, let's use centroid properties. $ AG = 42 $, so $ GD = 21 $ (since $ AG:GD = 2:1 $), but problem says $ DG = 14 $, so maybe my assumption is wrong. Alternatively, $ FC = 35 $, $ G $ is centroid, so $ GC $ is part of median $ BE $? Wait, maybe $ GC = \sqrt{...} $, no, better to use centroid ratio. Wait, maybe $ GC = 2 \times DG = 28 $? No, $ DG = 14 $, $ D $ is midpoint of $ AB $, $ G $ is centroid, so $ DG = \frac{1}{3}AD $, $ AG = \frac{2}{3}AD $. So $ AD = \frac{3}{2}AG = \frac{3}{2} \times 42 = 63 $, so $ DG = \frac{1}{3} \times 63 = 21 $, but problem says $ DG = 14 $, so maybe the diagram has different lengths. Alternatively, maybe $ GC = 2 \times DG = 28 $, but this is confusing. Wait, maybe the user expects the following:

19. $ PO = 21 $ (since $ PO = PN = 21 $)
20. $ LO = 27 $ (since $ LO = ML = 27 $)
21. $ PL = \sqrt{21^2 + 27^2} = \sqrt{1170} \approx 34.2 $ (or simplified $ 3\sqrt{130} $)
32. $ AC = 70 $ (since $ AC = 2 \times FC = 2 \times 35 $)
33. $ BG = 38 $ (since $ BG = \frac{2}{3}BF = \frac{2}{3} \times 57 $)
34. $ GC = 28 $ (assuming $ DG = 14 $, $ GC = 2 \times DG $)
35. $ AE = 70 $ (since $ AE $ is median, $ AE = AC = 70 $? No, $ AE $ is median, so $ AE = 2 \times DG \times 3 $? Wait, no, $ AG = 42 $, so $ AE = AG + GE = 42 + 21 = 63 $? No, this is getting messy. Maybe the intended answers are:

19. $ PO = 21 $
20. $ LO = 27 $
21. $ PL = \sqrt{21^2 + 27^2} = 3\sqrt{130} \approx 34.2 $ (or 34 if rounded)
32. $ AC = 70 $
33. $ BG = 38 $
34. $ GC = 28 $
35. $ AE = 63 $ (since $ AG = 42 $, $ GE = 21 $, so $ AE = 42 + 21 = 63 $)

But let's focus on the first three (19 - 21) as per the diagram.

# Answer:
19. $ PO = \boldsymbol{21} $
20. $ LO = \boldsymbol{27} $
21. $ PL = \boldsymbol{3\sqrt{130}} $ (or approximately $ \boldsymbol{34.2} $)
32. $ AC = \boldsymbol{70} $
33. $ BG = \boldsymbol{38} $
34. $ GC = \boldsymbol{28} $
35. $ AE = \boldsymbol{63} $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Recall Incenter Properties

Step2: Recall Angle Bisector and Segment

Step3: Use Pythagorean Theorem for \( PL \)

Step32: Centroid and Median

Step33: Centroid and Median \( BE \)

Question 19: \( PO \)

Question 20: \( LO \)

Question 21: \( PL \)

Answer: