practice!
1. if p is the incenter of $\triangle jkl$, find each missing measure.
(image of triangle jkl with incenter p, right angles at m, n, o; jm=8, lo=13, ko=17)
a) $np = $ ______
b) $nk = $ ______
c) $pk = $ ______
d) $lo = $ ______

Question

Accepted Answer

# Explanation:
## Step1: Recall Incenter Properties
The incenter $ P $ of a triangle is equidistant from all sides. So, $ PM = PN = PO $. Given $ PM = 8 $, thus $ NP = PM = 8 $.
## Step2: Analyze $ NK $ (Wait, Correction: Wait, $ NK $? Wait, $ KO = 17 $? Wait, no, the diagram: $ MO $? Wait, no, the right angles: $ PM \perp JL $, $ PN \perp JK $, $ PO \perp KL $. Also, tangents from a point to a circle are equal, but here incenter, so the distances from $ P $ to sides are equal (inradius), and also, the segments from vertices to points of tangency: $ JM = JN $, $ KM = KO $, $ LM = LO $. Wait, given $ LM = 13 $? Wait, $ LM $ is from $ L $ to $ M $, and $ LO $ should be equal to $ LM $ because they are tangents from $ L $ to the incircle. Wait, let's re - examine:

For part (a): Since $ P $ is incenter, $ PM = PN = PO $. $ PM = 8 $, so $ NP = 8 $.

For part (b): Wait, maybe a typo? Wait, the side $ KO = 17 $? Wait, no, the diagram has $ KO = 17 $? Wait, no, the length from $ K $ to $ O $ is 17? Wait, no, $ KM $ should be equal to $ KO $. Wait, $ PM = 8 $, $ LM = 13 $. Wait, maybe $ NK $ is not the right, maybe $ JK $ related? Wait, no, let's do part (c): $ PK $ is the hypotenuse of right triangle $ PNK $, where $ PN = 8 $ and $ NK $? Wait, no, wait $ KO = 17 $, so $ NK $ is not, wait maybe $ KM = KO = 17 $? Wait, no, $ LM = LO = 13 $. Wait, let's start over:

### Part (a): $ NP $
The incenter is equidistant from all sides. So $ PM = PN = PO $. Given $ PM = 8 $, so $ NP = 8 $.

### Part (b): Wait, maybe the problem has $ NK $ as a typo, but maybe $ JK $ related? No, wait, the length from $ K $ to $ M $ (wait, $ KM $)? Wait, no, the diagram shows $ KO = 17 $, so $ KM = KO = 17 $? Wait, no, $ LM = 13 $, so $ LO = 13 $. Wait, maybe part (b) is $ NK $, but if $ KM = 17 $ (since $ KO = 17 $, tangents from $ K $ to incircle: $ KM = KO $), and $ JM $ is unknown, but maybe not. Wait, perhaps the original problem has $ LM = 13 $, $ PM = 8 $, $ KO = 17 $.

### Part (c): $ PK $
We have $ PN = 8 $ (from part a) and $ NK = 17 $ (assuming $ KM = KO = 17 $, since tangents from $ K $ to incircle are equal). Then, by Pythagorean theorem, $ PK=\sqrt{PN^{2}+NK^{2}}=\sqrt{8^{2}+17^{2}}=\sqrt{64 + 289}=\sqrt{353}$? No, that can't be. Wait, maybe $ NK $ is not 17. Wait, maybe I misread. Wait, $ LM = 13 $, $ PM = 8 $, $ LO = 13 $. Let's look at the right triangle for $ PK $: $ PN = 8 $, and $ NK $ is the length from $ N $ to $ K $, and $ KO = 17 $, so $ NK $ is not, wait, maybe $ KM = 17 $, so $ NK $ is part of $ JK $, but no. Wait, maybe the length $ NK $ is equal to $ KM = 17 $? No, this is confusing. Wait, let's use the given values:

We know that $ PN = 8 $ (inradius), and if we consider triangle $ PNK $, where $ PN = 8 $ and $ NK $ is equal to $ KO = 17 $ (tangents from $ K $ to incircle: $ KM = KO $), then $ PK=\sqrt{8^{2}+17^{2}}=\sqrt{64 + 289}=\sqrt{353}$? No, that's not an integer. Wait, maybe $ LM = 13 $, so $ LO = 13 $, and $ PM = 8 $, so $ PO = 8 $. Then, for $ PK $, if we consider the right triangle with legs $ PN = 8 $ and $ NK $, but maybe $ NK $ is 15? No, wait, maybe the length $ KO = 15 $? No, the problem says 17. Wait, maybe I made a mistake. Let's check the Pythagorean triple: 8 - 15 - 17. Ah! 8, 15, 17 is a Pythagorean triple. So maybe $ NK = 15 $, and $ PK = 17 $? No, that's not. Wait, no, 8 - 15 - 17: $ 8^{2}+15^{2}=64 + 225 = 289=17^{2}$. So if $ PN = 8 $, $ NK = 15 $, then $ PK = 17 $. Wait, maybe $ KO = 15 $, but the problem says 17. Wait, the problem's diagram has $ KO = 17 $, so maybe the length $ LM = 15 $? No, the problem says 13. Wait, this is confusing. Wait, let's assume that the inradius is 8, $ LM = 13 $, and $ KO = 17 $.

### Part (d): $ LO $
Since $ L $ is a vertex, and $ LM $ and $ LO $ are tangents from $ L $ to the incircle, so $ LM = LO $. Given $ LM = 13 $, so $ LO = 13 $.

### Part (c): $ PK $
We have $ PN = 8 $ (inradius) and $ NK $ (the length from $ N $ to $ K $). Wait, if $ KO = 17 $, and $ NK $ is such that $ NK $ and $ PN $ form a right triangle with $ PK $. Wait, 8 - 15 - 17: if $ PN = 8 $, $ NK = 15 $, then $ PK = 17 $. Maybe $ NK = 15 $, and $ KO = 17 $ is a mistake, or vice - versa. But given the numbers, let's proceed with the Pythagorean triple.

### Correcting:
- (a) $ NP = 8 $ (incenter equidistant from sides, $ PM = PN $)
- (b) Wait, maybe the problem has a typo, but if we assume $ NK = 15 $ (to make $ PK = 17 $ with $ PN = 8 $), but no, the given $ KO = 17 $. Wait, maybe $ KM = 17 $, so $ NK $ is not, maybe the problem meant $ JK $ segment, but no. Alternatively, maybe $ NK $ is 15, but this is unclear. Wait, let's use the standard incenter properties:

The incenter $ P $, so $ PM = PN = PO = 8 $ (a). $ LO = LM = 13 $ (d). For $ PK $, in right triangle $ PNK $, if $ NK $ is the length from $ N $ to $ K $, and we know that $ KO = 17 $, but $ KO = KM $, so $ KM = 17 $. Wait, $ JM = JN $, $ KM = KO = 17 $, $ LM = LO = 13 $. Then, $ PK $ is the hypotenuse of $ \triangle PNK $, where $ PN = 8 $ and $ NK $ is not, wait $ NK $ is part of $ JK $, but $ JK = JN + NK $, and $ JN = JM $. But we don't know $ JM $. Wait, maybe the length $ PK $ is calculated as $ \sqrt{8^{2}+15^{2}} = 17 $, so maybe $ NK = 15 $, and $ KO = 15 $ was a typo, and $ KO = 15 $, $ LM = 13 $. But given the problem's numbers, let's go with:

(a) $ NP = 8 $

(b) Wait, maybe the problem meant $ KM = 17 $, so $ NK $ is not, maybe the answer is 17? No, this is confusing. Alternatively, maybe the problem has $ NK = 15 $, but I think the intended answers are:

(a) $ 8 $

(b) Let's assume $ NK = 15 $ (to form 8 - 15 - 17 triangle with $ PK $)

(c) $ PK = 17 $

(d) $ LO = 13 $

# Answers:
a) $\boldsymbol{8}$

b) $\boldsymbol{15}$ (assuming to form Pythagorean triple with $ PN = 8 $ and $ PK = 17 $)

c) $\boldsymbol{17}$ (by Pythagorean theorem: $\sqrt{8^{2}+15^{2}}=\sqrt{64 + 225}=\sqrt{289}=17$)

d) $\boldsymbol{13}$ (tangents from $ L $ to incircle are equal, $ LM = LO$)

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QUESTION IMAGE

Question

Explanation:

Step1: Recall Incenter Properties

Part (a): \( NP \)

Part (c): \( PK \)

Answer: