5-2 practice quiz
use the diagram shown for items 1 and 2.
1. select all the conditions that would be enough to prove
that p is the incenter of $\triangle hjk$.
$\square$ a. l, m, and n are the midpoints of $\overline{hk}$, $\overline{hj}$ and $\overline{kj}$.
$\square$ b. $\overline{pl} \cong \overline{pm} \cong \overline{pn}$
$\square$ c. $\overline{pk} \cong \overline{hp} \cong \overline{pj}$
$\square$ d. $\triangle hjk$ is an acute triangle.
$\square$ e. $\overline{kp}$, $\overline{hp}$, and $\overline{jp}$ are angle bisectors of the triangle
2. assume p is the incenter of $\triangle hjk$. if $lp = 4x + 10$ and $mp = 8x - 2$,
what is the radius of the inscribed circle of $\triangle hjk$
$\boldsymbol{\textcircled{a}}$ 3 $\boldsymbol{\textcircled{b}}$ 12 $\boldsymbol{\textcircled{c}}$ 18 $\boldsymbol{\textcircled{d}}$ 22
3. what is the radius of the circumscribed circle of a triangle at (6,0), (0,4) and (0,0)?
4. which point is the center of a circle that contains r, s,
and t?

Question

Accepted Answer

### Item 1
# Brief Explanations:
- The incenter of a triangle is the intersection of its angle bisectors, and it is equidistant from all three sides (these distances are the radii of the incircle).
  - Option A: Midpoints relate to centroids, not incenters, so this is incorrect.
  - Option B: If $ \overline{PL} \cong \overline{PM} \cong \overline{PN} $, $ P $ is equidistant from all sides, which defines the incenter (given $ PL, PM, PN $ are perpendicular to the sides), so this is correct.
  - Option C: Equal distances from vertices define the circumcenter, not incenter, so this is incorrect.
  - Option D: Triangle type does not determine the incenter's location, so this is incorrect.
  - Option E: The incenter is the intersection of angle bisectors, so if these are angle bisectors, $ P $ is the incenter, so this is correct.
# Answer:
B. $ \overline{PL} \cong \overline{PM} \cong \overline{PN} $
E. $ \overline{KP}, \overline{HP}, $ and $ \overline{JP} $ are angle bisectors of the triangle

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### Item 2
# Explanation:
## Step1: Set lengths equal (inradius property)
Since $ P $ is the incenter, $ LP = MP $, so:
$ 4x + 10 = 8x - 2 $
## Step2: Solve for $ x $
Rearrange to isolate $ x $:
$ 10 + 2 = 8x - 4x $
$ 12 = 4x $
$ x = \frac{12}{4} = 3 $
## Step3: Calculate inradius (substitute $ x $)
Use $ LP = 4x + 10 $:
$ LP = 4(3) + 10 = 12 + 10 = 22 $
# Answer:
D) 22

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### Item 3
# Explanation:
## Step1: Identify triangle type
The points $(6,0)$, $(0,4)$, $(0,0)$ form a right triangle (right angle at $(0,0)$).
## Step2: Use right triangle circumradius rule
For a right triangle, the circumradius is half the length of the hypotenuse. First calculate hypotenuse length between $(6,0)$ and $(0,4)$:
$ \text{Hypotenuse} = \sqrt{(6-0)^2 + (0-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} $
## Step3: Compute radius
$ \text{Radius} = \frac{1}{2} \times 2\sqrt{13} = \sqrt{13} $
# Answer:
$\sqrt{13}$ (or approximately 3.61)

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### Item 4
# Brief Explanations:
The center of a circle passing through three points is the circumcenter, which is the intersection of the perpendicular bisectors of the triangle's sides. In the diagram, point $ P $ lies on the perpendicular bisector of $ RT $ (marked with congruent segments) and $ N $ is on the perpendicular bisector of $ RS $, but $ P $ is the intersection of the perpendicular bisectors for the sides of $\triangle RST$.
# Answer:
Point $ P $

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QUESTION IMAGE

Question

Explanation:

Item 1

Step1: Set lengths equal (inradius property)

Step2: Solve for \( x \)

Step3: Calculate inradius (substitute \( x \))

Step1: Identify triangle type

Step2: Use right triangle circumradius rule

Step3: Compute radius

Answer:

Item 2