directions: if p is the incenter of $ \triangle jkl $, find each measure.
18. find $ m \angle jkp $.
19. find mp.
20. find pj.

Question

Accepted Answer

### 18. Find $ m\angle JKP$
# Explanation:
## Step1: Solve for $ x$
Since $ P$ is the in - center, the angle bisectors divide the angles. We know that $7x - 6=5x + 4$ (because the in - center's angle - bisecting property gives equal angles from the same vertex).
$$
\begin{align*}
7x-6&=5x + 4\
7x-5x&=4 + 6\
2x&=10\
x&=5
\end{align*}
$$
## Step2: Find $m\angle JKL$
First, find $m\angle J$ and $m\angle L$. $m\angle J=(5x + 4)^{\circ}$, substituting $x = 5$, $m\angle J=(5\times5+4)^{\circ}=29^{\circ}$, $m\angle L = 26^{\circ}$.
Using the angle - sum property of a triangle ($m\angle J+m\angle K+m\angle L=180^{\circ}$), we get $m\angle K=180-(29 + 26)=125^{\circ}$.
Since $KP$ is an angle bisector, $m\angle JKP=\frac{1}{2}m\angle JKL$.
$$m\angle JKP=\frac{1}{2}\times125^{\circ}=62.5^{\circ}$$

# Answer:
$62.5^{\circ}$

### 19. Find $MP$
# Explanation:
## Step1: Solve for $x$
Since $P$ is the in - center, the distances from the in - center to the sides of the triangle are equal. So $3x + 14=9x-34$ (because $PN = PO$ and by the property of the in - center, $MP = PN=PO$).
$$
\begin{align*}
3x+14&=9x - 34\
- 6x&=-48\
x&=8
\end{align*}
$$
## Step2: Find $MP$
Substitute $x = 8$ into $3x + 14$ (or $9x-34$). $MP=3\times8+14=24 + 14=38$

# Answer:
$38$

### 20. Find $PJ$
# Explanation:
## Step1: Solve for $x$
Since $P$ is the in - center, the distances from the in - center to the sides of the triangle are equal. So $2x + 3=22$ (because $PM = PN$ and by the property of the in - center).
$$
\begin{align*}
2x+3&=22\
2x&=19\
x&=\frac{19}{2}
\end{align*}
$$
## Step2: Find $PJ$
We know that $PJ$ can be found using the Pythagorean theorem (if we consider the right - triangle formed by the perpendicular from $P$ to $JK$ and $PJ$). But since $PM = PN = 22$ (from the in - center property, $PM$ is the length from the in - center to the side $JK$).
Another way: Since $2x+3$ (the length from $J$ to the point of tangency on $JK$) and using the fact that $PJ$ is related to the equal - distance property of the in - center. Substituting $x=\frac{19}{2}$ into $2x + 3$ gives $2\times\frac{19}{2}+3=19 + 3=22$ (but actually, using the property that $PJ$ is the hypotenuse of a right - triangle with one leg equal to the in - center's distance to the side ($22$) and the other leg related to the segment on the side. However, a more straightforward way is using the in - center's equal - distance property for the perpendiculars to the sides. Since $PN = 22$ and $PM = PN$ (in - center property), and if we assume the right - triangle $PJM$ (where $PM\perp JK$), and using the fact that $PJ$ is the hypotenuse of a right - triangle with $PM = 22$ (distance from in - center to side) and the segment from $J$ to the point of tangency. But actually, since $PM = PN$ (in - center property) and if we consider the congruence of right - triangles (formed by the in - center and the sides), $PJ=\sqrt{(2x + 3)^{2}+PM^{2}}$, but since $2x+3$ (the non - perpendicular segment from $J$ to the point of tangency) and $PM = 22$ (perpendicular from in - center to side). Wait, no, actually, since $PM = PN = 22$ (in - center property: distances from in - center to sides are equal). And using the fact that $PJ$ is the hypotenuse of a right - triangle with $PM = 22$ (one leg) and if we assume the other leg is $JM$. But from $2x+3$ (assuming $JM=2x + 3$ and $x=\frac{19}{2}$, $JM=22$). Then $PJ=\sqrt{22^{2}+22^{2}}=\sqrt{2\times22^{2}}=22\sqrt{2}\approx31.11$. But a better approach: Since $P$ is the in - center, $PM = PN$ (distances from in - center to sides). If we assume the right - triangle $PJM$ and $PNL$ (where $PM\perp JK$, $PN\perp KL$). And from $2x+3$ (let's assume $JM = 2x+3$) and $PN = 22$. Since $x=\frac{19}{2}$, $JM=22$. Then $PJ=\sqrt{22^{2}+22^{2}}=22\sqrt{2}\approx31.11$. But wait, another way: Since $P$ is the in - center, the lengths of the perpendiculars from $P$ to the sides are equal. Let $d = 22$ (the length of $PN$). If we consider the right - triangle formed by $PJ$, $PM$ (where $PM = d = 22$) and $JM$. If $JM=2x + 3$ and $x=\frac{19}{2}$, $JM = 22$. Then by Pythagoras $PJ=\sqrt{22^{2}+22^{2}}=22\sqrt{2}$

# Answer:
$22\sqrt{2}\approx31.1$

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QUESTION IMAGE

Question

Explanation:

18. Find \( m\angle JKP\)

Step1: Solve for \( x\)

Step2: Find \(m\angle JKL\)

Step1: Solve for \(x\)

Step2: Find \(MP\)

Step1: Solve for \(x\)

Step2: Find \(PJ\)

Answer:

19. Find \(MP\)