medians and altitudes
1. select all the statements that are true for △abc.
□ a. $ \overline { a h } $ is an altitude.
□ b. $ \overline { i h } $ is a median.
□ c. $ \overline { j c } $ is a median.
□ d. the medians and altitudes intersect at the same point.
□ e. the altitudes intersect outside of the triangle.
2. in △abc, the segment $ c j = 18 $. if $ c g = b g $, what is $ k j $?
(a) 3
(b) 6
(c) 9
(d) 12
for items 3 - 4, use the coordinates $ j ( 7,8 ), k ( 1,2 ) $ and $ l ( 5,2 ) $ for △jkl.
3. the centroid for △jkl is at point m. what is jm rounded to the nearest tenth?
(a) 2.4
(b) 4.3
(c) 4.8
(d) 7.2
4. the orthocenter for △jkl is at point n. what is kn rounded to the nearest tenth?
(a) 2.7
(b) 4.3
(c) 6.0
(d) 6.3
5. what is the relationship between a scalene triangle and the location of its centroid? explain.
(a) the centroid will always be inside the scalene triangle, since the centroid is always inside any type of triangle.
(b) the centroid will always be outside the scalene triangle, since the centroid is always outside any type of triangle.
(c) the centroid will always be on the scalene triangle, since the centroid is always on a triangle with no congruent side lengths.
(d) the centroid will be inside, on, or outside the scalene triangle, depending on whether it is acute, right, or obtuse, respectively. the number of congruent sides does not affect the location of the centroid.

Question

Accepted Answer

### 1.
# Brief Explanations:
- **Option A**: An altitude is a perpendicular segment from a vertex to the line containing the opposite side. Since $AH\perp BC$, $\overline{AH}$ is an altitude.
- **Option B**: A median is a segment from a vertex to the mid - point of the opposite side. $\overline{IH}$ does not connect a vertex to the mid - point of the opposite side.
- **Option C**: Since $CG = BG$, $G$ is the mid - point of $BC$. A median is a segment from a vertex to the mid - point of the opposite side. $\overline{JC}$ connects vertex $J$ to the mid - point $G$ of $BC$, so $\overline{JC}$ is a median.
- **Option D**: In a triangle, the medians intersect at the centroid and the altitudes intersect at the orthocenter. In an acute triangle (the given triangle appears to be acute), they do not intersect at the same point.
- **Option E**: In an acute triangle, the altitudes intersect inside the triangle.

# Answer:
A. $\overline{AH}$ is an altitude; C. $\overline{JC}$ is a median

### 2.
# Explanation:
## Step1: Recall the property of the centroid
The centroid of a triangle divides each median in a ratio of $2:1$. If $CJ$ is a median and $K$ is the centroid, then $CK: KJ=2:1$ and $CJ = CK + KJ$. Let $KJ=x$, then $CK = 2x$.
## Step2: Solve for $x$
Since $CJ=18$ and $CJ=CK + KJ$, we have $18=2x + x$. Combining like terms gives $18 = 3x$. Dividing both sides by $3$, we get $x=\frac{18}{3}=6$.

# Answer:
B. $6$

### 3.
# Explanation:
## Step1: Find the centroid formula
The centroid $M$ of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ has coordinates $M(\frac{x_1 + x_2+x_3}{3},\frac{y_1 + y_2 + y_3}{3})$. For $\triangle JKL$ with $J(7,8)$, $K(1,2)$, $L(5,2)$, the $x$ - coordinate of $M$ is $\frac{7 + 1+5}{3}=\frac{13}{3}\approx4.3$ and the $y$ - coordinate of $M$ is $\frac{8 + 2+2}{3}=4$.
## Step2: Use the distance formula
The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Here, $x_1 = 7,y_1 = 8,x_2=\frac{13}{3},y_2 = 4$. Then $JM=\sqrt{(7-\frac{13}{3})^2+(8 - 4)^2}=\sqrt{(\frac{21 - 13}{3})^2+16}=\sqrt{(\frac{8}{3})^2+16}=\sqrt{\frac{64}{9}+16}=\sqrt{\frac{64 + 144}{9}}=\sqrt{\frac{208}{9}}\approx4.8$

# Answer:
C. $4.8$

### 4.
# Explanation:
## Step1: Analyze the triangle $\triangle JKL$
Since $K(1,2)$ and $L(5,2)$ have the same $y$ - value, $KL$ is a horizontal line. The altitude from $J$ to $KL$ is a vertical line $x = 7$. The altitude from $K$ to $JL$: The slope of $JL$ is $m_{JL}=\frac{8 - 2}{7 - 5}=\frac{6}{2}=3$, so the slope of the altitude from $K$ to $JL$ is $-\frac{1}{3}$. The equation of the altitude from $K(1,2)$ with slope $-\frac{1}{3}$ is $y-2=-\frac{1}{3}(x - 1)$. The orthocenter $N=(7,2)$ (intersection of $x = 7$ and $y-2=-\frac{1}{3}(x - 1)$ when $x = 7$, $y-2=-\frac{1}{3}(7 - 1)=-2$, $y=0$ is wrong, actually for $\triangle JKL$ with $KL$ horizontal ($y = 2$) and altitude from $J$ is $x=7$, altitude from $L$ to $JK$: slope of $JK$ is $m_{JK}=\frac{8 - 2}{7 - 1}=1$, slope of altitude from $L$ is $-1$, equation $y - 2=-(x - 5)$, intersection of $x = 7$ and $y-2=-(x - 5)$ gives $y=0$ (error in previous thought, correct: since $KL$ is horizontal ($y = 2$) and $JK$ has slope $1$, altitude from $L$ to $JK$: $y-2=-(x - 5)$, altitude from $J$ is $x = 7$. Substitute $x = 7$ into $y-2=-(x - 5)$ gives $y=0$. The distance between $K(1,2)$ and $N(7,2)$ using the distance formula $d=\sqrt{(7 - 1)^2+(2 - 2)^2}=\sqrt{36}=6.0$

# Answer:
C. $6.0$

### 5.
# Brief Explanations:
The centroid of a triangle is the intersection of its medians. By the definition of the centroid (the point of intersection of the medians of a triangle),  the centroid of any triangle (acute, obtuse, right - angled, scalene, isosceles, equilateral) is always inside the triangle.

# Answer:
A. The centroid will always be inside the scalene triangle, since the centroid is always inside any type of triangle.

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QUESTION IMAGE

Question

Explanation:

1.

Step1: Recall the property of the centroid

Step2: Solve for \(x\)

Step1: Find the centroid formula

Step2: Use the distance formula

Answer:

2.