1. select all the statements that are true for $ \triangle a b c $.
$ \square $ a. $ \overline{a h} $ is an altitude.
$ \square $ b. $ \overline{i h} $ is a median.
$ \square $ c. $ \overline{j c} $ is a median.
$ \square $ d. the medians and altitudes intersect at the same point.
$ \square $ e. the altitudes intersect outside of the triangle.
2. in $ \triangle a b c $, the segment $ c j = 18 $. if $ c g = b g $, what is $ k j $?
(a) 3
(c) 9
(b) 6
(d) 12
for items 3 - 4, use the coordinates $ j ( 7, 8 ), k ( 1, 2 ) $ and $ l ( 5, 2 ) $ for $ \triangle j k l $.
3. the centroid for $ \triangle j k l $ is at point $ m $. what is $ j m $ rounded to the nearest tenth?
(a) 2.4
(c) 4.8
(b) 4.3
(d) 7.2
4. the orthocenter for $ \triangle j k l $ is at point $ n $. what is $ k n $ rounded to the nearest tenth?
(a) 2.7
(c) 6.0
(b) 4.3
(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.
- **A. $\overline{AH}$ is an altitude**:
    - An altitude is a perpendicular segment from a vertex to the line containing the opposite side. Since $AH\perp CB$, $\overline{AH}$ is an altitude. This statement is True.
- **B. $\overline{IH}$ is a median**:
    - A median is a segment from a vertex to the mid - point of the opposite side. There is no indication that $H$ is the mid - point of any side with respect to $I$. This statement is False.
- **C. $\overline{JC}$ is a median**:
    - A median is a segment from a vertex to the mid - point of the opposite side. Since $J$ is the mid - point of $AB$ (as $AJ = JB$ from the markings), $\overline{JC}$ is a median. This statement is True.
- **D. The medians and altitudes intersect at the same point**:
    - In a triangle, the centroid (intersection of medians) and the orthocenter (intersection of altitudes) are not the same point in a non - equilateral triangle. This statement is False.
- **E. The altitudes intersect outside of the triangle**:
    - In an acute triangle (which $\triangle ABC$ appears to be), the altitudes intersect inside the triangle. This statement is False.

### 2.
- Recall the property of the centroid of a triangle: The centroid divides each median in a ratio of $2:1$.
    - If $CJ$ is a median and $CJ = 18$, and the centroid $K$ divides $CJ$ such that $CK: KJ=2:1$.
    - Let $CK = 2x$ and $KJ=x$. Then $CJ=CK + KJ=2x + x=3x$.
    - Given $CJ = 18$, we have $3x = 18$, so $x=\frac{18}{3}=6$.

### 3.
- **Step1: Find the mid - point of $KL$**
    - The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. For $K(1,2)$ and $L(5,2)$, the mid - point $N=(\frac{1 + 5}{2},\frac{2+2}{2})=(3,2)$.
- **Step2: Use the centroid formula**
    - The centroid formula for a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is $M=(\frac{x_1 + x_2+x_3}{3},\frac{y_1 + y_2 + y_3}{3})$. For $J(7,8)$, $K(1,2)$ and $L(5,2)$, $M=(\frac{7 + 1+5}{3},\frac{8 + 2+2}{3})=( \frac{13}{3},\frac{12}{3})=( \frac{13}{3},4)$.
- **Step3: Use the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$**
    - For $J(7,8)$ and $M(\frac{13}{3},4)$, $x_1 = 7,y_1 = 8,x_2=\frac{13}{3},y_2 = 4$.
    - $JM=\sqrt{(7-\frac{13}{3})^2+(8 - 4)^2}=\sqrt{(\frac{21 - 13}{3})^2+16}=\sqrt{(\frac{8}{3})^2+16}=\sqrt{\frac{64}{9}+16}=\sqrt{\frac{64 + 144}{9}}=\sqrt{\frac{208}{9}}\approx4.8$.

### 4.
- **Step1: Find the equations of two altitudes**
    - The slope of $KL$ is $m_{KL}=\frac{2 - 2}{5 - 1}=0$. The altitude from $J$ to $KL$ is a vertical line $x = 7$.
    - The slope of $JK$ is $m_{JK}=\frac{8 - 2}{7 - 1}=1$. The slope of the altitude from $L$ to $JK$ is $m=-1$. The equation of the line passing through $L(5,2)$ with slope $-1$ is $y - 2=-(x - 5)$, or $y=-x + 7$.
- **Step2: Find the orthocenter $N$**
    - Substitute $x = 7$ into $y=-x + 7$, we get $y=0$. So $N=(7,0)$.
- **Step3: Use the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$**
    - For $K(1,2)$ and $N(7,0)$, $x_1 = 1,y_1 = 2,x_2 = 7,y_2 = 0$.
    - $KN=\sqrt{(7 - 1)^2+(0 - 2)^2}=\sqrt{36 + 4}=\sqrt{40}\approx6.3$.

### 5.
- **Brief Explanations**:
    - The centroid of a triangle is defined as the average of the coordinates of its vertices. By the formula $G=(\frac{x_A+x_B + x_C}{3},\frac{y_A+y_B + y_C}{3})$, since it is a weighted average of the interior points (vertices) of the triangle,  the centroid will always be inside any type of triangle (acute, obtuse, right - angled, scalene, isosceles, equilateral).

# Answer:
1. A. $\overline{AH}$ is an altitude (True), C. $\overline{JC}$ is a median (True)
2. B. $6$
3. C. $4.8$
4. D. $6.3$
5. A. The centroid will always be inside the scalene triangle, since the centroid is always inside any type of triangle.

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