Medians and Altitudes of Triangles | 264 plays | Quizizz - Free Printable

Problem Analysis:

The image shows a triangle \( \triangle ABC \) with the following information:
- \( AB = 5.1 \)
- \( BC = 8.2 \)
- \( CA = 9.4 \)

From vertex \( B \), three lines are drawn:
1. Altitude: A perpendicular line from \( B \) to side \( AC \).
2. Median: A line segment from \( B \) to the midpoint of side \( AC \).
3. Angle Bisector: A line segment from \( B \) that bisects \( \angle ABC \).

The task is to solve for specific properties or lengths related to these lines, but since no explicit question is provided in the image, I will outline the general approach to solving problems involving altitudes, medians, and angle bisectors.

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Step-by-Step Solution:

#### 1. Altitude from \( B \) to \( AC \):
The altitude from \( B \) to \( AC \) is a perpendicular line segment from \( B \) to \( AC \). Let the foot of the altitude be \( D \) on \( AC \). The length of the altitude can be found using the area of the triangle in two ways:
- Using Heron's formula to find the area of \( \triangle ABC \).
- Using the formula for the area of a triangle: \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \).

##### Step 1.1: Calculate the semi-perimeter \( s \):
\[
s = \frac{AB + BC + CA}{2} = \frac{5.1 + 8.2 + 9.4}{2} = 11.35
\]

##### Step 1.2: Use Heron's formula to find the area \( K \):
\[
K = \sqrt{s(s - AB)(s - BC)(s - CA)} = \sqrt{11.35(11.35 - 5.1)(11.35 - 8.2)(11.35 - 9.4)}
\]
\[
K = \sqrt{11.35 \times 6.25 \times 3.15 \times 1.95}
\]
Using a calculator:
\[
K \approx \sqrt{430.18} \approx 20.74
\]

##### Step 1.3: Find the altitude \( BD \):
The area can also be expressed as:
\[
K = \frac{1}{2} \times AC \times BD
\]
\[
20.74 = \frac{1}{2} \times 9.4 \times BD
\]
\[
BD = \frac{2 \times 20.74}{9.4} \approx 4.41
\]

#### 2. Median from \( B \) to \( AC \):
The median from \( B \) to \( AC \) is a line segment from \( B \) to the midpoint \( M \) of \( AC \). The coordinates of \( M \) are the averages of the coordinates of \( A \) and \( C \). However, without specific coordinates, we can use Apollonius's theorem to find the length of the median:
\[
BM^2 = \frac{2AB^2 + 2BC^2 - AC^2}{4}
\]
\[
BM^2 = \frac{2(5.1)^2 + 2(8.2)^2 - (9.4)^2}{4}
\]
\[
BM^2 = \frac{2(26.01) + 2(67.24) - 88.36}{4}
\]
\[
BM^2 = \frac{52.02 + 134.48 - 88.36}{4} = \frac{98.14}{4} = 24.535
\]
\[
BM \approx \sqrt{24.535} \approx 4.95
\]

#### 3. Angle Bisector from \( B \) to \( AC \):
The angle bisector from \( B \) to \( AC \) divides \( AC \) into segments proportional to the adjacent sides \( AB \) and \( BC \). Let the point where the angle bisector intersects \( AC \) be \( P \). By the angle bisector theorem:
\[
\frac{AP}{PC} = \frac{AB}{BC} = \frac{5.1}{8.2}
\]
Let \( AP = x \) and \( PC = 9.4 - x \). Then:
\[
\frac{x}{9.4 - x} = \frac{5.1}{8.2}
\]
\[
8.2x = 5.1(9.4 - x)
\]
\[
8.2x = 48.06 - 5.1x
\]
\[
13.3x = 48.06
\]
\[
x = \frac{48.06}{13.3} \approx 3.61
\]
So, \( AP \approx 3.61 \) and \( PC \approx 5.79 \).

The length of the angle bisector \( BP \) can be found using the angle bisector length formula:
\[
BP = \sqrt{AB \cdot BC \left(1 - \frac{AC^2}{(AB + BC)^2}\right)}
\]
\[
BP = \sqrt{5.1 \cdot 8.2 \left(1 - \frac{9.4^2}{(5.1 + 8.2)^2}\right)}
\]
\[
BP = \sqrt{5.1 \cdot 8.2 \left(1 - \frac{88.36}{18.3^2}\right)}
\]
\[
BP = \sqrt{5.1 \cdot 8.2 \left(1 - \frac{88.36}{334.89}\right)}
\]
\[
BP = \sqrt{5.1 \cdot 8.2 \left(1 - 0.264\right)}
\]
\[
BP = \sqrt{5.1 \cdot 8.2 \cdot 0.736}
\]
\[
BP \approx \sqrt{30.42 \cdot 0.736} \approx \sqrt{22.34} \approx 4.73
\]

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Final Answer:

The lengths of the altitude, median, and angle bisector from \( B \) are approximately:
\[
\boxed{4.41, 4.95, 4.73}
\]