Math Problem Statement

In triangle ABC, AB:AC = 5:8, and D and E are points on side BC such that AD is the angle bisector of ∠BAC and AE is the median of triangle ABC. If the area of triangle ABC is 104 cm², find the area of triangle ADE.

Solution

Let's break down the problem step by step to find the area of triangle ΔADE\Delta ADE.

Problem Summary:

In triangle ΔABC\Delta ABC:

  • The ratio of sides AB:AC=5:8AB:AC = 5:8.
  • DD and EE are points on side BCBC.
  • ADAD is the angle bisector of BAC\angle BAC.
  • AEAE is the median of ΔABC\Delta ABC.
  • The area of ΔABC=104cm2\Delta ABC = 104 \, \text{cm}^2.

We need to find the area of ΔADE\Delta ADE.

Step-by-Step Solution:

1. Use the Angle Bisector Theorem for ADAD:

The angle bisector theorem tells us that the angle bisector divides the opposite side in the ratio of the other two sides. Since AB:AC=5:8AB:AC = 5:8, the bisector ADAD will divide side BCBC in the ratio 5:8, i.e., BDDC=ABAC=58.\frac{BD}{DC} = \frac{AB}{AC} = \frac{5}{8}. Let the length of BC=xBC = x. Then: BD=513x,DC=813x.BD = \frac{5}{13}x, \quad DC = \frac{8}{13}x.

2. Use the Median AEAE:

The median AEAE divides triangle ΔABC\Delta ABC into two triangles of equal area. Therefore, the area of ΔABE\Delta ABE and the area of ΔACE\Delta ACE are both 1042=52cm2\frac{104}{2} = 52 \, \text{cm}^2.

3. Area of ΔADE\Delta ADE:

The triangle ΔADE\Delta ADE is part of both ΔABE\Delta ABE and ΔACE\Delta ACE. To find the area of ΔADE\Delta ADE, we observe that ADAD divides ΔABC\Delta ABC into two parts in the same ratio as the sides, i.e., 513\frac{5}{13} and 813\frac{8}{13}.

Thus, the area of ΔADE\Delta ADE is proportional to the area of ΔABC\Delta ABC. Since ΔADE\Delta ADE lies within both ΔABE\Delta ABE and ΔACE\Delta ACE, its area will be: Area of ΔADE=513×52=20cm2.\text{Area of } \Delta ADE = \frac{5}{13} \times 52 = 20 \, \text{cm}^2.

However, the area needs to be halved because we are looking at a smaller section formed by the angle bisector and median. This means: Area of ΔADE=202=10cm2.\text{Area of } \Delta ADE = \frac{20}{2} = 10 \, \text{cm}^2.

Thus, the area of ΔADE\Delta ADE is 10cm2\boxed{10 \, \text{cm}^2}.

Would you like further details or have any questions?


Here are 5 relative questions that build on this concept:

  1. How is the angle bisector theorem applied in general triangles?
  2. What is the significance of a median in dividing the area of a triangle?
  3. How can we calculate the area of a triangle using the sides and an angle?
  4. How do special points like the centroid and orthocenter relate to triangle areas?
  5. How does the ratio of sides affect the area distribution in triangles?

Tip: Understanding the role of the angle bisector theorem and median can help in solving many geometry problems involving proportionality in triangles.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Triangle Geometry
Area Ratios
Angle Bisector Theorem
Median in Triangles

Formulas

Angle Bisector Theorem: BD/DC = AB/AC
Area of triangle ADE = (Proportional Area) × (Total Area)

Theorems

Angle Bisector Theorem

Suitable Grade Level

Grades 9-11