Question

In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\). 

Hint:

For triangles sharing the same height, the ratio of their areas is simply the ratio of their bases. Similarly, for triangles sharing the same base, the ratio of their areas is the ratio of their heights.

Answer 1

Step 1: Understanding the Concept:
The ratio of the areas of two triangles with a common vertex and bases on the same straight line is equal to the ratio of their corresponding bases. This is because both triangles share the same height.

Step 2: Key Formula or Approach:
If \(\triangle ABD\) and \(\triangle ABC\) have a common vertex A and their bases BD and BC lie on the same line, their height 'h' from vertex A is the same.
The area of a triangle is given by \( \frac{1}{2} \times \text{base} \times \text{height} \).
Therefore, the ratio of their areas is: \[ \frac{A(\triangle ABD)}{A(\triangle ABC)} = \frac{\frac{1}{2} \times BD \times h}{\frac{1}{2} \times BC \times h} = \frac{BD}{BC} \]

Step 3: Detailed Explanation:
Given: \[\begin{array}{rl} \bullet & \text{BD = 7} \\ \bullet & \text{BC = 20} \\ \end{array}\] \(\triangle ABD\) and \(\triangle ABC\) share a common vertex A. Their bases BD and BC are on the same line. Let 'h' be the height of both triangles from the common vertex A to the line containing their bases.
Area of \(\triangle ABD = \frac{1}{2} \times BD \times h = \frac{1}{2} \times 7 \times h \).
Area of \(\triangle ABC = \frac{1}{2} \times BC \times h = \frac{1}{2} \times 20 \times h \).
Now, we find the ratio of their areas: \[ \frac{A(\triangle ABD)}{A(\triangle ABC)} = \frac{\frac{1}{2} \times 7 \times h}{\frac{1}{2} \times 20 \times h} \] The terms \( \frac{1}{2} \) and \( h \) cancel out. \[ \frac{A(\triangle ABD)}{A(\triangle ABC)} = \frac{7}{20} \]

Step 4: Final Answer:
The ratio \( \frac{A(\triangle ABD)}{A(\triangle ABC)} \) is \( \frac{7}{20} \).