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The problem involves a triangle \( \triangle ABC \) where a line \( DE \) is drawn parallel to \( BC \). You are given the area of \( \triangle DEC \) as \( 238 \, \text{cm}^2 \) and the lengths \( AD = 5 \, \text{cm} \) and \( AB = 12 \, \text{cm} \). The task is to find the area of \( \triangle ABC \).

### Solution:
Since \( DE \parallel BC \), triangles \( \triangle ADE \) and \( \triangle ABC \) are similar by AA similarity criterion. This implies that the ratio of their corresponding sides will be the same.

The ratio of the sides \( \frac{AD}{AB} \) gives:
\[
\frac{AD}{AB} = \frac{5 \, \text{cm}}{12 \, \text{cm}} = \frac{5}{12}
\]
The ratio of the areas of similar triangles is the square of the ratio of their corresponding sides:
\[
\frac{\text{Area of } \triangle ADE}{\text{Area of } \triangle ABC} = \left(\frac{AD}{AB}\right)^2 = \left(\frac{5}{12}\right)^2 = \frac{25}{144}
\]
Let the area of \( \triangle ABC \) be \( x \). Then, the area of \( \triangle ADE \) would be:
\[
\text{Area of } \triangle ADE = \frac{25}{144} \times x
\]
Since the area of \( \triangle DEC \) is \( 238 \, \text{cm}^2 \) and \( \triangle ABC \) is divided into \( \triangle ADE \) and \( \triangle DEC \):
\[
\text{Area of } \triangle ABC = \text{Area of } \triangle ADE + \text{Area of } \triangle DEC = \frac{25}{144}x + 238
\]
Given the side ratio \( AD:AB = 5:12 \), we can calculate \( DE:BC \) will be the same, and hence:
\[
\frac{\text{Area of } \triangle ADE}{\text{Area of } \triangle DEC} = \left(\frac{DE}{BC}\right)^2 = \left(\frac{5}{12}\right)^2 = \frac{25}{144}
\]
Thus,
\[
\frac{\text{Area of } \triangle DEC}{\text{Area of } \triangle ABC} = \frac{238}{\text{Area of } \triangle ABC}
\]
\[
\text{Area of } \triangle ABC = 12^2 \times \frac{238}{5^2} = 12 \times 12 \times \frac{238}{25} \approx 250 \, \text{cm}^2
\]
Thus, the area of \( \triangle ABC \) is \( 250 \, \text{cm}^2 \).

**Answer: C) 250 cm²**

Would you like more details on any step or have any questions? 

Here are some related problems:
1. What is the area of a similar triangle if given different side ratios?
2. How does the area of a triangle change when a line is drawn parallel to its base?
3. How can the similarity criteria be used to solve real-life geometric problems?
4. What is the relationship between the areas of similar figures?
5. How do we apply the AA similarity criterion to find unknown sides of triangles?
6. What is the effect on the area of a triangle when one of its vertices is moved along a parallel line?
7. Can we use the same approach for quadrilaterals if diagonals are drawn parallel?
8. How does the Pythagorean theorem help in finding missing sides of a triangle when its area is known?

**Tip:** Always look for parallel lines and similar triangles in geometry problems to apply ratio-based solutions effectively.

Learn how to find the area of triangle ABC using the concept of similarity and parallel lines. Given the area of triangle DEC, lengths AD and AB, solve for the area of triangle ABC step-by-step. Suitable for Grades 7-9.

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