Math Problem Statement

Solution

The problem involves a triangle ABC\triangle ABC where a line DEDE is drawn parallel to BCBC. You are given the area of DEC\triangle DEC as 238cm2238 \, \text{cm}^2 and the lengths AD=5cmAD = 5 \, \text{cm} and AB=12cmAB = 12 \, \text{cm}. The task is to find the area of ABC\triangle ABC.

Solution:

Since DEBCDE \parallel BC, triangles ADE\triangle ADE and ABC\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 ADAB\frac{AD}{AB} gives: ADAB=5cm12cm=512\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: Area of ADEArea of ABC=(ADAB)2=(512)2=25144\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 ABC\triangle ABC be xx. Then, the area of ADE\triangle ADE would be: Area of ADE=25144×x\text{Area of } \triangle ADE = \frac{25}{144} \times x Since the area of DEC\triangle DEC is 238cm2238 \, \text{cm}^2 and ABC\triangle ABC is divided into ADE\triangle ADE and DEC\triangle DEC: Area of ABC=Area of ADE+Area of DEC=25144x+238\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=5:12AD:AB = 5:12, we can calculate DE:BCDE:BC will be the same, and hence: Area of ADEArea of DEC=(DEBC)2=(512)2=25144\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, Area of DECArea of ABC=238Area of ABC\frac{\text{Area of } \triangle DEC}{\text{Area of } \triangle ABC} = \frac{238}{\text{Area of } \triangle ABC} Area of ABC=122×23852=12×12×23825250cm2\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 ABC\triangle ABC is 250cm2250 \, \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.

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Math Problem Analysis

Mathematical Concepts

Geometry
Similarity of triangles

Formulas

Area ratio of similar triangles

Theorems

AA similarity criterion

Suitable Grade Level

Grades 7-9