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In Delta ABC, D and E are the points on ...

In `Delta ABC, D` and E are the points on sides AB and BC respectively such that DE || AC. If AD : DB = 5 : 3, then what is the ratio of the area of `Delta BDE` to that of the trapezium ACED ?

A

`1 : 6`

B

`4 : 25`

C

`9 : 55`

D

`9 : 64`

Text Solution

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The correct Answer is:
To solve the problem, we need to find the ratio of the area of triangle \( BDE \) to the area of trapezium \( ACED \) given that \( DE \parallel AC \) and \( AD : DB = 5 : 3 \). ### Step-by-Step Solution: 1. **Understanding the Given Information:** - We have triangle \( ABC \) with points \( D \) on side \( AB \) and \( E \) on side \( BC \). - The line segment \( DE \) is parallel to \( AC \). - The ratio \( AD : DB = 5 : 3 \). 2. **Finding the Lengths:** - Let \( AD = 5x \) and \( DB = 3x \). - Therefore, the total length \( AB = AD + DB = 5x + 3x = 8x \). 3. **Using Similar Triangles:** - Since \( DE \parallel AC \), triangles \( BDE \) and \( ABC \) are similar by the Basic Proportionality Theorem (or Thales' theorem). - The ratio of corresponding sides of similar triangles is equal to the ratio of their areas. 4. **Calculating the Ratio of Areas:** - The ratio of the sides \( DB \) to \( AB \) is: \[ \frac{DB}{AB} = \frac{3x}{8x} = \frac{3}{8} \] - The ratio of the areas of triangles \( BDE \) and \( ABC \) is the square of the ratio of their corresponding sides: \[ \frac{\text{Area of } BDE}{\text{Area of } ABC} = \left(\frac{DB}{AB}\right)^2 = \left(\frac{3}{8}\right)^2 = \frac{9}{64} \] 5. **Finding the Area of Trapezium \( ACED \):** - The area of trapezium \( ACED \) can be found by subtracting the area of triangle \( BDE \) from the area of triangle \( ABC \): \[ \text{Area of } ACED = \text{Area of } ABC - \text{Area of } BDE \] - Let the area of triangle \( ABC \) be \( A \). Then: \[ \text{Area of } BDE = \frac{9}{64}A \] - Therefore: \[ \text{Area of } ACED = A - \frac{9}{64}A = \frac{64}{64}A - \frac{9}{64}A = \frac{55}{64}A \] 6. **Finding the Ratio of Areas:** - Now, we can find the ratio of the area of triangle \( BDE \) to the area of trapezium \( ACED \): \[ \frac{\text{Area of } BDE}{\text{Area of } ACED} = \frac{\frac{9}{64}A}{\frac{55}{64}A} = \frac{9}{55} \] ### Final Answer: The ratio of the area of triangle \( BDE \) to the area of trapezium \( ACED \) is \( 9 : 55 \).
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