In a triangle ABC, points D and E respectively lie on side AB and AC such that `DE||BC`. If `AD=6cm, DB=(12x-6)cm, AE=2x cm and CE=16-2xcm` then what is the value of x

Options

A. 2
B. 4
C. `sqrt(2)`
D. 8

Answer

To solve the problem, we will use the properties of similar triangles. Since DE is parallel to BC, triangles ABC and ADE are similar. This allows us to set up a proportion between the corresponding sides of the triangles.

### Step 1: Identify the sides of the triangles
In triangle ABC:
- AB = AD + DB = 6 cm + (12x - 6) cm = 12x cm
- AC = AE + CE = 2x cm + (16 - 2x) cm = 16 cm

### Step 2: Set up the proportion using the similarity of triangles
Since triangles ABC and ADE are similar, we can write the following proportion:
\[ \frac{AB}{AD} = \frac{AC}{AE} \]

Substituting the values we found:
\[ \frac{12x}{6} = \frac{16}{2x} \]

### Step 3: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ 12x \cdot 2x = 6 \cdot 16 \]
This simplifies to:
\[ 24x^2 = 96 \]

### Step 4: Solve for x
Now, divide both sides by 24:
\[ x^2 = \frac{96}{24} \]
This simplifies to:
\[ x^2 = 4 \]

Taking the square root of both sides, we find:
\[ x = 2 \]

### Conclusion
The value of \( x \) is \( 2 \).

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