A straight line parallel to the base BC of the triangle ABC intersects AB and AC at the points D and E respectively. If the area of the `Delta ABE` be 36 sq. cm, then the area of the `Delta ACD` is

Options

A. 18 sq. cm
B. 36 sq. cm
C. 18 cm
D. 36 cm

Answer

To solve the problem, we will use the properties of similar triangles and the relationship between the areas of triangles formed by a line parallel to the base of a triangle.

### Step-by-Step Solution:

1. **Understand the Configuration**:
We have triangle ABC with a line DE parallel to the base BC. Points D and E are on sides AB and AC respectively.

2. **Identify the Given Information**:
We are given that the area of triangle ABE is 36 sq. cm.

3. **Use the Property of Parallel Lines**:
Since DE is parallel to BC, triangles ABE and ACD are similar. This means that the ratios of their corresponding sides are equal.

4. **Set Up the Ratios**:
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Therefore, we can write:
\[ \frac{\text{Area of } \triangle ABE}{\text{Area of } \triangle ABC} = \frac{AD}{AB} = \frac{AE}{AC} \]
and
\[ \frac{\text{Area of } \triangle ACD}{\text{Area of } \triangle ABC} = \frac{CD}{CB} = \frac{CE}{CA} \]

5. **Establish the Relationship Between Areas**:
Since the triangles ABE and ACD are formed by the same height from point A to line BC, the areas of triangles ABE and ACD will be equal if the segments AD and CD are equal. Therefore:
\[ \text{Area of } \triangle ABE = \text{Area of } \triangle ACD \]

6. **Calculate the Area of Triangle ACD**:
Given that the area of triangle ABE is 36 sq. cm, we conclude:
\[ \text{Area of } \triangle ACD = 36 \text{ sq. cm} \]

### Final Answer:
The area of triangle ACD is **36 sq. cm**.

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