If D is the mid-point of the side BC of `DeltaABC`and the area of `DeltaABD` and the area of `DeltaACD` is `16cm^(2)`, then the area of `DeltaABC` is

Options

A. `16 cm^(2)`
B. `24 cm^(2)`
C. `32 cm^(2)`
D. `48 cm^(2)`

Answer

To find the area of triangle ABC given that D is the midpoint of side BC and the areas of triangles ABD and ACD are both 16 cm², we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Triangle Configuration**:
- We have triangle ABC with D as the midpoint of side BC. This means that BD = DC.

2. **Area of Triangles ABD and ACD**:
- We know that the area of triangle ABD = 16 cm² and the area of triangle ACD = 16 cm².

3. **Relationship Between Areas**:
- Since D is the midpoint of BC, triangles ABD and ACD are equal in area. Therefore, the area of triangle ABC can be expressed as the sum of the areas of triangles ABD and ACD.
- Area of triangle ABC = Area of triangle ABD + Area of triangle ACD.

4. **Calculate the Area of Triangle ABC**:
- Substitute the known areas into the equation:
\[ \text{Area of triangle ABC} = 16 \, \text{cm}^2 + 16 \, \text{cm}^2 = 32 \, \text{cm}^2. \]

5. **Final Result**:
- Thus, the area of triangle ABC is 32 cm².