In `Delta ABC, D` is the mid-point of BC . Length AD is 27 cm . N is a point in AD such that the length of DN is 12 cm . The distance of N from the centroid of `Delta ABC` is equal to

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have triangle ABC with D as the midpoint of side BC. The length of median AD is given as 27 cm. A point N lies on AD such that DN = 12 cm. We need to find the distance of N from the centroid G of triangle ABC.

### Step 2: Identify the properties of the centroid
The centroid G of a triangle divides each median in the ratio 2:1. This means that if AG is the segment from A to G and GD is the segment from G to D, then:
\[ AG : GD = 2 : 1 \]

### Step 3: Express AG and GD in terms of a variable
Let the length of GD be \( x \). Then, the length of AG will be \( 2x \). Since the total length of the median AD is given as 27 cm, we can write:
\[ AG + GD = AD \]
\[ 2x + x = 27 \]
\[ 3x = 27 \]
\[ x = 9 \]

### Step 4: Calculate the lengths of AG and GD
From our calculations:
- \( GD = x = 9 \) cm
- \( AG = 2x = 18 \) cm

### Step 5: Find the length of NG
We know that \( DN = 12 \) cm. Since \( GD = 9 \) cm, we can find \( NG \) as follows:
\[ NG = DN - GD \]
\[ NG = 12 - 9 = 3 \text{ cm} \]

### Conclusion
The distance of point N from the centroid G of triangle ABC is:
\[ \text{Distance of N from G} = 3 \text{ cm} \]

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