The areas of two similar triangles are `144cm^2 and 81cm^2.` If one median of the first triangleis 16 cm, length of corresponding median of the second triangle is

Options

A. 9 cm
B. 27 cm
C. 12 cm
D. 16 cm

Answer

To solve the problem step by step, we will use the properties of similar triangles and the relationship between their areas and corresponding medians.

### Step 1: Identify the areas of the triangles
Let the areas of the two triangles be:
- Area of Triangle 1 (A1) = 144 cm²
- Area of Triangle 2 (A2) = 81 cm²

### Step 2: Write the ratio of the areas
The ratio of the areas of the two triangles can be expressed as:
\[ \frac{A1}{A2} = \frac{144}{81} \]

### Step 3: Simplify the ratio of the areas
To simplify \(\frac{144}{81}\), we can divide both the numerator and the denominator by their greatest common divisor (GCD):
\[ \frac{144 \div 9}{81 \div 9} = \frac{16}{9} \]

### Step 4: Relate the ratio of the areas to the ratio of the medians
For similar triangles, the ratio of the areas is equal to the square of the ratio of corresponding lengths (including medians):
\[ \frac{A1}{A2} = \left(\frac{m1}{m2}\right)^2 \]
Where \(m1\) is the median of Triangle 1 and \(m2\) is the median of Triangle 2.

### Step 5: Substitute known values
We know that:
- \(m1 = 16\) cm (the median of the first triangle)
- \(A1/A2 = 16/9\)

Thus, we can write:
\[ \frac{16}{9} = \left(\frac{16}{m2}\right)^2 \]

### Step 6: Solve for \(m2\)
Taking the square root of both sides:
\[ \sqrt{\frac{16}{9}} = \frac{16}{m2} \]
\[ \frac{4}{3} = \frac{16}{m2} \]

Now, cross-multiplying gives:
\[ 4 \cdot m2 = 16 \cdot 3 \]
\[ 4m2 = 48 \]
\[ m2 = \frac{48}{4} = 12 \text{ cm} \]

### Final Answer
The length of the corresponding median of the second triangle is **12 cm**.

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