A thin uniform wire is bent to form the two equal sides AB and AC of triangle ABC, where AB=AC=5 cm. The third side BC, of length 6cm, is made from uniform wire of twice the density of the first. The distance of centre of mass from A is :

Options

A. `(34)/(11) cm`
B. `(11)/(34)cm`
C. `(34)/(9)cm`
D. `(11)/(45)cm`

Answer

To solve the problem of finding the distance of the center of mass from point A in triangle ABC, we will follow these steps:

### Step 1: Define the Geometry
We have a triangle ABC where:
- AB = AC = 5 cm (sides made of uniform wire)
- BC = 6 cm (made from a wire of twice the density of the first two sides)

### Step 2: Set Up Coordinates
Let's place the triangle in a coordinate system:
- Point B at the origin (0, 0)
- Point C at (6, 0) since BC = 6 cm
- To find the coordinates of point A, we can use the Pythagorean theorem. The distance from B to A and from C to A is 5 cm.

Using the coordinates:
- The x-coordinate of A can be found as the midpoint of BC, which is at (3, y) where y is to be determined.
- Using the Pythagorean theorem: \(AB^2 = x^2 + y^2\) gives \(5^2 = 3^2 + y^2\) → \(25 = 9 + y^2\) → \(y^2 = 16\) → \(y = 4\).
- Thus, the coordinates of A are (3, 4).

### Step 3: Identify the Masses
Now we need to find the masses of each segment:
- For AB and AC (both are 5 cm long and have the same density, ρ):
- Mass of AB, \(m_1 = ρ \times 5\)
- Mass of AC, \(m_2 = ρ \times 5\)

- For BC (6 cm long and has twice the density, 2ρ):
- Mass of BC, \(m_3 = 2ρ \times 6 = 12ρ\)

### Step 4: Calculate the Center of Mass
The coordinates of the center of mass (CM) can be calculated using the formula:
\[ x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} \]
\[ y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} \]

Where:
- \(x_1, y_1\) are the coordinates of A (3, 4)
- \(x_2, y_2\) are the coordinates of B (0, 0)
- \(x_3, y_3\) are the coordinates of C (6, 0)

Substituting the values:
- \(m_1 = 5ρ\), \(m_2 = 5ρ\), \(m_3 = 12ρ\)
- \(x_{cm} = \frac{(5ρ)(3) + (5ρ)(0) + (12ρ)(6)}{5ρ + 5ρ + 12ρ}\)
- \(y_{cm} = \frac{(5ρ)(4) + (5ρ)(0) + (12ρ)(0)}{5ρ + 5ρ + 12ρ}\)

### Step 5: Simplify the Equations
Calculating \(x_{cm}\):
\[ x_{cm} = \frac{15ρ + 0 + 72ρ}{22ρ} = \frac{87ρ}{22ρ} = \frac{87}{22} \approx 3.95 \]

Calculating \(y_{cm}\):
\[ y_{cm} = \frac{20ρ + 0 + 0}{22ρ} = \frac{20ρ}{22ρ} = \frac{20}{22} = \frac{10}{11} \approx 0.91 \]

### Step 6: Distance from A to CM
To find the distance from point A (3, 4) to the center of mass (3.95, 0.91):
Using the distance formula:
\[ d = \sqrt{(x_{cm} - x_A)^2 + (y_{cm} - y_A)^2} \]
Substituting the values:
\[ d = \sqrt{(3.95 - 3)^2 + (0.91 - 4)^2} \]
\[ = \sqrt{(0.95)^2 + (-3.09)^2} \]
\[ = \sqrt{0.9025 + 9.5561} = \sqrt{10.4586} \approx 3.23 \text{ cm} \]

### Final Answer
The distance of the center of mass from point A is approximately **3.23 cm**.