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 find the distance of the center of mass from point A in triangle ABC, where AB = AC = 5 cm and BC = 6 cm, we can follow these steps:

### Step 1: Understand the Geometry
We have triangle ABC with:
- AB = AC = 5 cm (the equal sides)
- BC = 6 cm (the base)

### Step 2: Determine the Coordinates of Points
Let's place the triangle in a coordinate system for easier calculations:
- Point A at (0, 0)
- Point B at (5, 0) (since AB = 5 cm)
- To find the coordinates of point C, we can use the properties of the triangle.

### Step 3: Calculate the Height of Triangle ABC
Using the Pythagorean theorem, we can find the height (h) from point A to side BC:
1. The midpoint D of BC can be found at:
- D = ((5 + 0)/2, 0) = (3, 0)
2. The length of AD (height) can be calculated using:
- AD = sqrt(AC² - CD²)
- CD = 3 cm (half of BC)
- AC = 5 cm
- AD = sqrt(5² - 3²) = sqrt(25 - 9) = sqrt(16) = 4 cm

Thus, point C is at (3, 4).

### Step 4: Calculate the Masses
1. The mass of wire AB (length = 5 cm) is m1 = 5ρ (where ρ is the density of the first wire).
2. The mass of wire AC (length = 5 cm) is m2 = 5ρ.
3. The mass of wire BC (length = 6 cm) is m3 = 6(2ρ) = 12ρ (since it has twice the density).

### Step 5: 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:
- For wire AB: \( (x_1, y_1) = (2.5, 0) \)
- For wire AC: \( (x_2, y_2) = (1.5, 4) \) (1.5 is the x-coordinate of point C)
- For wire BC: \( (x_3, y_3) = (3, 0) \)

### Step 6: Substitute Values
1. Calculate total mass:
\[ m_{total} = m_1 + m_2 + m_3 = 5\rho + 5\rho + 12\rho = 22\rho \]

2. Calculate \( x_{cm} \):
\[ x_{cm} = \frac{(5\rho)(2.5) + (5\rho)(1.5) + (12\rho)(3)}{22\rho} \]
\[ = \frac{12.5\rho + 7.5\rho + 36\rho}{22\rho} = \frac{56\rho}{22\rho} = \frac{56}{22} \approx 2.545 \text{ cm} \]

3. Calculate \( y_{cm} \):
\[ y_{cm} = \frac{(5\rho)(0) + (5\rho)(4) + (12\rho)(0)}{22\rho} \]
\[ = \frac{20\rho}{22\rho} = \frac{20}{22} \approx 0.909 \text{ cm} \]

### Step 7: Distance from Point A
The distance of the center of mass from point A can be calculated using the Pythagorean theorem:
\[ d = \sqrt{x_{cm}^2 + y_{cm}^2} = \sqrt{(2.545)^2 + (0.909)^2} \]
Calculating this gives:
\[ d \approx \sqrt{6.48 + 0.826} \approx \sqrt{7.306} \approx 2.7 \text{ cm} \]

### Final Answer
The distance of the center of mass from point A is approximately **2.7 cm**.

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