ABC is an isosceles triangle right angled at A. forces of magnitude `2sqrt(2),5 and 6` act along BC, CA and AB respectively. The magnitude of their resultant force is

Options

A. 4
B. 5
C. `11+2sqrt(2)`
D. 30

Explanation

To solve the problem step by step, we will analyze the forces acting on the isosceles right triangle ABC, where A is the right angle.

### Step 1: Understand the Triangle Configuration
Given that triangle ABC is isosceles and right-angled at A, we know:
- AB = AC
- Angles ABC and ACB are both 45 degrees.

### Step 2: Identify the Forces Acting on the Triangle
The forces acting along the sides of the triangle are:
- Along BC: \( F_{BC} = 2\sqrt{2} \)
- Along CA: \( F_{CA} = 5 \)
- Along AB: \( F_{AB} = 6 \)

### Step 3: Resolve the Forces into Components
Since the triangle is oriented such that:
- BC is horizontal,
- CA is vertical,
- AB is at a 45-degree angle,

We will resolve the forces into their horizontal (x) and vertical (y) components.

1. **Force along BC (horizontal)**:
- \( F_{BC,x} = 2\sqrt{2} \)
- \( F_{BC,y} = 0 \)

2. **Force along CA (vertical)**:
- \( F_{CA,x} = 0 \)
- \( F_{CA,y} = 5 \)

3. **Force along AB (at 45 degrees)**:
- \( F_{AB,x} = 6 \cos(45^\circ) = 6 \cdot \frac{1}{\sqrt{2}} = 3\sqrt{2} \)
- \( F_{AB,y} = 6 \sin(45^\circ) = 6 \cdot \frac{1}{\sqrt{2}} = 3\sqrt{2} \)

### Step 4: Calculate the Net Force Components
Now we will sum up the components in both the x and y directions.

**Net Horizontal Component (R_x)**:
\[ R_x = F_{BC,x} + F_{AB,x} = 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2} \]

**Net Vertical Component (R_y)**:
\[ R_y = F_{CA,y} + F_{AB,y} = 5 + 3\sqrt{2} \]

### Step 5: Calculate the Magnitude of the Resultant Force
The magnitude of the resultant force \( R \) can be found using the Pythagorean theorem:
\[ R = \sqrt{R_x^2 + R_y^2} \]
Substituting the values:
\[ R = \sqrt{(5\sqrt{2})^2 + (5 + 3\sqrt{2})^2} \]
Calculating each term:
\[ (5\sqrt{2})^2 = 50 \]
\[ (5 + 3\sqrt{2})^2 = 25 + 30\sqrt{2} + 18 = 43 + 30\sqrt{2} \]
Thus,
\[ R = \sqrt{50 + 43 + 30\sqrt{2}} = \sqrt{93 + 30\sqrt{2}} \]

### Final Result
The magnitude of the resultant force is:
\[ R = \sqrt{93 + 30\sqrt{2}} \]