Consider the lines
`L_(1): x+y=10, " " L_(2):x+y=60`
`L_(3):x=40, " " L_(4):y=40`.
`L_(1)` meets x - axis and y - axis at A and B respectively.
`L_(4)` meets y - axis at C and `L_(2)` at D
`L_(3)` meets `L_(2)` at E and x-axis at F.
Perimeter of the hexagon ABCDEF is

Options

A. `100+50sqrt(2)`
B. `50+40sqrt(2)`
C. 100
D. none of these

Answer

To find the perimeter of the hexagon ABCDEF formed by the intersection of the given lines, we will follow these steps:

### Step 1: Find the coordinates of point A
The line \( L_1: x + y = 10 \) meets the x-axis when \( y = 0 \).
Substituting \( y = 0 \) into the equation:
\[ x + 0 = 10 \implies x = 10 \]
Thus, the coordinates of point A are \( A(10, 0) \).

### Step 2: Find the coordinates of point B
The line \( L_1: x + y = 10 \) meets the y-axis when \( x = 0 \).
Substituting \( x = 0 \) into the equation:
\[ 0 + y = 10 \implies y = 10 \]
Thus, the coordinates of point B are \( B(0, 10) \).

### Step 3: Find the coordinates of point C
The line \( L_4: y = 40 \) meets the y-axis when \( x = 0 \).
Thus, the coordinates of point C are \( C(0, 40) \).

### Step 4: Find the coordinates of point D
The line \( L_2: x + y = 60 \) meets the y-axis when \( x = 0 \).
Substituting \( x = 0 \) into the equation:
\[ 0 + y = 60 \implies y = 60 \]
Thus, the coordinates of point D are \( D(0, 60) \).

### Step 5: Find the coordinates of point E
The line \( L_3: x = 40 \) intersects line \( L_2: x + y = 60 \).
Substituting \( x = 40 \) into the equation:
\[ 40 + y = 60 \implies y = 20 \]
Thus, the coordinates of point E are \( E(40, 20) \).

### Step 6: Find the coordinates of point F
The line \( L_3: x = 40 \) meets the x-axis when \( y = 0 \).
Thus, the coordinates of point F are \( F(40, 0) \).

### Step 7: Calculate the lengths of the sides of the hexagon
1. Length \( AB \):
\[ AB = \sqrt{(10 - 0)^2 + (0 - 10)^2} = \sqrt{10^2 + (-10)^2} = \sqrt{100 + 100} = 10\sqrt{2} \]

2. Length \( BC \):
\[ BC = \sqrt{(0 - 0)^2 + (10 - 40)^2} = \sqrt{0 + (-30)^2} = 30 \]

3. Length \( CD \):
\[ CD = \sqrt{(0 - 0)^2 + (40 - 60)^2} = \sqrt{0 + (-20)^2} = 20 \]

4. Length \( DE \):
\[ DE = \sqrt{(0 - 40)^2 + (60 - 20)^2} = \sqrt{(-40)^2 + (40)^2} = \sqrt{1600 + 1600} = 40\sqrt{2} \]

5. Length \( EF \):
\[ EF = \sqrt{(40 - 40)^2 + (20 - 0)^2} = \sqrt{0 + 20^2} = 20 \]

6. Length \( FA \):
\[ FA = \sqrt{(40 - 10)^2 + (0 - 0)^2} = \sqrt{(30)^2 + 0} = 30 \]

### Step 8: Calculate the perimeter of hexagon ABCDEF
The perimeter \( P \) is the sum of all the lengths:
\[ P = AB + BC + CD + DE + EF + FA \]
Substituting the lengths we found:
\[ P = 10\sqrt{2} + 30 + 20 + 40\sqrt{2} + 20 + 30 \]
Combining like terms:
\[ P = (10\sqrt{2} + 40\sqrt{2}) + (30 + 20 + 20 + 30) = 50\sqrt{2} + 100 \]

### Final Answer
The perimeter of the hexagon ABCDEF is:
\[ P = 50\sqrt{2} + 100 \]