An equilateral triangle has side length 8. The area of the region containing all points outside the triangle but not more than 3 units from a point on the triangle is :

Options

A. `9(8+pi)`
B. `8(9+pi)`
C. `9(8+pi/2)`
D. `8(9+pi/2)`

Answer

To find the area of the region containing all points outside an equilateral triangle of side length 8 but not more than 3 units from a point on the triangle, we can break the problem into several steps.

### Step 1: Calculate the area of the equilateral triangle.
The formula for the area \( A \) of an equilateral triangle with side length \( a \) is given by:

\[ A = \frac{\sqrt{3}}{4} a^2 \]

Substituting \( a = 8 \):

\[ A = \frac{\sqrt{3}}{4} \times 8^2 = \frac{\sqrt{3}}{4} \times 64 = 16\sqrt{3} \]

### Step 2: Determine the area of the rectangles formed outside the triangle.
Each side of the triangle will have a rectangle extending outward with a width of 3 units. The length of each rectangle is equal to the length of the side of the triangle, which is 8.

The area \( A_r \) of one rectangle is:

\[ A_r = \text{length} \times \text{width} = 8 \times 3 = 24 \]

Since there are 3 sides to the triangle, the total area of the rectangles \( A_{rect} \) is:

\[ A_{rect} = 3 \times A_r = 3 \times 24 = 72 \]

### Step 3: Calculate the area of the sectors formed at each vertex.
At each vertex of the triangle, there is a sector of a circle with a radius of 3 units and a central angle of \( 120^\circ \) (since the angles of an equilateral triangle are \( 60^\circ \) each, and the external angle is \( 120^\circ \)).

The area \( A_s \) of one sector is given by the formula:

\[ A_s = \frac{\theta}{360^\circ} \times \pi r^2 \]

Substituting \( \theta = 120^\circ \) and \( r = 3 \):

\[ A_s = \frac{120}{360} \times \pi \times 3^2 = \frac{1}{3} \times \pi \times 9 = 3\pi \]

Since there are 3 vertices, the total area of the sectors \( A_{sector} \) is:

\[ A_{sector} = 3 \times A_s = 3 \times 3\pi = 9\pi \]

### Step 4: Combine the areas to find the final answer.
The total area of the region outside the triangle but within 3 units from it is the sum of the area of the rectangles and the area of the sectors:

\[ \text{Total Area} = A_{rect} + A_{sector} = 72 + 9\pi \]

Thus, the area of the region containing all points outside the triangle but not more than 3 units from a point on the triangle is:

\[ \text{Total Area} = 72 + 9\pi \]

### Final Answer:
The area of the region is \( 72 + 9\pi \).

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