Question

select the correct answer from each drop-down menu.
the hexagonal seating section in an auditorium has a small entrance at the front. a walkway leading from the entrance to the front of the stage is 30 ft long. what is the approximate area of the entrance and seating section combined?
image of a hexagon with a 30 ft segment
each side of the hexagon is about drop-down ft long. the area of the seating section is about drop-down ft². the area of the entrance is about drop-down ft². the area of the entrance and seating section combines is about drop-down ft².
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Explanation:

Answer:

To solve the problem, we analyze the hexagonal seating section:

Step 1: Side Length of Hexagon

A regular hexagon can be divided into 6 equilateral triangles. The walkway (30 ft) is likely the distance between two opposite sides (diameter of the circumscribed circle) or the length of a diagonal. For a regular hexagon, the side length \( s \) is related to the distance between opposite sides (height) \( h \) by \( h = s\sqrt{3} \). If the walkway is the distance between opposite sides, \( s=\frac{30}{\sqrt{3}}\approx17.32 \) ft (but more simply, a regular hexagon with side \( s \) has area \( \frac{3\sqrt{3}}{2}s^2 \)). However, the problem states “each side of the hexagon is about [ ] ft long.” For a regular hexagon, if the walkway (30 ft) is the length of a side (if the hexagon is divided into two trapezoids or rhombuses), the side length is 15 ft? Wait, no—if the hexagon is split into two congruent parts by a 30-ft diagonal, each side is \( \frac{30}{2}=15 \) ft (since a regular hexagon’s side equals the radius of its circumscribed circle; the distance between two opposite vertices is \( 2s \), so \( 2s = 30 \implies s = 15 \) ft).

Step 2: Area of Seating Section (Hexagon)

A regular hexagon with side \( s = 15 \) ft has area \( A_{\text{hexagon}}=\frac{3\sqrt{3}}{2}s^2 \). Plugging \( s = 15 \):

$$ A_{\text{hexagon}}=\frac{3\sqrt{3}}{2}(15)^2=\frac{3\sqrt{3}}{2}(225)\approx\frac{3(1.732)}{2}(225)\approx2.598\times225\approx584.55\approx585\ \text{ft}^2 $$

Step 3: Area of Entrance (Assumed Shape)

The entrance is likely a smaller shape (e.g., a triangle or trapezoid) or complementary to the seating. If the combined area is the sum, but the problem’s drop-downs suggest:

• Side length of hexagon: 15 ft (since \( 2s = 30 \implies s = 15 \)).
• Area of seating (hexagon): ~585 ft².
• Area of entrance: If the combined area is, say, 707 ft² (but likely, the entrance is a triangle or smaller hexagon). However, rechecking: A regular hexagon with side 15 ft has area \( \frac{3\sqrt{3}}{2}(15)^2\approx584.5 \approx 585 \) ft². The entrance area would be smaller, but the problem’s “combined area” is the sum.

Final Answers (Filling Drop-Downs):

• Each side of the hexagon: \( \boldsymbol{15} \) ft.
• Area of seating section: \( \boldsymbol{585} \) ft².
• Area of entrance: (Assume entrance is a triangle or smaller; if combined area is, e.g., 707, but likely the seating area is ~585, entrance ~122, combined ~707. But the key is:

• Side length: \( \boldsymbol{15} \) ft (since \( 30 \div 2 = 15 \), as a regular hexagon’s side equals the radius, and the diagonal is \( 2s \)).
• Seating area: \( \boldsymbol{585} \) ft² (using \( \frac{3\sqrt{3}}{2}(15)^2 \approx 585 \)).
• Entrance area: If entrance is a triangle with base 30 and height 15 (matching hexagon’s height), area \( \frac{1}{2}(30)(15)=225 \) ft². Combined area: \( 585 + 225 = 810 \)? No, confusion. Alternatively, the hexagon is split into two trapezoids, each with bases 15 and 30, height \( \frac{15\sqrt{3}}{2}\approx12.99 \). Area of one trapezoid: \( \frac{(15 + 30)}{2}(12.99)\approx22.5\times12.99\approx292.28 \), so two trapezoids (seating) would be \( 2\times292.28\approx584.56 \approx 585 \) ft².

Thus, the side length of the hexagon is 15 ft, the area of the seating section is 585 ft², and the combined area (entrance + seating) would depend, but the key filled values are:

• Side length: \( \boldsymbol{15} \)
• Seating area: \( \boldsymbol{585} \)

(Note: The problem’s drop-downs likely have options like 15, 585, 225, 810, etc. Adjust based on standard hexagon area formulas.)