Question

a circle is inscribed in a regular hexagon with side length 10 feet. what is the area of the shaded region? recall that in a (30^circ - 60^circ - 90^circ) triangle, if the shortest leg measures (x) units, then the longer leg measures (xsqrt{3}) units and the hypotenuse measures (2x) units. (\bigcirc (150sqrt{3}-75pi) \text{ft}^2) (\bigcirc (300 - 75pi) \text{ft}^2) (\bigcirc (150sqrt{3}-25pi) \text{ft}^2) (\bigcirc (300 - 25pi) \text{ft}^2)

Explanation:

Step1: Find area of regular hexagon

A regular hexagon can be divided into 6 equilateral triangles, but also, when a circle is inscribed, we can use the formula for the area of a regular polygon: \( A = \frac{1}{2} \times perimeter \times apothem \). The side length \( s = 10 \) ft, so perimeter \( P = 6 \times 10 = 60 \) ft. For a regular hexagon, the apothem (distance from center to a side, which is the radius of the inscribed circle) can be found using the 30 - 60 - 90 triangle. The side of the hexagon is 10, so half of the side is 5 (shortest leg of the 30 - 60 - 90 triangle). Then the apothem (longer leg) is \( 5\sqrt{3} \)? Wait, no, wait. Wait, in a regular hexagon, the side length is equal to the radius of the circumscribed circle, but the apothem (inscribed circle radius) is \( \frac{\sqrt{3}}{2} \times side \). Wait, actually, when we take a triangle from the center to a side, the central angle is \( 60^\circ \) (since 360/6 = 60), but when we drop a perpendicular from the center to the side, we split the triangle into two 30 - 60 - 90 triangles. The side of the hexagon is 10, so half of the side is 5 (opposite 30°? Wait, no: the central angle is 60°, so the triangle from center to two vertices is equilateral, so all sides are 10. Then when we drop a perpendicular (apothem) to the side, we split the equilateral triangle into two 30 - 60 - 90 triangles, where the hypotenuse is 10, the shorter leg is 5 (half of the side), and the longer leg (apothem) is \( 5\sqrt{3} \)? Wait, no, 30 - 60 - 90: hypotenuse is 2x, longer leg is \( x\sqrt{3} \), shorter leg is x. Wait, if the hypotenuse is 10 (the side of the equilateral triangle), then x (shorter leg) is 5, longer leg (apothem) is \( 5\sqrt{3} \)? Wait, no, that's incorrect. Wait, actually, in the regular hexagon, the apothem (r) is related to the side length (s) by \( r = \frac{\sqrt{3}}{2}s \). So for s = 10, \( r = 5\sqrt{3} \)? Wait, no, \( \frac{\sqrt{3}}{2} \times 10 = 5\sqrt{3} \)? Wait, no, \( \frac{\sqrt{3}}{2} \times 10 = 5\sqrt{3} \)? Wait, no, that's not right. Wait, let's recast: the area of a regular hexagon is also \( 6 \times \) area of equilateral triangle with side 10. The area of an equilateral triangle is \( \frac{\sqrt{3}}{4}s^2 \), so 6 times that is \( 6 \times \frac{\sqrt{3}}{4} \times 10^2 = 6 \times \frac{\sqrt{3}}{4} \times 100 = 150\sqrt{3} \) square feet. Yes, that's correct. So area of hexagon \( A_{hex} = 150\sqrt{3} \) ft².

Step2: Find area of inscribed circle

The radius of the inscribed circle (apothem) can be found from the equilateral triangle. In the equilateral triangle (side 10), the height (which is the apothem) is \( \frac{\sqrt{3}}{2} \times 10 = 5\sqrt{3} \)? Wait, no, wait: the height of an equilateral triangle with side 10 is \( \frac{\sqrt{3}}{2} \times 10 = 5\sqrt{3} \)? Wait, no, \( \frac{\sqrt{3}}{2} \times 10 = 5\sqrt{3} \)? Wait, no, \( \frac{\sqrt{3}}{2} \times 10 = 5\sqrt{3} \), but wait, when we have the inscribed circle, the radius is the apothem, which is the height of the equilateral triangle? Wait, no, the apothem is the distance from center to a side, which is the height of the equilateral triangle? Wait, no, the equilateral triangle has side 10, height \( h = \frac{\sqrt{3}}{2} \times 10 = 5\sqrt{3} \). But wait, in the 30 - 60 - 90 triangle, if we split the equilateral triangle into two 30 - 60 - 90 triangles, the hypotenuse is 10 (side of hexagon), the shorter leg is 5 (half of the base), and the longer leg (height) is \( 5\sqrt{3} \). Wait, but then the radius of the inscribed circle is \( 5\sqrt{3} \)? Wait, no, that can't be, because when we calculate the area of the circle, let's check the answer options. The answer options have 75π or 25π. So maybe my mistake is here. Wait, wait, maybe the side length of the hexagon is equal to the radius of the circumscribed circle, but the inscribed circle's radius (apothem) is \( \frac{\sqrt{3}}{2} \times side \), but wait, let's re - examine the 30 - 60 - 90 triangle. The problem says "Recall that in a 30° - 60° - 90° triangle, if the shortest leg measures x units, then the longer leg measures \( x\sqrt{3} \) units and the hypotenuse measures 2x units." So in the regular hexagon, when we take a triangle from the center to a side, we split the side into two equal parts (length 5, since side is 10). So the shortest leg (opposite 30°) is 5, so the longer leg (apothem, which is the radius of the inscribed circle) is \( 5\sqrt{3} \)? Wait, no, wait: the central angle for each side is 60°, so the triangle formed by the center and two adjacent vertices is equilateral (all sides 10, all angles 60°). When we drop a perpendicular from the center to the side, we get two 30 - 60 - 90 triangles, where the angle at the center is 30° (since we split the 60° angle into two 30° angles). So the hypotenuse of each 30 - 60 - 90 triangle is the radius of the circumscribed circle (which is equal to the side length of the hexagon, 10). Wait, that makes sense: hypotenuse is 10, so the shortest leg (opposite 30°) is \( \frac{10}{2}=5 \), and the longer leg (apothem, radius of inscribed circle) is \( 5\sqrt{3} \). Wait, but then the area of the circle would be \( \pi r^2=\pi(5\sqrt{3})^2 = 75\pi \). Ah! That's where the 75π comes from. So area of circle \( A_{circle}=\pi r^2=\pi(5\sqrt{3})^2 = 75\pi \) ft².

Step3: Find area of shaded region

The shaded region is the area of the hexagon minus the area of the circle. So \( A_{shaded}=A_{hex}-A_{circle}=150\sqrt{3}-75\pi \) ft².

Answer:

\( (150\sqrt{3}-75\pi) \text{ ft}^2 \) (corresponding to the first option)