Calculate Shaded Area A Circle Inscribed In Hexagon

by qnaftunila 52 views
Iklan Headers

Hey guys! Today, let's dive into an exciting geometry problem that involves calculating the shaded area when a circle is inscribed in a regular hexagon. This is a classic problem that combines our understanding of hexagons, circles, and some special right triangles. So, grab your thinking caps, and let’s get started!

Problem Statement

The problem we’re tackling today is: A circle is inscribed in a regular hexagon with a side length of 10 feet. What is the area of the shaded region?

This problem is a fantastic exercise in applying geometric principles and problem-solving skills. We’ll break it down step by step to make it super easy to understand. Let's make sure that area calculation is as clear as day for everyone!

Understanding the Geometry

Before we jump into calculations, let’s make sure we understand the shapes we're dealing with. A regular hexagon is a six-sided polygon where all sides are of equal length, and all interior angles are equal. Think of it as six equilateral triangles perfectly pieced together around a central point. The circle inscribed in the hexagon is a circle that fits snugly inside the hexagon, touching each side at exactly one point. This point of tangency is crucial because it gives us a direct link between the circle's radius and the hexagon's dimensions.

Key Properties of a Regular Hexagon

Understanding the properties of a regular hexagon is crucial for solving this problem. A regular hexagon can be divided into six congruent equilateral triangles. If you draw lines from the center of the hexagon to each vertex, you'll see this division clearly. Each of these triangles has equal sides and equal angles. Since the side length of our hexagon is 10 feet, each side of these equilateral triangles is also 10 feet. The interior angles of a regular hexagon are all 120 degrees.

Inscribed Circle and Tangency

Now, let's focus on the inscribed circle. The center of this circle is the same as the center of the hexagon. The circle touches each side of the hexagon at its midpoint. This point of tangency is where a radius of the circle meets the side of the hexagon at a 90-degree angle. This is a critical piece of information because it helps us form a right triangle, which we can then use to find the radius of the circle. This tangency property is key to linking the dimensions of the hexagon and the circle, making our calculations smoother.

30βˆ˜βˆ’60βˆ˜βˆ’90∘30^{\circ}-60^{\circ}-90^{\circ} Triangles

Our problem also mentions a special type of right triangle: the 30βˆ˜βˆ’60βˆ˜βˆ’90∘30^{\circ}-60^{\circ}-90^{\circ} triangle. This triangle has a unique property: if the shortest leg (opposite the 30∘30^{\circ} angle) measures x units, then the longer leg (opposite the 60∘60^{\circ} angle) measures x3x \sqrt{3} units, and the hypotenuse (opposite the 90∘90^{\circ} angle) measures 2x units. This relationship is essential for our calculations, so let’s keep it in mind. Recognizing these triangles within the hexagon and circle setup allows us to use their special properties to find crucial lengths, especially the radius of the inscribed circle. We'll see how this comes into play as we proceed with the solution.

Solving for the Radius

To find the shaded area, we first need to determine the radius of the inscribed circle. This involves a bit of geometric insight and applying the properties of our 30βˆ˜βˆ’60βˆ˜βˆ’90∘30^{\circ}-60^{\circ}-90^{\circ} triangles.

Forming the Right Triangle

Imagine drawing a line from the center of the hexagon to the midpoint of one of its sides. This line is a radius of the inscribed circle and is perpendicular to the side of the hexagon. Now, draw a line from the center of the hexagon to one of the vertices that is connected to this side. What you've created is a right triangle! This triangle is half of one of the equilateral triangles we discussed earlier. This step is crucial because it connects the side length of the hexagon to the radius of the circle, allowing us to use the properties of right triangles to find the radius.

Identifying the 30βˆ˜βˆ’60βˆ˜βˆ’90∘30^{\circ}-60^{\circ}-90^{\circ} Triangle

This right triangle isn't just any right triangle; it’s a 30βˆ˜βˆ’60βˆ˜βˆ’90∘30^{\circ}-60^{\circ}-90^{\circ} triangle. Here’s why: The angle at the center of the hexagon is half of the 120-degree interior angle of the hexagon, which is 60 degrees. The angle at the point of tangency is 90 degrees (since the radius is perpendicular to the side). That leaves the third angle to be 30 degrees (since the angles in a triangle add up to 180 degrees). Recognizing this as a 30βˆ˜βˆ’60βˆ˜βˆ’90∘30^{\circ}-60^{\circ}-90^{\circ} triangle is a game-changer because it allows us to use the special ratios between the sides.

Applying the 30βˆ˜βˆ’60βˆ˜βˆ’90∘30^{\circ}-60^{\circ}-90^{\circ} Triangle Properties

Now comes the fun part! We know the side length of the hexagon is 10 feet. This means the base of our 30βˆ˜βˆ’60βˆ˜βˆ’90∘30^{\circ}-60^{\circ}-90^{\circ} triangle (the side along the hexagon) is half of that, which is 5 feet. This side is opposite the 30∘30^{\circ} angle, making it the shorter leg (x) in our triangle. According to the properties of a 30βˆ˜βˆ’60βˆ˜βˆ’90∘30^{\circ}-60^{\circ}-90^{\circ} triangle, the longer leg (opposite the 60∘60^{\circ} angle) is x3x \sqrt{3}. The longer leg is also the radius (r) of our inscribed circle. So, we have: r=53r = 5 \sqrt{3} feet. This radius calculation is a key step in solving the problem, as it directly impacts the area of the circle and, consequently, the shaded region.

Calculating Areas

With the radius in hand, we can now calculate the areas we need: the area of the hexagon and the area of the inscribed circle. Subtracting the circle’s area from the hexagon’s area will give us the shaded region. Let's jump into some area calculations!

Area of the Hexagon

A regular hexagon can be divided into six equilateral triangles. The area of an equilateral triangle with side s is given by the formula A=s234A = \frac{s^2 \sqrt{3}}{4}. Since our hexagon has a side length of 10 feet, the area of one equilateral triangle is: A=10234=10034=253A = \frac{10^2 \sqrt{3}}{4} = \frac{100 \sqrt{3}}{4} = 25 \sqrt{3} square feet. Because there are six such triangles in the hexagon, the total area of the hexagon is: 6Γ—253=15036 \times 25 \sqrt{3} = 150 \sqrt{3} square feet. Understanding how to decompose the hexagon into equilateral triangles is key here. It simplifies the area calculation and leverages a well-known formula, making the process much more manageable.

Area of the Circle

The area of a circle is given by the formula A=Ο€r2A = \pi r^2, where r is the radius. We found earlier that the radius of our inscribed circle is 535 \sqrt{3} feet. Plugging this into the formula, we get: A=Ο€(53)2=Ο€(25Γ—3)=75Ο€A = \pi (5 \sqrt{3})^2 = \pi (25 \times 3) = 75 \pi square feet. This circle's area represents the portion of the hexagon that is not shaded, and it's a crucial component in determining the final shaded area. A solid grasp of the circle area formula is essential here, as it directly influences the final answer.

Area of the Shaded Region

Finally, to find the area of the shaded region, we subtract the area of the circle from the area of the hexagon: Shaded Area = Area of Hexagon - Area of Circle Shaded Area = 1503βˆ’75Ο€150 \sqrt{3} - 75 \pi square feet. This subtraction gives us the area that lies within the hexagon but outside the circle. It's the part of the figure that's