Question

What is the value of the ratio $\frac{\text{Area of the hexagon}}{\text{Area of the grey triangle}}$?

Hint:

For geometry problems involving regular polygons, decomposing the shape into smaller, simpler shapes (like equilateral triangles in a hexagon) is a very effective strategy. Also, using a coordinate system can simplify area and length calculations.

Answer 1

Correct Answer: 3

Step 1: Understanding the Concept: 
The problem requires finding the ratio of the area of a regular hexagon to the area of a specific triangle inscribed within it. A regular hexagon can be divided into six identical equilateral triangles meeting at its center. This property is key to solving the problem. 
Step 2: Key Formula or Approach: 
Let the side length of the regular hexagon be \(s\). 
The area of a regular hexagon is given by the area of six equilateral triangles with side \(s\). 
Area of one equilateral triangle = \(\frac{\sqrt{3}}{4}s^2\). 
Area of Hexagon = \(6 \times \frac{\sqrt{3}}{4}s^2 = \frac{3\sqrt{3}}{2}s^2\). 
We will find the area of the grey triangle by identifying its base and height in terms of \(s\). 
Step 3: Detailed Explanation: 


Identify the vertices of the triangle: Let the vertices of the hexagon be A, B, C, D, E, F in a counter-clockwise direction. Based on the drawing, the vertices of the grey triangle can be identified as A, B, and D. 
Calculate the area of the grey triangle (ABD): 

Let's place the hexagon in a coordinate system with its center at the origin (0,0) and vertex D at (-s, 0). 
The coordinates of the vertices would be: D(-s, 0), E(-s/2, -s\(\sqrt{3}\)/2), F(s/2, -s\(\sqrt{3}\)/2), A(s, 0), B(s/2, s\(\sqrt{3}\)/2), C(-s/2, s\(\sqrt{3}\)/2). 
The base of the triangle ABD is the line segment AD, which is the main diagonal of the hexagon. Its length is \(2s\). 
The height of the triangle with respect to the base AD is the perpendicular distance from vertex B to the line AD (the x-axis). This is the y-coordinate of B, which is \(\frac{s\sqrt{3}}{2}\). 
The area of triangle ABD is: \[ \text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2s) \times \left(\frac{s\sqrt{3}}{2}\right) = \frac{s^2\sqrt{3}}{2} \] 

Calculate the ratio: \[ \text{Ratio} = \frac{\text{Area of the hexagon}}{\text{Area of the grey triangle}} = \frac{\frac{3\sqrt{3}}{2}s^2}{\frac{\sqrt{3}}{2}s^2} = 3 \] 
Alternatively, the area of the grey triangle (\(\frac{s^2\sqrt{3}}{2}\)) is equal to the area of two of the small equilateral triangles that make up the hexagon ( \(2 \times \frac{\sqrt{3}}{4}s^2\) ). Since the hexagon is made of 6 such triangles, the ratio is \(6/2 = 3\). 
Step 4: Final Answer: 
The ratio of the area of the hexagon to the area of the grey triangle is 3.