Question

The area of the square is `1` square unit.

The area of one triangle is 1/4.

Determine as many other areas as you can.

Answer 1

The areas are: square (1 unit), isosceles right triangle
`(\( (1)/(8) \)`units), circle
`(\( (\pi)/(4) \) units),` rectangle (1 unit), and equilateral triangle
`(\( (√(3))/(4) \) units).`

Let's denote the side length of the square as \(s\). Since the area of the square is 1 square unit, we have:

`\[s^2 = 1\]`

So,
`\(s = 1\)` (since the side length cannot be negative).

Now, for one triangle, let's consider an isosceles right triangle with legs of length
`\(s/2\).`The area of a triangle is given by the formula:

`\[A_{\text{triangle}} = (1)/(2) * \text{base} * \text{height}\]`

In this case, the base and height are both \(s/2\), so:

`\[A_{\text{triangle}} = (1)/(2) * (s)/(2) * (s)/(2) = (s^2)/(8) = (1)/(8)\]`

So, the area of one triangle is
`\((1)/(8)\)` square units.

Now, let's consider other shapes:

1. Circle:

The area of a circle is given by the formula
`\(A_{\text{circle}} = \pi r^2\),`where \(r\) is the radius. Since the square's side length is \(s = 1\), the radius of the circle that can fit inside the square is
`\(r = (s)/(2) = (1)/(2)\).`Therefore:

`\[A_{\text{circle}} = \pi * \left((1)/(2)\right)^2 = (\pi)/(4)\]`

2. Rectangle:

Let's consider a rectangle with dimensions
`\(2s\) and \(s/2\)`. The area of the rectangle is given by
`\(A_{\text{rectangle}} = \text{length} * \text{width}\)`

`\[A_{\text{rectangle}} = 2s * (s)/(2) = s^2 = 1\]`

3. Equilateral Triangle:

The area of an equilateral triangle with side length \(s\) is given by
`\(A_{\text{equilateral triangle}} = (√(3))/(4) * s^2\)`

`\[A_{\text{equilateral triangle}} = (√(3))/(4) * 1^2 = (√(3))/(4)\]`

These are a few examples of shapes with different areas that can be related to the original square and triangle.