3.7k views
5 votes
sides and of equilateral triangle are tangent to a circle at points and respectively. what fraction of the area of lies outside the circle?

User Haytham
by
7.9k points

1 Answer

3 votes

The fraction of the area of the triangle that lies outside the circle is: 1 - (2√3 / 3π) * (r² / s²) .

Given that the sides of the equilateral triangle are tangent to the circle at points A and B, we can see that the triangle is divided into two parts: a sector of the circle and the remaining area of the triangle.

To find the fraction of the area that lies outside the circle, we need to calculate the area of the sector and subtract it from the area of the triangle.

Step 1: Calculate the area of the sector

The area of a sector is given by the formula:

A_sector = (θ/360) * πr²,

where:

θ is the central angle of the sector (in degrees)

π is the mathematical constant pi (approximately 3.14159)

r is the radius of the circle

In this case, the central angle of the sector is equal to 60 degrees, as each angle of an equilateral triangle measures 60 degrees.

Therefore, the area of the sector is:

A_sector = (60/360) * πr² = (1/6) * πr²

Step 2: Calculate the area of the triangle

The area of an equilateral triangle is given by the formula:

A_triangle = √3 / 4 * s²,

where:

s is the side length of the triangle

Step 3: Calculate the area outside the circle

The area outside the circle is the difference between the area of the triangle and the area of the sector:

A_outside = A_triangle - A_sector

A_outside = √3 / 4 * s² - (1/6) * πr²

Step 4: Express the area outside the circle as a fraction of the area of the triangle

To express the area outside the circle as a fraction of the area of the triangle, we can divide both sides of the equation by the area of the triangle:

A_outside / A_triangle = (√3 / 4 * s²) / (√3 / 4 * s²) - (1/6) * πr² / (√3 / 4 * s²)

A_outside / A_triangle = 1 - (1/6) * πr² / (√3 / 4 * s²)

A_outside / A_triangle = 1 - (2√3 / 3π) * (r² / s²)

Therefore, the fraction of the area of the triangle that lies outside the circle is:

1 - (2√3 / 3π) * (r² / s²)

User MagikMaker
by
8.0k points

No related questions found