Problem 17 - Entrance Test

An equilateral triangle has a side length of s. Three circles are drawn with centers at the vertices of the triangle, each with radius s/2. Find the area of the region inside the triangle but outside all three circles.

A. A) (sqrt(3)/4 - pi/8)s^2B. B) (sqrt(3)/4 - pi/6)s^2C. C) (sqrt(3)/4 - pi/4)s^2D. D) (sqrt(3)/4 - pi/2)s^2E. E) (sqrt(3)/2 - pi/8)s^2

Correct: A

The area of the region inside the triangle but outside all three circles can be found by subtracting the area of the circular sectors within the triangle from the total area of the equilateral triangle. 1. Area of the equilateral triangle: The formula for the area of an equilateral triangle with side length s is (sqrt(3)/4)s^2. 2. Area of the circular sectors: Each circle is centered at a vertex of the equilateral triangle, and its radius is r = s/2. The angle at each vertex of an equilateral triangle is 60 degrees. Therefore, each circular sector inside the triangle forms a 60-degree sector of a circle with radius s/2. The area of one such sector is given by (theta / 360) * pi * r^2, where theta is the angle in degrees. Area of one sector = (60 / 360) * pi * (s/2)^2 Area of one sector = (1/6) * pi * (s^2/4) Area of one sector = (pi * s^2) / 24. Since there are three such sectors (one at each vertex) and they do not overlap within the triangle (as the radius s/2 means the circles only touch at the midpoints of the sides, they don't 'cross over' into each other's sectors near the center of the triangle), The total area covered by the three sectors = 3 * (pi * s^2) / 24 = (pi * s^2) / 8. 3. Area of the desired region: Area (inside triangle, outside circles) = Area (triangle) - Total Area (three sectors) = (sqrt(3)/4)s^2 - (pi/8)s^2 = (sqrt(3)/4 - pi/8)s^2. This expression matches option A.