arabella wrote:

What is the greatest possible area of a triangular region with one side that corresponds with the diameter of a circle with radius 6, and the other vertex of the triangle on the circle?

(A) 24

(B) 36

(C) 40

(D) 48

(E) 72

One side of this triangle, its base, IS the diameter. Hence it is a right triangle inscribed in a semicircle, no matter where the third vertex is placed on the circle.

For any right triangle, isoceles right triangle has the maximum area. Maximum area is achieved with maximum height. Maximum height, as becomes clear from inscribing the right triangle in a circle, bisects and is perpendicular to the diameter. That maximum height is the radius.

Imagine two right triangles in a semicircle. The first has a vertex on the circle that, on a face clock, would be at 12 o'clock.

The second triangle has a vertex on the circle that, on a face clock, would be at 10 o'clock.

Drop altitudes from both vertices. The first has a height of radius = 6.

The second has a height that is not the radius, and is < 6. Base for both = 12

Area =

\(\frac{b*h}{2}\)First ∆:

\(\frac{(12)(6)}{2}\) = 36

Second ∆:

\(\frac{(12)(<6)}{2}\) < 36

The first triangle has the greater area, with height = radius. No height of any inscribed right triangle can be greater than the radius. Stated differently, the altitude that is the radius yields the maximum height, which yields the maximum area.

First triangle has the maximum area every time. (And it's an isosceles right triangle.)

From above, that triangle's area, where base = diameter = 12, and h = radius = 6, is 36.

Answer B

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