Question

what is the diameter of the inscribed circle of the triangle?
the diameter of the inscribed circle of the triangle is
(type a whole number.)
image of a triangle with angle bisectors and segments labeled 2x - 3 and x + 2, and right angles indicating tangent segments

Explanation:

Step1: Set equal segments

In a triangle with an inscribed circle, the two tangent segments from a vertex to the incircle are equal. So, we set \(2x - 3=x + 2\).

Step2: Solve for x

Subtract \(x\) from both sides: \(2x - x-3=x - x+ 2\), which simplifies to \(x - 3=2\). Then add 3 to both sides: \(x=2 + 3=5\).

Step3: Find the radius

Substitute \(x = 5\) into \(x + 2\) (or \(2x - 3\)) to find the radius. \(x+2=5 + 2=7\)? Wait, no, wait. Wait, the segments \(2x - 3\) and \(x + 2\) are the radii? Wait, no, in a right triangle? Wait, the triangle has right angles, so it's a right triangle. The inradius \(r\) of a right triangle is given by \(r=\frac{a + b - c}{2}\), but also, the tangent segments from the right angle to the incircle are equal to \(r\). Wait, maybe I misread. Wait, the two segments \(2x - 3\) and \(x + 2\) are both radii? Wait, no, in a triangle, the lengths from the vertex to the point of tangency are equal. So if the triangle is a right triangle, the inradius \(r\) satisfies that the two legs are \(a\) and \(b\), hypotenuse \(c\), and \(r=\frac{a + b - c}{2}\), and also, the segments from the right angle to the incircle are \(r\), and from the other vertices are \(a - r\) and \(b - r\). But in the diagram, the two segments \(2x - 3\) and \(x + 2\) are both equal to the inradius? Wait, no, maybe they are the lengths of the tangent segments from the same vertex? Wait, no, the diagram shows two right angles, so those are the radii? Wait, no, the inradius is the distance from the incenter to the sides, so the segments \(2x - 3\) and \(x + 2\) are both equal to the inradius \(r\). So set \(2x - 3=x + 2\), solve for \(x\): \(2x - x=2 + 3\), so \(x = 5\). Then \(r=x + 2=5 + 2=7\)? Wait, no, that can't be. Wait, no, maybe the diameter is \(2r\). Wait, no, wait, let's re - examine. If \(2x - 3\) and \(x + 2\) are the radii, then \(2x-3=x + 2\) gives \(x = 5\), so \(r=x + 2=7\), then diameter is \(2r = 14\)? Wait, no, that seems too big. Wait, maybe I made a mistake. Wait, no, maybe the segments \(2x - 3\) and \(x + 2\) are the lengths of the tangent segments, not the radius. Wait, no, in a triangle, the length from the vertex to the point of tangency is equal to \(s - a\), where \(s=\frac{a + b + c}{2}\). But in a right triangle, \(s=\frac{a + b + c}{2}\), and \(r=s - c\). Wait, maybe the two segments \(2x - 3\) and \(x + 2\) are equal to \(r\). So if \(2x-3=x + 2\), then \(x = 5\), so \(r=x + 2=7\), diameter is \(14\)? Wait, no, that can't be. Wait, maybe the triangle is a right triangle with legs \(a\) and \(b\), hypotenuse \(c\), and the inradius \(r=\frac{a + b - c}{2}\), and also, the segments from the right angle to the incircle are \(r\), and from the other vertices are \(a - r\) and \(b - r\). But in the diagram, the two segments \(2x - 3\) and \(x + 2\) are both equal to \(r\). So \(2x-3=x + 2\) gives \(x = 5\), so \(r=7\), diameter \(d = 2r=14\)? Wait, no, that seems incorrect. Wait, maybe I messed up the equation. Wait, let's start over.

Wait, the two segments \(2x - 3\) and \(x + 2\) are the lengths of the tangent segments from a vertex to the incircle. In a triangle, the lengths of the tangent segments from a single vertex to the incircle are equal. So if those two segments are from the same vertex, then \(2x - 3=x + 2\), so \(x = 5\). Then the length of each segment is \(x + 2=7\). But in a right triangle, the inradius \(r\) is equal to the length of the tangent segment from the right angle to the incircle. So if the triangle is a right triangle, then the inradius \(r\) is equal to those segments. So the radius \(r = 7\)? No, that can't be, because for a right triangle with legs \(a\) and \(b\), hypotenuse \(c\), \(r=\frac{a + b - c}{2}\), and \(a=r + (c - b)\), \(b=r+(c - a)\). Wait, maybe the triangle is not a right triangle? Wait, the diagram has right angles, so it's a right triangle. Wait, maybe the segments \(2x - 3\) and \(x + 2\) are the radii, so the diameter is \(2r\). Wait, if \(r=x + 2\) and \(r=2x - 3\), then \(x = 5\), \(r=7\), diameter \(14\). But that seems large. Wait, maybe I made a mistake in the equation. Wait, let's check again. \(2x-3=x + 2\), subtract \(x\) from both sides: \(x-3 = 2\), add 3: \(x = 5\). Then \(2x-3=10 - 3=7\), \(x + 2=7\). So the radius is 7? Then the diameter is \(2\times7 = 14\)? Wait, but maybe the segments are not the radius. Wait, in a triangle, the inradius is the distance from the incenter to the side, so the length of the tangent from the vertex to the incircle is equal to \(s - a\), where \(s\) is the semi - perimeter. But in a right triangle, \(s=\frac{a + b + c}{2}\), and \(r=s - c=\frac{a + b - c}{2}\). Also, the two tangent segments from the right angle to the incircle are both equal to \(r\). So if the two segments \(2x - 3\) and \(x + 2\) are those tangent segments (from the right angle), then they are equal to \(r\), so \(r=7\), diameter \(14\).

Answer:

14