Question

express x in terms of a, b, and c.

Explanation:

Step1: Identify Similar Triangles

The two right triangles are similar (by AA similarity, as both have a right angle and share the same acute angle). So, the ratios of corresponding sides are equal.
Let the larger triangle have base \(a + b\) and height \(c\), and the smaller triangle (with the segment of length \(a\)) have base \(a\) and height \(x\). The ratio of base to height for similar triangles is equal: \(\frac{a + b}{c}=\frac{a}{x}\)? Wait, no, wait. Wait, actually, the smaller triangle (the one with height \(x\)) has base \(a\)? Wait, no, looking at the diagram: the larger triangle has base \(a + b\) and height \(c\), and the smaller triangle (the one with the vertical side \(x\)) has base \(a\)? Wait, no, maybe I got the correspondence wrong. Wait, the two right triangles: one has legs \(b\) and \(c\), and the other has legs \(a\) and \(x\)? No, wait, the triangles are similar because they are both right-angled and share the same angle at the vertex. So the triangle with height \(c\) has base \(a + b\), and the triangle with height \(x\) has base \(a\)? Wait, no, maybe the correct correspondence is: the big triangle (with base \(a + b\) and height \(c\)) and the small triangle (with base \(a\) and height \(x\))? Wait, no, actually, the triangle with the vertical side \(x\) and horizontal side \(a\) is similar to the triangle with vertical side \(c\) and horizontal side \(a + b\). So the ratio of vertical to horizontal should be equal. So \(\frac{x}{a}=\frac{c}{a + b}\)? Wait, no, that doesn't seem right. Wait, maybe the other way: the triangle with base \(b\) and height \(c - x\)? No, the diagram shows a right triangle with a smaller right triangle inside, sharing the same angle at the left vertex. So the larger triangle has base \(a + b\) and height \(c\), and the smaller triangle (the one with the vertical segment \(x\)) has base \(a\) and height \(x\)? Wait, no, the vertical side of the larger triangle is \(c\), and the vertical side of the smaller triangle is \(x\), and the horizontal sides are \(a + b\) and \(a\) respectively? Wait, no, the horizontal side of the larger triangle is \(a + b\) (from the left vertex to the right end), and the horizontal side of the smaller triangle is \(a\) (from the left vertex to the point below \(x\)). So the two triangles are similar, so the ratio of height to base is equal. So \(\frac{x}{a}=\frac{c}{a + b}\)? Wait, no, that would mean \(x=\frac{ac}{a + b}\), but that doesn't seem right. Wait, maybe the other triangle: the triangle with base \(b\) and height \(c - x\) is similar? No, the correct approach is: the two right triangles are similar, so corresponding sides are proportional. Let's denote the left triangle (with height \(x\) and base \(a\)) and the right triangle (with height \(c\) and base \(a + b\))? No, the right triangle has base \(b\) and height \(c\), and the left triangle has base \(a\) and height \(x\), and they are similar because they share the same angle. Wait, actually, the two triangles are similar, so \(\frac{x}{a}=\frac{c}{a + b}\) is incorrect. Wait, maybe the correct proportion is \(\frac{x}{a}=\frac{c}{a + b}\)? No, let's think again. The larger triangle has legs \((a + b)\) and \(c\), and the smaller triangle has legs \(a\) and \(x\). Since they are similar, \(\frac{x}{a}=\frac{c}{a + b}\), so \(x=\frac{ac}{a + b}\)? Wait, no, that can't be. Wait, maybe the other way: the triangle with base \(b\) and height \(c - x\) is similar to the triangle with base \(a + b\) and height \(c\). So \(\frac{c - x}{b}=\frac{c}{a + b}\). Then solving for \(x\): \(c - x=\frac{bc}{a + b}\), so \(x = c-\frac{bc}{a + b}=\frac{c(a + b)-bc}{a + b}=\frac{ac + bc - bc}{a + b}=\frac{ac}{a + b}\). Wait, that's the same as before. Wait, maybe the diagram is a right triangle with a line parallel to the vertical side, creating a smaller similar triangle. So the original triangle has base \(a + b\) and height \(c\), and the smaller triangle has base \(a\) and height \(x\). So by similarity, \(\frac{x}{a}=\frac{c}{a + b}\), so \(x=\frac{ac}{a + b}\). Wait, but that seems odd. Wait, maybe the correct proportion is \(\frac{x}{b}=\frac{c}{a + b}\)? No, that would be if the base of the smaller triangle is \(b\). Wait, the diagram shows: on the horizontal axis, from left to right, there's a segment of length \(a\), then \(b\), so total \(a + b\). The vertical side on the right is \(c\), and the vertical side above the segment \(a\) is \(x\). So the two triangles (the big one with base \(a + b\) and height \(c\), and the small one with base \(a\) and height \(x\)) are similar. So the ratio of height to base is equal: \(\frac{x}{a}=\frac{c}{a + b}\), so \(x=\frac{ac}{a + b}\). Wait, but maybe I mixed up the base and height. Alternatively, the triangle with base \(b\) and height \(c - x\) is similar to the triangle with base \(a + b\) and height \(c\). So \(\frac{c - x}{b}=\frac{c}{a + b}\), so \(c - x=\frac{bc}{a + b}\), so \(x = c - \frac{bc}{a + b}=\frac{c(a + b)-bc}{a + b}=\frac{ac + bc - bc}{a + b}=\frac{ac}{a + b}\). So that's the same result. So the answer is \(x=\frac{ac}{a + b}\)? Wait, no, maybe the correct proportion is \(\frac{x}{c}=\frac{a}{a + b}\), so \(x=\frac{ac}{a + b}\). Yes, that makes sense. So the key is recognizing similar triangles and setting up the proportion.

Step1: Recognize Similar Triangles

The two right triangles are similar (AA similarity: right angle and common acute angle). Let the larger triangle have base \(a + b\) and height \(c\), and the smaller triangle have base \(a\) and height \(x\).

Step2: Set Up Proportion

For similar triangles, the ratio of corresponding sides is equal:
\(\frac{x}{a}=\frac{c}{a + b}\)

Step3: Solve for \(x\)

Cross - multiply to get:
\(x(a + b)=ac\)
Divide both sides by \(a + b\):
\(x = \frac{ac}{a + b}\)

Answer:

\(x=\dfrac{ac}{a + b}\)