Question

segment \\( \overline{bd} \\) is a median. solve for \\( x \\). round to the nearest tenth, if necessary. (image not necessarily to scale.)

Explanation:

Step1: Identify the median property

Since \( BD \) is a median of \( \triangle ABC \), \( D \) is the midpoint of \( AC \), so \( AD = DC \). Wait, no, wait. Wait, the sides: \( AB = 17 \), \( BC = 17 \), so \( \triangle ABC \) is isoceles with \( AB = BC \). Then the median from \( B \) to \( AC \) should also be the altitude? Wait, no, wait the given sides: \( AC \) is split into \( 14 \) and \( x \)? Wait, no, the diagram: \( AC \) is a side, \( D \) is on \( AC \), \( BD \) is the median, so \( AD = DC \)? Wait, no, the lengths: \( DC = 14 \), \( AD = x \), and \( AB = 17 \), \( BC = 17 \). So \( \triangle ABC \) has \( AB = BC = 17 \), so it's isoceles with \( AB = BC \). Then the median from \( B \) to \( AC \) (i.e., \( BD \)) should bisect \( AC \), so \( AD = DC \)? Wait, but \( DC = 14 \), so \( AD = 14 \)? No, that can't be. Wait, maybe I misread. Wait, \( AB = 17 \), \( BC = 17 \), \( AC \) is composed of \( AD = x \) and \( DC = 14 \). Wait, no, in an isoceles triangle with \( AB = BC \), the median from \( B \) to \( AC \) should make \( AD = DC \). Wait, but \( BC = 17 \), \( AB = 17 \), so \( \triangle ABC \) is isoceles with \( AB = BC \), so base is \( AC \), and median \( BD \) is also the altitude. Wait, no, let's use the formula for the length of a median. The median formula: in a triangle with sides \( a, b, c \), the length of the median to side \( a \) is \( m_a = \frac{1}{2}\sqrt{2b^2 + 2c^2 - a^2} \). But here, \( AB = BC = 17 \), so \( b = c = 17 \), \( a = AC = x + 14 \). Wait, no, maybe \( D \) is the midpoint, so \( AD = DC \), so \( x = 14 \)? But that seems too simple. Wait, no, the diagram: \( AB = 17 \), \( BC = 17 \), \( BD \) is the median, so \( AD = DC \), so \( x = 14 \)? But that can't be, because then \( AC = 28 \), and using Pythagoras, \( BD^2 + AD^2 = AB^2 \), so \( BD^2 + 14^2 = 17^2 \), \( BD^2 = 289 - 196 = 93 \), \( BD = \sqrt{93} \approx 9.6 \). But the question is to solve for \( x \). Wait, maybe I misread the sides. Wait, maybe \( AB = 17 \), \( BD \) is the median, \( AD = x \), \( DC = 14 \), and \( BC = 17 \). So \( \triangle ABD \) and \( \triangle CBD \) have \( AB = BC = 17 \), \( BD = BD \), and \( AD = DC \) (since median), so by SSS, they are congruent. Wait, but if \( AD = DC \), then \( x = 14 \). But that seems too easy. Wait, maybe the diagram is different. Wait, maybe \( AB = 17 \), \( BC = 17 \), \( AC \) is not the base. Wait, no, let's re-express. Wait, the problem says "Segment \( BD \) is a median. Solve for \( x \)". So \( D \) is the midpoint of \( AC \), so \( AD = DC \). Wait, but \( DC = 14 \), so \( AD = x = 14 \)? But that seems too simple. Wait, maybe I made a mistake. Wait, no, in a triangle, a median connects a vertex to the midpoint of the opposite side. So \( BD \) is the median from \( B \) to \( AC \), so \( D \) is the midpoint of \( AC \), so \( AD = DC \). Therefore, \( x = 14 \). But that seems too easy. Wait, maybe the sides are \( AB = 17 \), \( BC = 17 \), \( AC \) is split into \( AD = x \) and \( DC = 14 \), so \( x = 14 \). So the answer is \( 14 \).

Wait, no, maybe I misread the lengths. Wait, the diagram: \( AB = 17 \), \( BC = 17 \), \( DC = 14 \), \( AD = x \). So since \( BD \) is the median, \( AD = DC \), so \( x = 14 \).

Step1: Recall median definition

A median of a triangle connects a vertex to the midpoint of the opposite side. So \( D \) is the midpoint of \( AC \), meaning \( AD = DC \).

Step2: Set up equation

Given \( DC = 14 \) and \( AD = x \), we have \( x = DC \) (since \( D \) is the midpoint).

Thus, \( x = 14 \).

Wait, but that seems too simple. Wait, maybe the triangle is not isoceles. Wait, no, \( AB = BC = 17 \), so it is isoceles. So the median from \( B \) to \( AC \) must bisect \( AC \), so \( AD = DC \). Therefore, \( x = 14 \).

Answer:

\( \boxed{14} \)