Question

in the diagram, dc is 10 units and bc is 6 units. what is the length of segment ac? 6 units 8 units 12 units 16 units

Explanation:

Step1: Identify triangle type

Triangle \(DBC\) is right - angled at \(B\) (given by the right - angle symbol at \(B\) in \(\triangle DBC\)). We know \(DC = 10\) units and \(BC = 6\) units. We can use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) (where \(c\) is the hypotenuse and \(a,b\) are the legs of a right - triangle) to find the length of \(DB\). Let \(DB=x\), \(BC = 6\), \(DC = 10\). Then \(x^{2}+6^{2}=10^{2}\).

Step2: Solve for \(DB\)

\[ \begin{align*} x^{2}+36&=100\\ x^{2}&=100 - 36\\ x^{2}&=64\\ x&=\sqrt{64}=8 \end{align*} \]
From the diagram, we can see that \(B\) is the mid - point of \(AC\) (since the segments \(AB\) and \(BC\) have the same marking, indicating they are equal in length) and also \(B\) is the mid - point of \(DE\) (but we focus on \(AC\) here). Also, triangles \(ABD\) and \(CBD\) seem to be congruent (by SAS, as \(AB = BC\), \(\angle ABD=\angle CBD = 90^{\circ}\), and \(DB\) is common). So \(AB = BC\) and also from the right - triangle \(DBC\), we found \(DB = 8\), but more importantly, since \(B\) is the mid - point of \(AC\) and we can also see that \(AC = 2\times AB\), and also from the right - triangle, we can infer that \(AB = DB = 8\)? Wait, no. Wait, actually, since \(B\) is the mid - point of \(AC\) (the markings on \(AB\) and \(BC\) show they are equal) and in the right - triangle \(DBC\), we can find \(DB\) as 8, and also, since the diagram shows that \(AB = DB\) (maybe due to some congruency or symmetry), but actually, the key is that \(B\) is the mid - point of \(AC\) and also, from the right - triangle \(DBC\), we can find \(DB = 8\), and since \(AB = DB\) (assuming congruent triangles or perpendicular bisector), but actually, the main point is that \(AC=2\times AB\), and also, since \(B\) is the mid - point, \(AC = 2\times BC\)? No, wait, the markings on \(AB\) and \(BC\) are the same, so \(AB = BC\)? Wait, no, the markings on \(AB\) and the other segment (the one from \(B\) to the other side) - wait, looking at the diagram, the two segments from \(B\) on \(AC\) are marked equal, so \(AB = BC\)? No, wait, the right - angle is at \(B\) between \(DB\) and \(AC\), and the markings on \(AB\) and \(BC\) are the same, so \(B\) is the mid - point of \(AC\), so \(AC = 2\times AB\). Also, in right - triangle \(DBC\), \(DC = 10\), \(BC = 6\), so \(DB=\sqrt{10^{2}-6^{2}}=\sqrt{100 - 36}=\sqrt{64}=8\). And since \(DB\) is perpendicular to \(AC\) and \(AB = BC\) (mid - point), and also, the triangles \(ABD\) and \(CBD\) are congruent (SAS: \(AB = BC\), \(\angle ABD=\angle CBD = 90^{\circ}\), \(DB = DB\)), so \(AD = DC\), but we need \(AC\). Wait, actually, since \(B\) is the mid - point of \(AC\) and \(DB\) is perpendicular to \(AC\), \(AC = 2\times AB\), and also, from the right - triangle \(DBC\), we found \(DB = 8\), but also, since \(AB = DB\) (maybe because of the congruency or the way the diagram is drawn, with the markings on \(AB\) and the segment equal to \(DB\)), but actually, the key is that \(AC=2\times AB\), and since \(AB = DB = 8\)? No, wait, no. Wait, the length of \(AC\): since \(B\) is the mid - point, \(AC = 2\times BC\)? No, \(BC = 6\), that would be 12, but that's not right. Wait, no, I made a mistake. Let's re - examine. The right - triangle is \(DBC\), with \(DC = 10\), \(BC = 6\), so \(DB=\sqrt{10^{2}-6^{2}} = 8\). Now, looking at the diagram, the segment \(AB\) is equal to \(DB\) (because of the markings, maybe the two segments with the same tick marks are equal), and \(B\) is the mid - point of \(AC\), so \(AC=2\times AB\). Since \(AB = DB = 8\)? No, wait, \(AB\) and \(BC\) have the same tick marks, so \(AB = BC\)? No, the tick marks on \(AB\) and the segment from \(B\) to \(E\) (or maybe \(AB\) and \(DB\))? Wait, the correct approach: since \(DB\perp AC\) and \(AB = BC\) (mid - point), triangle \(ADC\) is isoceles with \(AD = DC\). But we need \(AC\). Wait, no, the problem is that \(B\) is the mid - point of \(AC\), so \(AC = 2\times AB\), and also, in right - triangle \(DBC\), \(DB = 8\), and since \(AB = DB\) (because of the congruency of triangles \(ABD\) and \(CBD\), as \(AB = BC\), \(\angle ABD=\angle CBD = 90^{\circ}\), \(DB = DB\)), so \(AB = 8\), then \(AC=2\times AB = 16\)? No, that's not right. Wait, no, I think I messed up. Wait, the length of \(AC\): since \(B\) is the mid - point, \(AC = 2\times BC\) only if \(AB = BC\), but \(BC = 6\), that would be 12, but that's not matching. Wait, no, the right - triangle is \(DBC\), \(DC = 10\), \(BC = 6\), so \(DB = 8\). Now, the segment \(AC\): if we look at the diagram, \(AB = DB = 8\) and \(BC = 6\)? No, that can't be. Wait, maybe the diagram shows that \(B\) is the mid - point of \(AC\), so \(AC = 2\times AB\), and also, \(DB\) is perpendicular to \(AC\), and triangle \(ABD\) is congruent to triangle \(CBD\), so \(AB = BC\)? No, that would mean \(AC = 2\times BC=12\), but then \(DB\) would be calculated as \(\sqrt{DC^{2}-BC^{2}}=\sqrt{100 - 36}=8\), and if \(AB = BC = 6\), then \(AC = 12\). Wait, that makes sense. Because if \(B\) is the mid - point, \(AC = 2\times BC\), and \(BC = 6\), so \(AC = 12\)? But then \(DB = 8\), which is a different length. Wait, maybe the markings on \(AB\) and \(BC\) are the same, so \(AB = BC\), so \(AC=AB + BC=2\times BC = 12\) units.

Step3: Conclusion

Since \(B\) is the mid - point of \(AC\) (indicated by the equal markings on \(AB\) and \(BC\)) and \(BC = 6\) units, then \(AC=2\times BC = 12\) units. Wait, but earlier when we calculated \(DB = 8\), how does that fit? Maybe the diagram has \(DB\) perpendicular to \(AC\), and \(AB = BC\), so \(AC = 12\) units.

Answer:

12 units