Question

1. similar triangles are shown

in the diagram ( ae = 12 ), ( ec = 26 ), and ( bd = 6 ). what is the length of ( dc )?
____ units

Explanation:

Step1: Identify similar triangles

Triangles \( AEC \) and \( BDC \) are similar (given as similar triangles). So, their corresponding sides are proportional. Let \( DC = x \), then \( EC = ED + DC = 26 + x \)? Wait, no, looking at the diagram, \( EG \) (maybe a typo, should be \( EC \)) is 26? Wait, the diagram has \( E \) to \( C \) as 26? Wait, the labels: \( AE = 12 \), \( BD = 6 \), \( AE \) and \( BD \) are the heights, \( EC \) is the base of the larger triangle, and \( DC \) is the base of the smaller triangle. Wait, actually, for similar triangles, the ratio of corresponding sides is equal. So \( \frac{AE}{BD} = \frac{EC}{DC} \)? Wait, no, \( AE \) is the height of the larger triangle, \( BD \) is the height of the smaller triangle. The bases are \( EC \) (larger) and \( DC \) (smaller)? Wait, no, maybe \( E \) to \( D \) is some length, but the problem says \( EG = 26 \) (maybe \( EC = 26 \))? Wait, the problem states \( AE = 12 \), \( EG = 26 \) (probably \( EC = 26 \)), and \( BD = 6 \). Let's correct: Let the larger triangle be \( AEC \) with height \( AE = 12 \) and base \( EC \), and the smaller triangle be \( BDC \) with height \( BD = 6 \) and base \( DC \). Since they are similar, \( \frac{AE}{BD} = \frac{EC}{DC} \). Wait, but \( EC = ED + DC \)? No, maybe the diagram has \( E \) to \( D \) as part of the base, but the problem says \( EG = 26 \), maybe \( EC = 26 \)? Wait, the problem says "In the diagram \( AE = 12 \), \( EG = 26 \), and \( BD = 6 \). What is the length of \( DC \)?" Wait, maybe \( EG \) is a typo for \( EC \), and \( E \) to \( D \) is some length, but no, let's re-express. Actually, for similar triangles, the ratio of heights is equal to the ratio of bases. So \( \frac{AE}{BD} = \frac{EC}{DC} \). Wait, \( AE = 12 \), \( BD = 6 \), so the ratio of heights is \( \frac{12}{6} = 2 \). So the ratio of bases should also be 2. So \( \frac{EC}{DC} = 2 \), which means \( EC = 2 \times DC \). But \( EC = ED + DC \)? No, maybe \( EC \) is the base of the larger triangle, and \( DC \) is the base of the smaller triangle. Wait, maybe the diagram has \( E \) to \( C \) as 26, but that can't be. Wait, no, the problem must have \( EC = 26 \)? Wait, no, the user's diagram: "E to C is 26" (the curved line with 26). So \( EC = 26 \), \( AE = 12 \) (height of larger triangle), \( BD = 6 \) (height of smaller triangle). So similar triangles: \( \triangle AEC \sim \triangle BDC \), so \( \frac{AE}{BD} = \frac{EC}{DC} \). So \( \frac{12}{6} = \frac{26}{DC} \)? Wait, no, that would be wrong. Wait, maybe the base of the larger triangle is \( EC \), and the base of the smaller is \( DC \), but actually, the larger triangle's base is \( EC \), and the smaller's is \( DC \), but the height of the larger is \( AE = 12 \), height of the smaller is \( BD = 6 \). So similarity ratio is \( 12/6 = 2 \), so \( EC = 2 \times DC \). Wait, but \( EC = 26 \)? Then \( DC = 26 / 2 = 13 \)? No, that doesn't make sense. Wait, maybe I got the triangles reversed. Maybe the larger triangle is \( AEC \) with base \( EC \) and height \( AE \), and the smaller is \( BDC \) with base \( DC \) and height \( BD \), but actually, the base of the larger triangle is \( EC \), and the base of the smaller is \( DC \), but the horizontal segment from \( E \) to \( D \) is part of the base. Wait, no, the correct proportion for similar triangles (right triangles, since \( AE \perp EC \) and \( BD \perp EC \)) is \( \frac{AE}{BD} = \frac{EC}{DC} \). Wait, \( AE = 12 \), \( BD = 6 \), so \( 12/6 = 2 \), so \( EC = 2 \times DC \). But if \( EC = 26 \), then \( DC = 13 \). Wait, but let's check: \( 12/6 = 2 \), so \( EC = 2 \times DC \), so \( DC = EC / 2 = 26 / 2 = 13 \). Wait, but that seems correct. Wait, maybe the problem is that \( EC \) is the entire base, and \( DC \) is the base of the smaller triangle, so the ratio of heights is 12/6 = 2, so the ratio of bases is also 2, so \( EC = 2 \times DC \), hence \( DC = EC / 2 = 26 / 2 = 13 \). Wait, but let's do it properly. Let \( DC = x \). Then, since \( \triangle AEC \sim \triangle BDC \), \( \frac{AE}{BD} = \frac{EC}{DC} \). So \( \frac{12}{6} = \frac{26}{x} \)? No, that would be \( 12x = 6 \times 26 \), \( x = (6 \times 26)/12 = 13 \). Yes, that's correct. So \( x = 13 \).

Step1: Set up proportion for similar triangles

Since \( \triangle AEC \sim \triangle BDC \), the ratios of corresponding sides are equal.
\( \frac{AE}{BD} = \frac{EC}{DC} \)
Given \( AE = 12 \), \( BD = 6 \), \( EC = 26 \). Substitute:
\( \frac{12}{6} = \frac{26}{DC} \)

Step2: Solve for \( DC \)

Simplify \( \frac{12}{6} = 2 \), so:
\( 2 = \frac{26}{DC} \)
Multiply both sides by \( DC \):
\( 2 \times DC = 26 \)
Divide both sides by 2:
\( DC = \frac{26}{2} = 13 \)

Answer:

13