Question

1. if ( \triangle abcsim\triangle dec ), find the values of ( x ) and ( y ).

Explanation:

Step1: Use the property of similar triangles (corresponding sides are proportional)

Since \(\triangle ABC\sim\triangle DEC\), we have \(\frac{AB}{DE}=\frac{BC}{DC}=\frac{AC}{EC}\).
We know \(AB = 6\), \(DE=21\), \(BC = 10\), \(DC = 28\), \(AC=x\), \(EC = y\).
From \(\frac{AB}{DE}=\frac{BC}{DC}\), we get \(\frac{6}{21}=\frac{10}{28}\) (checking the ratio for setup, but the main ratios for calculation are \(\frac{AB}{DE}=\frac{AC}{EC}\) and \(\frac{BC}{DC}=\frac{AC}{EC}\))
From \(\frac{AB}{DE}=\frac{AC}{EC}\) and \(\frac{BC}{DC}=\frac{AC}{EC}\)
First, use \(\frac{AB}{DE}=\frac{BC}{DC}\) to confirm the similarity ratio. The ratio of similarity \(k=\frac{AB}{DE}=\frac{6}{21}=\frac{2}{7}\) (or \(k = \frac{BC}{DC}=\frac{10}{28}=\frac{5}{14}\), no, correct ratio: \(\frac{AB}{DE}=\frac{6}{21}=\frac{2}{7}\), \(\frac{BC}{DC}=\frac{10}{28}=\frac{5}{14}\) - wrong approach. Correct: \(\frac{AB}{DE}=\frac{BC}{DC}=\frac{AC}{EC}\)
\(\frac{AB}{DE}=\frac{AC}{EC}\) and \(\frac{BC}{DC}=\frac{AC}{EC}\)
\(\frac{6}{21}=\frac{x}{28}\) and \(\frac{10}{28}=\frac{6}{21}=\frac{x}{28}\) (no, correct: \(\frac{AB}{DE}=\frac{BC}{DC}=\frac{AC}{EC}\)
\(\frac{AB}{DE}=\frac{AC}{EC}\): \(\frac{6}{21}=\frac{x}{28}\), cross - multiply \(21x=6\times28\)
\(\frac{BC}{DC}=\frac{AB}{DE}\): \(\frac{10}{28}=\frac{6}{21}\) (check, \(10\times21 = 210\), \(6\times28=168\) - wrong. Correct: \(\triangle ABC\sim\triangle DEC\) (order of similarity: \(A\) corresponds to \(D\), \(B\) corresponds to \(E\), \(C\) corresponds to \(C\))
So \(\frac{AB}{DE}=\frac{BC}{EC}=\frac{AC}{DC}\)
\(\frac{6}{21}=\frac{10}{y}=\frac{x}{28}\)
From \(\frac{6}{21}=\frac{x}{28}\), cross - multiply: \(21x=6\times28\), \(x=\frac{6\times28}{21}\)
From \(\frac{6}{21}=\frac{10}{y}\), cross - multiply: \(6y = 21\times10\)

Step2: Solve for \(x\)

For \(x\):

$$
LATEXBLOCK0
$$

Step3: Solve for \(y\)

For \(y\):

$$
LATEXBLOCK1
$$

Answer:

\(x = 8\) and \(y = 35\)