Question

in the diagram below, $\angle cab \cong \angle cde$. solve for $x$. round your answer to the nearest tenth if necessary.

(diagram: triangle with points c, a, b; d on ca, e on cb. cd = 5.1, da = 6.9, eb = 5.1, ce = x)

answer attempt 1 out of 2
$x = \square$ submit answer

Explanation:

Step1: Use the property of similar triangles

Since \(\angle CAB\cong\angle CDE\) and \(\angle C\) is common to both \(\triangle CAB\) and \(\triangle CDE\), by the AA (Angle - Angle) similarity criterion, \(\triangle CDE\sim\triangle CAB\).
For similar triangles, the ratios of corresponding sides are equal. So, \(\frac{CD}{CA}=\frac{CE}{CB}\).
We know that \(CA = CD + DA=5.1 + 6.9 = 12\), \(CB=CE + EB=x + 5.1\), and \(CD = 5.1\), \(CE=x\).
Substituting these values into the proportion \(\frac{CD}{CA}=\frac{CE}{CB}\), we get \(\frac{5.1}{12}=\frac{x}{x + 5.1}\).

Step2: Cross - multiply

Cross - multiplying the equation \(\frac{5.1}{12}=\frac{x}{x + 5.1}\) gives \(5.1(x + 5.1)=12x\).
Expand the left - hand side: \(5.1x+26.01 = 12x\).

Step3: Solve for \(x\)

Subtract \(5.1x\) from both sides: \(26.01=12x - 5.1x\).
Combine like terms: \(26.01 = 6.9x\).
Divide both sides by \(6.9\): \(x=\frac{26.01}{6.9}=3.8\).

Answer:

\(x = 3.8\)