Question

1. △cde ~ △fge; find ce

Explanation:

Step1: Use the property of similar triangles

Since \(\triangle CDE\sim\triangle FGE\), the ratios of corresponding sides are equal. That is \(\frac{CD}{FG}=\frac{CE}{FE}\). Here \(CD = 6x + 3\), \(FG=21\), \(CE=8x - 1\), and \(FE = 21+17=38\). So we have the equation \(\frac{6x + 3}{21}=\frac{8x - 1}{38}\).

Step2: Cross - multiply

Cross - multiplying the equation \(\frac{6x + 3}{21}=\frac{8x - 1}{38}\) gives \((6x + 3)\times38=21\times(8x - 1)\).
Expanding both sides:

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Step3: Solve for \(x\)

Subtract \(168x\) from both sides: \(228x-168x + 114=168x-168x - 21\), which simplifies to \(60x+114=- 21\).
Subtract \(114\) from both sides: \(60x=-21 - 114=-135\).
Divide both sides by \(60\): \(x =-\frac{135}{60}=-\frac{9}{4}\).

Step4: Find \(CE\)

Substitute \(x =-\frac{9}{4}\) into \(CE = 8x-1\).
\(CE=8\times(-\frac{9}{4})-1=-18 - 1=-19\). But length cannot be negative. There is a mistake. Let's re - check the ratio.

The correct ratio for similar triangles \(\triangle CDE\sim\triangle FGE\) is \(\frac{CD}{FG}=\frac{CE}{GE}\) (corresponding sides). So \(\frac{6x + 3}{21}=\frac{8x - 1}{17}\).
Cross - multiply: \((6x + 3)\times17=21\times(8x - 1)\).
Expand: \(102x+51 = 168x-21\).
Subtract \(102x\) from both sides: \(51=168x - 102x-21\).
\(51 + 21=66x\).
\(72 = 66x\).
\(x=\frac{72}{66}=\frac{12}{11}\).
Now substitute \(x = \frac{12}{11}\) into \(CE=8x - 1\).
\(CE=8\times\frac{12}{11}-1=\frac{96}{11}-\frac{11}{11}=\frac{96 - 11}{11}=\frac{85}{11}\approx7.73\).

Answer:

\(CE=\frac{85}{11}\)