Question

1. \\( \triangle uvw \sim \triangle dcw \\) 16. \\( \triangle efg \sim \triangle eqr \\) ( cw = \\) ( eq = \\)

Explanation:

Step1: For \(\triangle UVW\sim\triangle DCW\)

Since the triangles are similar, the ratios of corresponding sides are equal. \(\frac{UV}{DC}=\frac{VW}{CW}=\frac{UW}{DW}\). We know \(UV = 96\), \(DC = 40\), \(UW=48\), \(DW = 20\), and \(VW=84\). Let \(CW=x\). Using the ratio \(\frac{UW}{DW}=\frac{VW}{CW}\), we have \(\frac{48}{20}=\frac{84}{x}\).
Cross - multiply: \(48x=20\times84\).

Step2: Solve for \(x\)

\(48x = 1680\), then \(x=\frac{1680}{48}=35\).

Step3: For \(\triangle EFG\sim\triangle EQR\)

Since the triangles are similar, the ratios of corresponding sides are equal. \(\frac{EF}{EQ}=\frac{EG}{ER}=\frac{FG}{QR}\). Let \(EQ = y\). We know \(EF = 40\), \(FG = 44\), \(QR = 11\), \(ER=13\), \(EG = 13+(52 - 13)=52\). Using the ratio \(\frac{FG}{QR}=\frac{EF}{EQ}\), we have \(\frac{44}{11}=\frac{40}{y}\).
Cross - multiply: \(44y=11\times40\).

Step4: Solve for \(y\)

\(44y = 440\), then \(y = 10\).

Answer:

1. \(CW = 35\)
2. \(EQ=10\)