$\Delta PQR \sim \Delta DEF$ for the correspondence $PQR \leftrightarrow DEF$. If $3 PQ = 2 DE$,$EF = 6$,and $PR = 8$,find $QR$ and $DF$.
(N/A) Given that $\Delta PQR \sim \Delta DEF$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{PQ}{DE} = \frac{QR}{EF} = \frac{PR}{DF}$.
From the given equation $3 PQ = 2 DE$,we get $\frac{PQ}{DE} = \frac{2}{3}$.
Now,using the ratio $\frac{PQ}{DE} = \frac{QR}{EF}$:
$\frac{2}{3} = \frac{QR}{6} \implies QR = \frac{2 \times 6}{3} = 4$.
Next,using the ratio $\frac{PQ}{DE} = \frac{PR}{DF}$:
$\frac{2}{3} = \frac{8}{DF} \implies DF = \frac{8 \times 3}{2} = 12$.
Thus,$QR = 4$ and $DF = 12$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{PQ}{DE} = \frac{QR}{EF} = \frac{PR}{DF}$.
From the given equation $3 PQ = 2 DE$,we get $\frac{PQ}{DE} = \frac{2}{3}$.
Now,using the ratio $\frac{PQ}{DE} = \frac{QR}{EF}$:
$\frac{2}{3} = \frac{QR}{6} \implies QR = \frac{2 \times 6}{3} = 4$.
Next,using the ratio $\frac{PQ}{DE} = \frac{PR}{DF}$:
$\frac{2}{3} = \frac{8}{DF} \implies DF = \frac{8 \times 3}{2} = 12$.
Thus,$QR = 4$ and $DF = 12$.
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View SolutionWhich of the following correctly matches the information in Part $I$ and Part $II$?
Part $I$ Part $II$ $1.$ In $\Delta ABC$ and $\Delta PQR, \angle A \cong \angle P$ and $\angle C \cong \angle Q$ $a.$ Correspondence $ABC \leftrightarrow RQP$ is a similarity. $2.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{QR} = \frac{BC}{PQ}$ and $\angle B \cong \angle Q$ $b.$ Correspondence $ABC \leftrightarrow QPR$ is a similarity. $3.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{BC}{PR} = \frac{CA}{QR}$ $c.$ Correspondence $ABC \leftrightarrow PQR$ is a similarity. $4.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{CA}{PR}$ and $\angle A \cong \angle P$ $d.$ Correspondence $ABC \leftrightarrow PRQ$ is a similarity.
| Part $I$ | Part $II$ |
| $1.$ In $\Delta ABC$ and $\Delta PQR, \angle A \cong \angle P$ and $\angle C \cong \angle Q$ | $a.$ Correspondence $ABC \leftrightarrow RQP$ is a similarity. |
| $2.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{QR} = \frac{BC}{PQ}$ and $\angle B \cong \angle Q$ | $b.$ Correspondence $ABC \leftrightarrow QPR$ is a similarity. |
| $3.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{BC}{PR} = \frac{CA}{QR}$ | $c.$ Correspondence $ABC \leftrightarrow PQR$ is a similarity. |
| $4.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{CA}{PR}$ and $\angle A \cong \angle P$ | $d.$ Correspondence $ABC \leftrightarrow PRQ$ is a similarity. |
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