A) \[\frac{EF}{PR}=\frac{DF}{PQ}\]
B) \[\frac{DE}{PQ}=\frac{EF}{RP}\]
C) \[\frac{DE}{QR}=\frac{DF}{PQ}\]
D) \[\frac{EF}{RP}=\frac{DE}{QR}\]
Correct Answer: B
Solution :
[b] Given, in \[\Delta DEF\] and \[\Delta PQR\], |
\[\angle D=\angle Q,\,\,\,\angle R=\angle E\] |
\[\therefore \,\,\,\Delta DEF\tilde{\ }\Delta QRP\](By AA similarity criterion) |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\angle F=\angle P\] |
(Corresponding angles of similar triangles) |
\[\therefore \,\,\,\,\,\,\,\,\,\,\frac{DF}{QP}=\frac{ED}{RQ}=\frac{FE}{PR}\] |
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