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  • If in two triangles DEF and PQR, angle D = angle Q and angle R = angle E, then which of the following is not true? (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}

If in two triangles DEF and PQR, \angle D = \angle Q \; \text {and} \; \angle R = \angle E, then which of the following is not true?

(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}

Answers (1)

Answer : [B]

In \Delta DEF and \Delta PQR

\angle D=\angle Q and \angle R=\angle E

Now draw two triangles with the help of given conditions:-

Here \angle D=\angle Q and \angle E=\angle R  {Given}

\therefore \; \; \; \; \Delta DEF\sim \Delta QRP  {by AA similarity}

\Rightarrow \angle F=\angle P  {corresponding angles of similar triangles}

\therefore \frac{DF}{QP}=\frac{ED}{RQ}=\frac{FE}{PR}\; \; \; \; \; \; \; \; ....(1)

Here option (A) \frac{EF}{PR}=\frac{DF}{PQ}  is true because both the terms are derived from equation (1)

Here option (B) \frac{DE}{PQ}=\frac{EF}{RP} is not true because the first term is not derived from equation (1)

Here option (C) \frac{DE}{QR}=\frac{DF}{PQ}  is true because both the terms are derived from equation (1)

Here option D i.e. is also true because both terms are derived from equation (1)

Hence option (B) is not true.

Posted by

infoexpert23

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