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Asked by 13th February 2009, 10:33 PM
Answered by Expert

Draw triangle PQR,  with PQ=PR= 5 cm.9 for convenience sake take the base as RQ.)

Draw a ray  RT making an  acute angle QRS with the side QR( below QR)

With the pointed end of the compass on R mark 7 points at equal distances on RS.

Mark the 7th point  from R as T.

 join TQ.

Mark the 6th point   from R as S.

Thru' S ,  construct  a line segment  parallel to TQ.

Let this line segment meet the side RQ at A .

Thru' A construct a line segment AB parallel to  PQ meeting RP in A.

ABR will be the triangle similar to triangle  PQR.


Answered by Expert 14th February 2009, 5:48 PM
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