Asked by | 13th Feb, 2009, 10:33: PM
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.
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 | 14th Feb, 2009, 05:48: PM
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