Q. 55.0( 3 Votes )

Ramu pradhan drew a parallelogram ABCD where AB=6cm, BC=9 cm. and ABC=60°. Let’s select two points P and Q on diagonal BD such that BP=PQ=QD.

Let’s write the nature of the quadrilateral APCQ joining the point A,P;P,C;C,Q and Q,A.

Answer :

First let us construct the parallelogram ABCD

Step1: Draw segment BC of length 9 cm



Step2: Draw a line segment BA of length 6 cm at 60° to BC from point B



Now we will draw line parallel to AB from point C


Step3: Take any distance in compass keep the needle on point B and mark an arc intersecting AB and BC at X and Y. Keeping the distance in the compass same keep the needle on point C and mark an arc intersecting BC extended at S




Step4: Take distance XY in compass keep the needle on point S and mark arc intersecting the arc with centre C at R. Draw a line CD of length 6 cm passing through R. Join AD. Join BD



Now we have to take points P and Q on BD such that BP = PQ = QD which means we have to divide BD in 3 equal parts


We first draw a ray at any angle to the given line which is to be divided. Here the line to be divided is BD let AD be the line which will act as the ray for BD.


Step5: Take any distance in compass and keep the needle of compass on point D and draw an arc intersecting AD. Name the intersection point as X1. Keeping the distance same in compass keep the needle on point X1 and mark an arc intersecting AD at X2. Draw 3 such parts that is upto X3. By doing this we are making 3 equal parts on AD



Step6: Join BX3 and draw lines parallel to BX3 from X2 and X1 intersecting BD at P and Q respectively



Step7: Join AP, AQ, CP and CQ. Using scale measure lengths AP, AQ, CP and CQ



As the lengths AP = QC and AQ = PC which are opposite sides of quadrilateral APCQ.


As opposite sides of quadrilateral are equal quadrilateral APCQ is a parallelogram


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