Q. 125.0( 2 Votes )

Construct a triangle ABC, similar to a given isosceles triangle PQR, with QR = 2.8 cm, PQ = 2.5 cm, such that each of its side th of the corresponding sides of the PQR. Also draw the circumcircle of PBC.

Answer :

Note: we have to construct triangle PBC and not ABC


Step1: Draw PQ = 2.5 cm



As ΔPQR is isosceles QR = PR = 2.8 cm


Step2: Take distance 2.8 cm in compass, keep the needle on point P and mark an arc above PQ. Keeping the distance in the compass same keep the needle on point Q and mark an arc intersecting the previous arc. Mark intersection point as R



Step3: Join PR and QR and draw a ray from point P below PQ



Step4: Take any distance in compass and keeping the needle of the compass on point P cut an arc on ray constructed in step3 and name that point X1. Keeping the distance in compass same keep the needle of the compass on point X1 and cut an arc on the same ray and mark that point as X2. Draw 7 such parts (greater of 6 and 7 in 6/7), i.e. by repeating this process mark points upto X7



Step5: join X7 and Q and from X6 (6 being smaller in ) draw a line parallel to X7Q intersecting PQ at B



Step6: Draw a line parallel to QR from point B intersecting PR at C and ΔPBC is ready



Now to construct the circumcircle of ΔPBC. The centre of circumcircle is the intersection of perpendicular bisectors, and the radius is the distance from the centre to any vertex of the triangle. We will draw perpendicular bisector of PC and BC


Step7: Take any distance approximately greater than half of BC in compass. Keep the needle of the compass on point B and mark arcs to both sides of BC.



Step8: Keeping the distance in the compass same keep the needle on point C and mark arcs intersecting arcs drawn in step7. Draw a line between these intersecting arcs



Step9: Repeat step7 and step8 to draw perpendicular bisector for PC and mark the intersection point of both perpendicular bisectors as O



Step10: Keep the needle of the compass on point O and draw circle taking radius OC. Circumcentre of ΔPBC is ready



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