Question

Construct a $\Delta$ ABC, in which BC = 6 cm, AB = 9 cm and $\angle$ABC = 60${}^{\circ }$

i. Construct the locus of all points inside $\Delta$ ABC, which are equidistant from B and C.

ii. Construct the locus of the vertices of the triangle with BC as base, which are equal in area to triangle ABC.

iii. Mark the point Q in your construction, which would make $\Delta$ QBC equal in area to $\Delta$ ABC, and isosceles.

iv. Measure and record the length of CQ.

Answer 1

A line XY is drawn and BC equal to 6 cm is taken on it. $\angle$ ABC = 60${}^{\circ }$ is drawn with arm AB = 9 cm. A and C are joined to get the required $\Delta$ ABC.

i. AD perpendicular to BC is drawn.

ii. A line X’Y’ is drawn perpendicular to AD (i.e. parallel to XY and passing through A). X’Y’ is the required locus.

iii. PQ, the perpendicular bisector of BC, is drawn meeting X’Y’ at Q. Q is the required point such that area ( $\Delta$ QBC) = area( $\Delta$ ABC).

iv. CQ = 8.4 cm.