Steps of construction:
(i) (1) Mark a horizontal line XY on your paper and take BC = 6 cm on it.
(2) Construct ∠ABC = 60 ̊ with arm AB = 9 cm.
(3) Join A and C to get the required ΔABC.
(iii) (1) Draw AD ⊥ BC.
(2) Construct a line X’Y’, perpendicular to AD, parallel to XY and passing through A.
(3) X’Y’ is the required locus of the vertices of Δ’s with base BC and area to ΔABC.
[∵ Δ’s having same base and height an equal in area ]
(iii) (1) Draw right bisector PQ of BC, meeting X’Y; in Q.
(2) Then Q is the point such that ΔQBC is an isosceles triangle and area (ΔQBC) = area(ΔABC).
(iv) On measuring, we find CQ = 8.4 cm.