1How can we construct \( \triangle DBC \), equal in area to \( \triangle ABC \) having \( \angle ABC = 60^\circ \), BC = 4.4 cm and AB = 5.5 cm, and having an angle of \( 120^\circ \)?
2How can we construct \( \triangle DBC \), equal in area to the same \( \triangle ABC \), having one side 6.2 cm?
3In the construction with the \( 120^\circ \) angle, the area of \( \triangle DBC \) is equal to the area of \( \triangle ABC \). How?
4In the construction with the 6.2 cm side, the area of \( \triangle BCD \) is equal to the area of \( \triangle ABC \). How?
5Why is \( \angle BAS \) made equal to \( \angle ABC \) when drawing the parallel line through A?
6If A is about 4.8 cm above BC, why can no \( \triangle DBC \) of equal area have BD = 4 cm?
7In the \( 120^\circ \) construction, show that ABCD is a parallelogram.
8How would you construct \( \triangle DBC \), equal in area to \( \triangle ABC \) with BC = 5 cm, \( \angle ABC = 45^\circ \) and AB = 6 cm, having \( \angle BCD = 90^\circ \)?
9If the 6.2 cm arc cuts ST at two points, which one should be taken as D?