Question

Construct a trapezium ABCD in which AB = 3.5 cm, BC = 6 cm, CD = 3.2 cm, AD = 4.5 cm and AD || BC.

Answer 1

Given: AB = 3.5 cm, BC = 6 cm, CD = 3.2 cm, AD = 4.5 cm, with AD || BC.

Step-wise calculation and construction:

1. Draw the base BC = 6 cm.

2. On BC, from B measure and mark point P so that BP = AD = 4.5 cm.

Compute PC = BC – BP

= 6 – 4.5

= 1.5 cm

3. We will construct triangle DPC (SSS), because in the parallelogram construction DP = AB, PC is known 1.5 cm and DC = 3.2 cm.

With centre C and radius CD = 3.2 cm, draw an arc.

With centre P and radius DP = AB = 3.5 cm, draw an arc. The intersection of these two arcs (above or below BC) gives point D.

Check: 3.5, 3.2 and 1.5 satisfy triangle inequalities, so the triangle exists.

4. Locate A by the intersection of two circles:

With centre B and radius AB = 3.5 cm, draw an arc.

With centre D and radius AD = 4.5 cm, draw an arc. Their intersection gives point A.

5. Join A to B, B to C, C to D and D to A to complete the trapezium ABCD.

6. Reason for AD || BC: because we set BP = AD and DP = AB, the quadrilateral ABPD is a parallelogram, so AD is parallel to BP, which lies on BC; hence AD || BC. This is the standard construction method for a trapezium when four sides are given and one pair is parallel.