Question

Construct a parallelogram ABCD in which AB = 5.5 cm, BC = 4.2 cm and AC = 7 cm.

Answer 1

Given: Parallelogram ABCD with AB = 5.5 cm, BC = 4.2 cm and diagonal AC = 7 cm.

Step-wise calculation (Construction + Numerical check):

1. Check feasibility

For triangle ABC we need AB + BC > AC etc.

Here 5.5 + 4.2 = 9.7 > 7.

So, a triangle with sides 5.5, 4.2 and 7 cm exists.

2. Construction (compass and straightedge)

Draw segment AB = 5.5 cm.

With centre A and radius 7 cm, draw an arc.

With centre B and radius 4.2 cm, draw an arc. Let the intersection (above AB be C). This constructs triangle ABC from AC = 7 and BC = 4.2. This is the standard method for “two adjacent sides and one diagonal given.”

Through A draw a line parallel to BC use a ruler and construction of parallel through a point.

Through C draw a line parallel to AB.

Let those two parallel lines meet at D. Join AD and CD. ABCD is the required parallelogram (AB || CD and BC || AD).

3. Optional coordinate calculation (Verification)

Place A at (0, 0) and B at (5.5, 0).

Let C = (x, y).

Solve x² + y² = 7² 

= 49

(x – 5.5)² + y²

= 4.2²

= 17.64 

Subtracting gives

11x – 30.25 = 31.36 

⇒ `x = 61.61/11`

= 5.601 cm 

Then y² = 49 – x²

= 49 – 31.3712

= 17.6288 

⇒ y = 4.198 cm.

So, C = (5.601, 4.198).

Then D is obtained as A + (C − B)

= (0, 0) + (5.601 – 5.5, 4.198) 

= (0.101, 4.198)

Distances: AD = distance A ↔ D

= `sqrt(0.101^2 + 4.198^2)`

= 4.200

= BC

CD = distance C ↔ D 

= 5.5 

= AB (exact by construction) AC = 7 by construction.

These confirm the opposite sides are equal and AC is the given diagonal.

The parallelogram ABCD constructed by the steps above satisfies AB = 5.5 cm, BC = 4.2 cm and AC = 7 cm.

The practical construction is draw AB = 5.5 cm, locate C by intersecting arcs of radii 7 cm (From A) and 4.2 cm (From B), then draw parallels through A and C to get D.

The method is the standard construction for two adjacent sides and a diagonal.