Question

Construct a parallelogram ABCD in which AC = 7 cm, BD = 8 cm and the angle between these diagonals is 60°.

Answer 1

Given: Parallelogram ABCD with diagonals AC = 7 cm, BD = 8 cm and angle between the diagonals = 60°.

Step-wise calculation:

1. Let O be the intersection of the diagonals.

Since the diagonals of a parallelogram bisect each other.

AO = OC

= `7/2`

= 3.5 cm

And BO = OD

= `8/2`

= 4 cm   ...(Recall: Diagonals bisect each other.)

2. Find AB (triangle AOB):

In ΔAOB:

AO = 3.5

BO = 4

∠AOB = 60°

By the law of cosines:

AB² = AO² + BO² – 2 × AO × BO × cos 60° 

= `3.5^2 + 4^2 - 2 xx 3.5 xx 4 xx 1/2` 

= 12.25 + 16 – 14

= 14.25

So `AB = sqrt(14.25)`

= `sqrt(57/4)`

= `sqrt(57)/2` 

= 3.775 cm

3. Find AD (triangle AOD):

Note: ∠AOD is the supplement of ∠AOB.

So, ∠AOD = 180° – 60°

= 120°

In ΔAOD:

AO = 3.5

OD = 4

∠AOD = 120°

By the law of cosines:

AD² = AO² + OD² – 2 × AO × OD × cos 120° 

= `3.5^2 + 4^2 − 2 xx 3.5 xx 4 xx (-1/2)` 

= 12.25 + 16 + 14

= 42.25

So `AD = sqrt(42.25)`

= 6.5 cm

Construction steps (compass and straightedge):

1. Draw segment AC = 7 cm and mark its midpoint O so AO = 3.5 cm and CO = 3.5 cm.

2. At O, construct a line through O making 60° with AC. This gives the direction of BD.

3. On that line mark B at distance OB = 4 cm from O on one side and D at distance OD = 4 cm on the opposite side of O.

4. Join A to B, B to C, C to D and D to A. The figure ABCD is the required parallelogram. Diagonals bisecting at O ensure that opposite sides are parallel and equal.

Side lengths found:

`AB = (sqrt57)/2`

= 3.775 cm

And AD = 6.5 cm

The construction above produces a parallelogram ABCD with AC = 7 cm, BD = 8 cm and the diagonals meeting at 60°.