Question

Construct a rectangle ABCD in which AC = 5 cm and acute angle between its diagonals is 45°.

Answer 1

Given:

AC = 5 cm.

Acute angle between the diagonals ∠AOB where O is intersection = 45°.

Step-wise calculation and construction:

1. Compute half-diagonals:

AO = CO

= `(AC)/2`

= `5/2`

= 2.5 cm

In a rectangle the diagonals are equal.

So, BD = AC = 5 cm and BO = DO = 2.5 cm.

2. Construction steps (geometric):

a. Draw segment AC = 5 cm.

b. Construct the perpendicular bisector of AC or otherwise locate the midpoint and mark the midpoint O so that AO = OC = 2.5 cm.

c. At O construct a straight line l making an angle of 45° with AC this line is the line of the other diagonal BD because the angle between diagonals is 45°.

d. On line l mark points B and D on opposite sides of O so that OB = OD = 2.5 cm. Use compass with radius 2.5 cm centered at O.

e. Join A to B, B to C, C to D and D to A to form quadrilateral ABCD.

3. Justification:

The two diagonals AC and BD meet at O and bisect each other we placed B and D so BO = DO and AO = OC, so ABCD is a parallelogram.

Since AC = BD, both 5 cm. The parallelogram has equal diagonals; therefore, it is a rectangle.

The construction follows the standard method for constructing a parallelogram from one diagonal and the angle between diagonals, adapted here with equal diagonals to get a rectangle.

Optional: side lengths (calculation)

Let the rectangle sides be L (longer) and W (shorter).

From diagonal d = 5 and angle φ = 45° between diagonals,

`cos "φ" = (L^2 - W^2)/(L^2 + W^2)` 

⇒ With φ = 45° gives `(L^2 - W^2)/(L^2 + W^2) = sqrt(2/2)`. 

Solving yields `L/W = 1 + sqrt(2)`

And `W = (5/2) xx sqrt(2) - sqrt(2)`

= 1.9129 cm

`L = (5/2) xx sqrt(2) + sqrt(2)`

= 4.6179 cm

The rectangle ABCD constructed by the steps above has diagonal AC = 5 cm and the acute angle between its diagonals 45°. The side lengths are AB = 4.618 cm and BC = 1.913 cm exact forms given above.