Question

Construct a rhombus ABCD in which AB = 5.2 cm and height is 4.5 cm.

Answer 1

Given: AB = 5.20 cm, height distance between a pair of opposite sides = 4.50 cm.

Step-wise calculation:

1. For a rhombus with side a and height h, h = a·sinθ, where θ is an interior acute angle.

`sin θ = 4.50/5.20`

= 0.8653846

θ = arcsin (0.8653846) ≈ 59.9°

The other interior angle

= 180° – 59.9°

= 120.1°

2. The horizontal projection distance between the foot of the altitude and the adjacent vertex

= `sqrt(a^2 - h^2)`

a² = 5.20²

= 27.04

h² = 4.50²

= 20.25

Projection = `sqrt(27.04 - 20.25)`

= `sqrt(6.79)`

= 2.605 cm

This value is useful when placing the top side parallel to AB at a distance of 4.5 cm.

Construction steps (ruler + compass):

1. Draw segment AB = 5.20 cm.
2. At B, construct a perpendicular to AB. On this perpendicular mark point P with BP = 4.50 cm.
3. Through P draw line l parallel to AB so l is at distance 4.50 cm from AB.
4. With centre A and radius 5.20 cm, draw an arc to cut line l; label the intersection D. There will be two symmetric choices; pick one.
5. With centre B and radius 5.20 cm, draw an arc to cut line l; label the intersection C that lies on the same side of AB as D.
6. Join AD, DC and CB. ABCD is the required rhombus with two mirror possibilities depending on the choice of intersections.

ABCD, constructed as above, is a rhombus with a side length of 5.20 cm and a height of 4.50 cm. The acute interior angle ≈ 59.9° (obtuse ≈ 120.1°) and the horizontal projection between adjacent vertices ≈ 2.605 cm.