Question

Construct a rhombus ABCD in which AC = 6.4 cm and height is 4 cm.

Answer 1

Given:

Diagonal AC = 6.4 cm.

Height distance between a pair of opposite sides h = 4 cm.

Step-wise calculation:

1. Let side = s and the other diagonal BD = x.

Area A of the rhombus can be expressed in two ways:

A = side × height

= s × h

`A = (AC xx BD)/2`

= `(6.4 xx x)/2`

= 3.2x 

Therefore, s × 4 = 3.2x 

⇒ `x = (4/3.2)s`

= 1.25s

2. In a rhombus, the diagonals are perpendicular and their half-lengths with the side form a right triangle:

`s^2 = ((AC)/2)^2 + ((BD)/2)^2`

= `3.2^2 + (x/2)^2`

Substitute x = 1.25s:

`s^2 = 3.2^2 + ((1.25s)/2)^2`

= 10.24 + (0.625s)²

= 10.24 + 0.390625s²

Rearranging:

s² – 0.390625s² = 10.24

0.609375s² = 10.24s²

= `10.24/0.609375` 

= `655.36/39` 

= 16.8082s

= `sqrt(16.8082)`

= 4.10 cm

3. Now BD = x = 1.25s

= 1.25 × 4.0998

= 5.1248 cm (≈ 5.12 cm)

So half-diagonal

`OB = (BD)/2`

= 2.5624 cm (≈ 2.56 cm)

Construction steps (practical compass-and-straightedge):

1. Draw segment AC = 6.4 cm. Mark its midpoint O (OA = OC = 3.2 cm).

2. Through O draw the perpendicular to AC.

3. On this perpendicular mark points B and D.

So, that OB = OD

= `(BD)/2`

= 2.562 cm

Place one point on each side of AC.

4. Join A to B, B to C, C to D and D to A. ABCD is the required rhombus.