Perimeter and Area of Rhombus

Here we will discuss about the perimeter and area of a rhombus and some of its geometrical properties.

$Perimeter and Area of Rhombus$

Perimeter of a rhombus (P) = 4 × side = 4a

Area of a rhombus (A) = $\frac{1}{2}$ (Product of the diagonals)

= $\frac{1}{2}$ × d$_{1}$ × d$_{2}$

Some geometrical properties of a rhombus:

In the rhombus PQRS,

PR ⊥ QS, OP = OR, OQ = OS,

PQ$^{2}$ = OP$^{2}$ + OQ$^{2}$

QR$^{2}$ = OQ$^{2}$ + OR$^{2}$

RS$^{2}$ = OR$^{2}$ + OS$^{2}$

SP$^{2}$ = OS$^{2}$ + OP$^{2}$

Solved Example Problem on Perimeter and Area of Rhombus:

1. The diagonals of a rhombus measure 8 cm and 6 cm. Find the area and the perimeter of the rhombus.

$Problem on Perimeter and Area of Rhombus$

Solution:

In the rhombus PQRS, QS = 8 cm and PR = 6 cm.

Then, area of the rhombus = $\frac{1}{2}$ × d$_{1}$ × d$_{2}$

= $\frac{1}{2}$ × QS × PR

= $\frac{1}{2}$ × 8 × 6 cm$^{2}$

= 24 cm$^{2}$

Now, OP = $\frac{1}{2}$ PR = $\frac{1}{2}$ × 6 cm = 3 cm and,

OQ = $\frac{1}{2}$ QS = $\frac{1}{2}$ × 8 cm = 4 cm.

Also, ∠POQ = 90°.

So by Pythagoras’ theorem, PQ$^{2}$ = OP$^{2}$ + OQ$^{2}$

= (3$^{2}$ + 4$^{2}$) cm$^{2}$

= (9 + 16) cm$^{2}$

= 25 cm$^{2}$

Therefore, PQ = 5 cm

Therefore, perimeter of a rhombus (P) = 4 × side

= 4 × 5 cm

= 20 cm

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