# Perimeter and Area of Rhombus

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

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.

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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