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

9th Grade Math

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