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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