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Perimeter and Area of a Rectangle

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

Perimeter and Area of a Rectangle

Perimeter of a rectangle (P) = 2(length + breadth) = 2(l + b)

Area of a rectangle (A) = length × breadth = l × b

Diagonal of a rectangle (d) = (length)2+(breadth)2

                                       = l2+b2

Length of a rectangle (l) = areabreadth=Ab

Breadth of a rectangle (b) = arealength=Al


Some geometrical properties of a rectangle:

Geometrical Properties of a Rectangle

In the rectangle PQRS,

PQ = SR, PS = QR, QS = PR;

OP = OR = OQ = OD;

∠PSC = ∠QRS = ∠RQP = ∠qps = 90°.

Also, PR2 = PS2 + SR2; [by Pythagoras’ theorem)

and QS2 = QR2 + SR2; [by Pythagoras’ theorem)

Area of the ∆PQR = Area of the ∆PSQ = Area of the ∆QRS = Are of the ∆PSR

                          = 12 (Area of the rectangle PQRS).


Solved Examples on Perimeter and Area of a Rectangle:

1. The area of a rectangle whose sides are in the ratio 4:3 is 96 cm2. What is the perimeter of the square whose each side is equal in length to the diagonal of the rectangle?

Solution:

As the sides og the rectangle are in the ratio 4:3, let the sides be 4x and 3x respectively.

Then, the area of the rectangle = 4x ∙ 3x = 96 cm2

Therefore, 12x2 = 96 cm2

or, x2 = 8 cm2

Therefore, x = 2√2 cm

Now, the length of a diagonal of the square = (4x)2+(3x)2

                                                                = 25x2

                                                                = 5x

Therefore, the perimeter of the square = 4 × side

                                                        = 4 × 5x

                                                        = 20x

                                                        = 20 × 2√2 cm

                                                        = 40√2 cm

                                                        = 40 × 1.41 cm

                                                        = 56.4 cm






9th Grade Math

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