A rectangle is a quadrilateral which is a four sided figure which has opposite sides equal and four equal angles. Each angle is $\frac{360^o}{4}$ = 90° and so every angle of a rectangle is a right angle. Opposite sides of a rectangle are parallel. A rectangle is called as an equiangular quadrilateral. Since in a rectangle we do not have the adjacent sides equal, we have two different dimensions in a rectangle. The length is the distance between the two adjacent vertices which join to make the longest side (generally the horizontal side as shown in the figure) of the rectangle. The width or breadth is the distance between the two adjacent vertices which join to make the shorter side (generally the vertical side as shown in the figure) of the rectangle. A rectangle with vertices or corners as A, B, C and D would be denoted  by $\square$ABCD.
Rectangle Picture
In this figure on the left A, B, C and D are the vertices or corners of $\square$ABCD. AB, BC, CD and AD are the four sides of $\square$ABCD. The line segments AC and BD are the diagonals of $\square$ABCD.

Also, AB = CD, AD = BC and AC = BD.

•  Opposite sides are congruent and parallel.  
•  All angles are 90° and so equiangular.  
•  The two diagonals bisect each other.
•  The two diagonals are congruent.
•  Each diagonal divides the rectangle into two congruent right triangles.
Like for most quadrilaterals, the area of a quadrilateral is the length times its perpendicular height. The area of the rectangle is given by length times its width. The formula is given by
Area of a rectangle = $l \times w$, where $l$ is its length and $w$ is its width.
In Geometry, the Perimeter of any polygon is the total distance around the outside of a 2-dimensional shape. So, the perimeter of a rectangle is twice the sum of its length and width as the opposite sides are equal. The formula is given by
Perimeter of a rectangle=2(l+w), where l is its length and w is its width.The length of diagonal of a rectangle can be found by using the Pythagoras theorem and is given by the formula
d = $\sqrt{l^2 + w^2}$, where $l$ is its length and $w$ is its width.

Solved Examples

Question 1: A rectangle is 5m long and 4 m wide. What is its area?
Solution:
 
Length of the rectangle, l = 5m.
  Width of the rectangle, w = 4m.
  Area of the rectangle = length $\times$ width = 5 $\times$ 4 = 20$m^2$.
 

Question 2: A rectangle is 5m long and 4 m wide. What is its perimeter?
Solution:
 
Length of the rectangle, l = 5m.
  Width of the rectangle, w = 4m.
       Perimeter of the rectangle = 2(l +w) = 2(5+4) = 2$\times$9 = 18 m.
 

Question 3: A rectangle is 4m long and 3 m wide. What is the length of its diagonal?
Solution:
 
Length of the rectangle, l = 4m.
 Width of the rectangle, w = 3m.
 The length of its diagonal, d = $\sqrt{l^2 + w^2}$, where l is its length and w is its width.
                                Diagonal length, d = $\sqrt{4^2 + 3^2}$ = $\sqrt{16 + 9}$ = $\sqrt{25}$ = 5m.
 

Question 4: A rectangle has a length which is 4 less than 5 times the width. If the area of this rectangle is 9 units, find the dimensions of this rectangle.
Solution:
 
Let the width of the rectangle = w.
    Given, the length of the rectangle is 4 less than 5 times its width. Then, l = 5w – 4.
    The area of the rectangle = 9 units.                        …………………(1)
    Also by the definition of area of rectangle, Area of rectangle = l $\times$ w = (5w - 4)$\times$ w = 5w$^2$ – 4w   ……(2)
    Equating (1) and (2),
                        5w$^2$ - 4w = 9
                5w$^2$ - 4w - 9 = 0
     5w$^2$ + 5w - 9w - 9 = 0
5w(w + 1) -9 (w + 1) = 0
           (5w - 9)(w + 1) = 0
Using the zero product rule, we have,  5w – 9 = 0 or w + 1 = 0
                 5w = 9 or w = -1
Discarding the negative values as width is a measurement, we have,

Width of the rectangle, w = $\frac{9}{5}$ = 1.8 units
   
Length of the rectangle, l = 5w – 4 = 5(1.8) - 4 = 9 - 4 = 5 units
 

Question 5: In the diagram as shown below, a rectangle's length is 3x + 2 and its width is 3x – 2. If its area is 32 cm$^2$, what are the rectangle's dimensions and what is its perimeter?
Solving Rectangle Problems
Solution:
 
Length of the rectangle, l = 3x + 2.
  Width of the rectangle, w = 3x - 2.
                   The area of the rectangle = 32 $cm^2$.                        ………………….(1)
 Also by the definition of area of rectangle, Area of rectangle = l $\times$ w = (3x + 2)$\times$(3x - 2)  …………(2)
Equating (1) and (2),
          (3x + 2)$\times$(3x - 2) = 32  
                9x$^2$ – 4 = 32
                        9x$^2$ = 36

                           x$^2$ = $\frac{36}{9}$ = 4

                             x = $\sqrt{4}$ = ±2  
Considering  x = 2 and discarding x = -2 as it gives negative dimensions, we have
Length of the rectangle, l = 3x + 2 = 3(2) + 2 = 6 + 2 = 8 cms.
Width of the rectangle, w = 3x - 2 = 3(2) – 2 = 6 - 2 = 4 cms.
Perimeter of the rectangle = 2(l + w) = 2(8 + 4) = 2(12) = 24 cms.
 

Question 6: A rectangle has a width of 12 units  and a diagonal of 13 units. Find the area and perimeter of this rectangle.
Solution:
 
Let the length of the rectangle, l = l units.
 The width of the rectangle = 12 units.
 The length of its diagonal, d = $\sqrt{l^2 + w^2}$, where l is its length and w is its width.
                   Diagonal length, d = $\sqrt{l^2 + 12^2}$ = $\sqrt{l^2 + 144}$.  
But given diagonal, d = 13 units, so we have,
13 = $\sqrt{l^2 + 144}$  

13$^2$  = l$^2$ + 144
l$^2$ = 169 – 144 = 25
Thus the length of the rectangle, l = $\sqrt{25}$  = 5 units.
Area of the rectangle = length $\times$ width = 5 $\times$ 12 = 60 units$^2$.
Perimeter of the rectangle = 2(l + w) = 2(5 + 12) = 2 $\times$ 17 = 34 units.