What is a Rectangle?
In very simple words, a rectangle is a four-sided, two-dimensional plane figure, and is a four-sided polygon. A rectangle is also a parallelogram with all angles of equal measure. It’s one of the quadrilaterals in which all four angles are 90 degrees or right angles, the other being a square.
Properties of a Rectangle
Relating to the sides and diagonals of a rectangle:
The opposite sides of a rectangle are parallel and equal in length to each other.
The diagonals of a rectangle are equal in length and they bisect each other.
The diagonal of a rectangle also makes two congruent halves with the sides.
If the sides of a rectangle are l and w, then the length of each diagonal is:
\(d~=~\sqrt{l^2~+~w^2}\)
Diagonal of Rectangle, \(d~=~\sqrt{(l^2~+~w^2)}\)
The diameter of a rectangle’s circumcircle is diagonal.
Relating to the Interior Angles of a Rectangle
The interior angles formed by the intersection of two sides of a rectangle is 90 degrees.
The sum of all interior angles of a rectangle is equal to 360 degrees and the sum of all the exterior angles of a rectangle is also 360 degrees.
Relating to the Perimeter and Area of a Rectangle
The perimeter of a rectangle with length l and width w is:
\(P~=~(2~\times~l)~+~(2~\times~w)\)
Or,
\(P~=~2\times~(l~+~w)\)
The area of a rectangle with length l and width w is
\(A~=~l\times~w\)
Relationship with other Shapes
The relationship between multiple types of quadrilaterals or four sided polygons is demonstrated in the following diagram. We can see that from the square the characteristics of these polygons keep altering until none of the sides are equal or parallel. But the conclusion is that rectangles form a part of a family of four-sided polygons whose interior angles add up to 360 degrees.
Rapid Recall
Perimeter of a rectangle:
\(P~=~(2~\times~l)~+~(2~\times~w)\)
Area of a rectangle:
\(A~=~l~\times~w\)
Where l is length of the rectangle and, w is width of the rectangle.
Pertaining to the diagonals:
AO = OD
BO = OC
AD = BC
Or, in other words,
\(d_1=d_2\)
Also, the length of both the diagonals can be found using the formula:
\(d~=~\sqrt{l^2~+~w^2}\)
Solved Examples
Example 1:
Find the perimeter of the following rectangle
Solution:
Perimeter of a rectangle:
\(P~=~(2~\times~l)~+~(2~\times~w)\)
\(P~=~(2~\times~5)~+~(2~\times~3)\) Substitute the values
\(P~=~(10)~+~(6)\)
\(P~=~16\)
Hence, the perimeter is 16 cm.
Example 2:
Find the area of the following rectangle.
Area of a rectangle = l × w
= \(~\frac{1}{3}~\times~\frac{2}{16}\) Substitute the values
= \(~\frac{2~\div~2}{48~\div~2}\) Divide the numerator and denominator by 2
= \(~\frac{1}{24}~in^2\)
Hence the area of the rectangle is \(\frac{1}{24}\) squared inches.
Example 3:
A playground needs to be fenced with a length of \(\frac{4}{5}\) km and breadth of \(\frac{5}{7}\) km. Find the length of material required to fence the playground. After the playground is fenced it also needs to be layered with sand. Find the area of land that needs to be covered with sand.
Solution:
To calculate length of the material required to fence the playground we need to find the perimeter :
\(P~=~(2~\times~l)~+~(2~\times~w)\)
\(~=~\left ( 2~\times~\frac{4}{5} \right )~+~\left ( 2~\times~\frac{5}{7} \right )\) Substitute the values
\(~=~\left ( \frac{8}{5} \right )~+~\left (\frac{10}{7} \right )\) Add
\(~=~\left ( \frac{56+50}{35} \right )\) As the LCM of 5 and 7 is 35
\(~=~ \frac{106}{35} \) km or 3.02 km
Therefore, the material required to fence the playground is 3.02 km.
Now, to find the area of land that needs to be covered with sand, we must find the area of the playground:
\(A~=~l~\times~w\)
\(~=~\frac{4}{5}~\times~\frac{5}{7}\)
\(~=~\frac{20~\div~5}{35~\div~5}\) Divide the numerator and denominator by 5.
\(~=~\frac{4}{7}~km^2\)
Therefore we require \(\frac{4}{7}~km^2\) of sand to cover the playground.
Example 4:
If ABCD is a rectangle then, find the unknowns in the figures shown:
1) Find BD and CD
2) Find ∠BDO
3) Find BC
4) Find the radius of the circle.
Solution:
1)
If AB = 4 cm then, CD = 4 cm and,
if AC = 3 cm then, BD = 3 cm.
This is because the opposite sides of a rectangle are equal in length to each other.
2) 
All the angles of a rectangle are 90 degrees, therefore,
∠BDC = 90°
Now, clearly angles BDC and BDO are supplementary as they form a linear pair, this suggests that,
∠BDC + ∠BDO = 180°
90° + ∠BDO = 180° Substitute the values
90° + ∠BDO – 90° = 180° – 90° Subtract 90° from both sides
∠BDO = 90°
Hence, ∠BDO is 90 degrees.
3) 
To find BC(d), length(l) = 4 cm and width(w) = 3 cm,
therefore,
\(d~=~\sqrt{l^2~+~w^2}\)
\(d~=~\sqrt{4^2~+~3^2}\) Substitute the values
\(d~=~\sqrt{16~+~9}\) As, \(3^2~=~9\) and \(4^2~=~16\)
\(d~=~\sqrt{25}\) or \(d~=~5\)
Hence, \(BC~(d)~=~5\) cm.
4) 
The diameter of a rectangle’s circumcircle is diagonal.
Clearly, the diameter of the circle shown above is BC, which is also the diagonal of the rectangle ABCD.
In part 3 we found \(BC~(d)~=~5\) cm, using the formula:
\(d~=~\sqrt{l^2~+~w^2}\)
Also, the radius BO is half the diameter BD, therefore,
\(BO~=~\frac{BC}{2}\)
\(BO~=~\frac{5}{2}\) or \(BO~=~\frac{BC}{2}\) cm
Hence, the radius of the circle \(BO~=~2.5\) cm.
All the angles formed by the intersection of two of its sides add up to 360 degrees.
The diagonals of a rectangle are always equal in length.
The length property of a rectangle specifies that the rectangle’s opposite sides are equal in length.
If we draw a rectangle,
- The length of the diagonals is the same.
- Diagonals bisect each other, or cut each other into halves.
- All the four halves of the diagonals are equal to each other.
A. Perimeter of a rectangle:
\(P~=~(2~\times~l)~+~(2~\times~w)\)
B. Area of a rectangle:
\(A~=~l~\times~w\)
C. Pertaining to the diagonals:
Or, in other words,
\(d_1=d_2\)
\(d~=~\sqrt{l^2~+~w^2}\)
Where, l is length of the rectangle and, w is width of the rectangle.