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Introduction to Diagonals
We know that a polygon is a closed shape formed by joining the adjacent vertices.
For example, a square has \(4\) sides, a pentagon has \(5\) sides, and a hexagon has \(6\) sides, and so on.
Here, we can see that every two adjacent vertices are joined by a line segment.
These line segments are known as the "sides" of the polygon.
But, can we join the non-adjacent vertices by line segments?
If yes, what are such line segments called?
Yes, we can join non-adjacent vertices by line segments.
Such line segments are called diagonals.
Let us study more about diagonals.
What is a Diagonal (with illustration)?
"A diagonal of a polygon is a line segment that is obtained by joining any two non-adjacent vertices."
Example 1:
A quadrilateral has two pairs of non-adjacent vertices.
Hence, it has two diagonals.
Let us see some quadrilaterals here.
In each of these quadrilaterals, \(\overline{AC}\) and \(\overline{BD}\) are diagonals.
Example 2:
A hexagon has \(9\) diagonals.
Here, all the \(9\) red-colored lines inside the hexagon are the diagonals.
Shape of a Diagonal
The shape of a diagonal is a "line segment" and it follows from the definition of a diagonal.
We can see the diagonals of different types of polygons using the following illustration.
- How many diagonals does a circle have?
Tip: Think about the number of vertices of a circle.
Diagonals of a Square
In a square, there are two pairs of non-adjacent vertices.
By joining the vertices of each such pair, we get two diagonals.
In this square \(ABCD\), the pairs of non-adjacent vertices are \(A\) and \(C\), and \(B\) and \(D\).
Hence, \(\overline{AC}\) and \(\overline{BD}\) are diagonals.
The number of diagonals of a square = 2 |
Diagonals of a Triangle
Let's look at the following triangle \(ABC\).
Do you see any two vertices of this triangle that are NOT adjacent?
You are right!
No vertices in a triangle are non-adjacent.
It means that there are no line segments which can form diagonals.
The number of diagonals of a triangle = 0 |
Number of Diagonals: Formula (with illustration)
Can you think of the number of diagonals of a heptagon, octagon, nonagon, etc?
We can manually see the number of diagonals of different polygons using the following illustration.
As you can see, it is difficult to draw the diagonals in a polygon with a large number of sides.
As hard as it is to draw them, counting them is equally hard. Don't you agree?
To make this simple, we will use a formula to find the number of diagonals.
The number of diagonals of a polygon with \(\mathbf{n}\) sides is:
\(\mathbf{\dfrac{n(n-3)}{2}}\) |
We can use this formula to find the number of diagonals of any polygon without actually drawing them.
Example:
Find the number of diagonals of a decagon.
Solution:
The number of sides of a decagon is, \(n=10\)
The number of diagonals of a decagon is calculated using:
\[ \begin{aligned} \dfrac{n(n-3)}{2} &= \dfrac{10(10-3)}{2}\\[0.2cm]&=\dfrac{10(7)}{2}\\[0.2cm]&= \dfrac{70}{2}\\[0.2cm]&=35 \end{aligned} \]
The number of diagonals of a decagon= 35 |
The following table shows the number of diagonals of some polygons which is calculated using this formula.
Shape | Number of sides, \(\mathbf{n}\) | Number of Diagonals Formula |
---|---|---|
Triangle | \(3\) |
\(\dfrac{3(3-3)}{2}=\color{darkblue}{\boxed{\mathbf{0}}}\) |
Quadrilateral | \(4\) |
\(\dfrac{4(4-3)}{2}=\color{darkblue}{\boxed{\mathbf{2}}}\) |
Pentagon | \(5\) |
\(\dfrac{5(5-3)}{2}=\color{darkblue}{\boxed{\mathbf{5}}}\) |
Hexagon | \(6\) |
\(\dfrac{6(6-3)}{2}=\color{darkblue}{\boxed{\mathbf{9}}}\) |
Heptagon | \(7\) |
\(\dfrac{7(7-3)}{2}=\color{darkblue}{\boxed{\mathbf{14}}}\) |
Octagon | \(8\) |
\(\dfrac{8(8-3)}{2}=\color{darkblue}{\boxed{\mathbf{20}}}\) |
Nonagon | \(9\) |
\(\dfrac{9(9-3)}{2}=\color{darkblue}{\boxed{\mathbf{27}}}\) |
Decagon | \(10\) |
\(\dfrac{10(10-3)}{2}=\color{darkblue}{\boxed{\mathbf{35}}}\) |
Hendecagon | \(11\) |
\(\dfrac{11(11-3)}{2}=\color{darkblue}{\boxed{\mathbf{44}}}\) |
Dodecagon | \(12\) |
\(\dfrac{12(12-3)}{2}=\color{darkblue}{\boxed{\mathbf{54}}}\) |
Length of a Diagonal
We have learnt how to find the number of diagonals.
Do you know how to find the length of a diagonal?
Though there are no formulas available to find the length of diagonals for all types of polygons, we have formulas for some specific shapes.
Length of a Diagonal of a Square
In a square, the length of both the diagonals are the same.
The length of a diagonal (\(d\)) of a square of side length \(x\) is calculated by the Pythagoras theorem.
Using Pythagoras theorem,
\[ \begin{aligned} d &= \sqrt{x^2+x^2}\\[0.2cm]&= \sqrt{2x^2}\\[0.2cm]&= \sqrt{2} \cdot \sqrt{x^2}\\[0.2cm] &=\sqrt{2}\,\,x \end{aligned} \]
Length of a diagonal of a square = \(\mathbf{\sqrt{2}\,\,x }\) |
Length of a Diagonal of a Rectangle
Similar to a square, the length of both the diagonals in a rectangle are the same.
The length of a diagonal (\(d\)) of a rectangle whose length is \(l\) and whose breadth is \(b\) is calculated by the Pythagoras theorem.
Using Pythagoras theorem,
\[ \begin{aligned} d &= \sqrt{l^2+b^2} \end{aligned} \]
Length of a diagonal of a rectangle = \(\mathbf{\sqrt{l^2+b^2}}\) |
Length of a Diagonal of a Cube
Consider a cube of length \(x\)
A cube has \(6\) faces.
Each face of a cube is a square.
Thus each face has two diagonals.
Hence, the length of each such diagonal is the same as the length of a diagonal of a square.
Length of each face diagonal of cube = \(\mathbf{\sqrt{2}\,\,x }\) |
Apart from the diagonals on the faces, there are \(4\) other diagonals (main diagonals or body diagonals) that pass through the center of the square.
Let us assume that the length of each such diagonal is \(d\).
The formula for the length of the diagonal of a cube is derived in the same way as we derive the length of the diagonal of a square.
Length of a diagonal of a cube = \(\mathbf{\sqrt{3}\,\,x }\) |
Length of a Diagonal of a Cuboid (Rectangular Prism)
Consider a cuboid of length \(l\), width \(w\) and height \(h\).
Let us assume that its main diagonal (or body diagonal) that passes through the center of the cuboid is \(d\).
The number of diagonals of a polygon with \( \mathbf{n}\) sides is
\(\mathbf{\dfrac{n(n-3)}{2}}\)
Length of a diagonal of a cuboid = \(\mathbf{\sqrt{l^2+w^2+h^2} }\) |
- The number of diagonals of a polygon with \( \mathbf{n}\) sides is \(\mathbf{\dfrac{n(n-3)}{2}}\)
- The length of diagonal of a square of side \(\mathbf{x}\) is \(\mathbf{\sqrt{2}\,\,x }\)
- The length of diagonal of a rectangle of sides \(\mathbf{l}\) and \(\mathbf{b}\) is \(\mathbf{\sqrt{l^2+b^2}\ }\)
- The length of diagonal of a cube of side \(\mathbf{x}\) is \(\mathbf{\sqrt{3}\,\,x }\)
- The length of diagonal of a cuboid of sides \(\mathbf{l, w}\) and \(\mathbf{h}\) is \(\mathbf{\sqrt{l^2+w^2+h^2} }\)
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Solved Examples
Example 1 |
If a polygon has \(90\) diagonals, how many sides does it have?
Solution:
Let us assume that the number of sides of the given polygon is \(n\).
The number of diagonals = \(90\)
\[ \begin{aligned} \dfrac{n(n-3)}{2}&= 90 \\[0.2cm] n(n-3) &=180\\[0.2cm] n^2-3n-180&=0\\[0.2cm] (n-15)(n+12)&=0\\[0.2cm] n=15; & n=-12 \end{aligned} \]
Since \(n\) cannot be negative, the value of \(n\) is \(15\).
\(\therefore\) Sides of the given polygon = \(15\) |
Example 2 |
A cube has a wall area of \(121 \text{ cm}^2\).
What is the length of the main diagonal of the cube?
Solution:
Let us assume that \(x\) cm is the side of the given cube.
Its wall is a square.
Thus, the area of the wall = \(x^2 \text{ cm}^2\)
Since the area of the wall is \(121 \text{ cm}^2\),
\[ \begin{aligned} x^2 &=121\\[0.2cm] x&=11 \end{aligned}\]
Let us assume that the length of the diagonal of the cube is \(d\) cm.
This implies,
\[ d = \sqrt{3} \,\, x = \sqrt{3} \,\, (11) = 11 \sqrt{3} \text { cm} \]
\(\therefore\) Length of the main diagonal = \(\ 11 \sqrt{3} \text { cm}\) |
Example 3 |
The size of a television screen is nothing but the length of its diagonal.
If the dimensions of a television are \(16\) inches and \(40\) inches, find the size of the television.
Solution:
We know that a television is in the shape of a rectangle.
Length \(l = 40\) inches
Breadth \(b =16\) inches
The size of the television is the length of its diagonal length and we will assume it to be \(d\).
We know that,
\[ \begin{aligned} d&= \sqrt{l^2+b^2}\\[0.2cm] &= \sqrt{40^2+16^2}\\[0.2cm] &= \sqrt{1856}\\[0.2cm] & \approx 43.08 \text{ inches} \end{aligned} \]
\(\therefore\) Size of the television = \(43.08 \text { inches}\) |
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Practice Questions
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Select/Type your answer and click the "Check Answer" button to see the result.
Maths Olympiad Sample Papers
IMO (International Maths Olympiad) is a competitive exam in Mathematics conducted annually for school students. It encourages children to develop their math solving skills from a competition perspective.
You can download the FREE grade-wise sample papers from below:
- IMO Sample Paper Class 1
- IMO Sample Paper Class 2
- IMO Sample Paper Class 3
- IMO Sample Paper Class 4
- IMO Sample Paper Class 5
- IMO Sample Paper Class 6
- IMO Sample Paper Class 7
- IMO Sample Paper Class 8
- IMO Sample Paper Class 9
- IMO Sample Paper Class 10
To know more about the Maths Olympiad you can click here
Frequently Asked Questions (FAQs)
1. What is the angle of a diagonal in a square?
Each diagonal in a square divides the angle at each vertex into two equal parts.
Hence, the angle between any side and any diagonal is 45 degrees.
2. What do you mean by diagonals?
A diagonal of a polygon is a line segment that is obtained by joining any two non-adjacent vertices.
For more information, you can refer to "What is a Diagonal?" section of this page.
3. What does a diagonal look like?
The shape of a diagonal is a line segment.
4. How many sides does a diagonal have?
A diagonal itself is a line segment.
Thus, a diagonal cannot have any sides.