Ungrouped Data: When a data collection is vast, a frequency distribution table is frequently used to arrange the data. A frequency distribution table provides the...
Ungrouped Data: Know Formulas, Definition, & Applications
December 11, 2024Quadrilateral Formulas: In Euclidean geometry, a quadrilateral is a four-sided polygon with the sum of interior angles equal to 360°. The word quadrilateral is derived from two Latin words, ‘quadri’ meaning four and ‘latus’ side. When trying to distinguish quadrilaterals from other polygons, it is important to identify their properties. A regular quadrilateral is one with equal sides.
Quadrilaterals can be classified as squares, rectangles, parallelograms, rhombuses, kites and trapezoids. Different formulas can be used to calculate their area. In this article, we will discuss how to calculate the quadrilateral area, perimeter, and cyclic quadrilateral.
Let \(P, Q, R\) and \(S\) be four points in a plane such that no three are collinear and the line segments \(PQ, QR, RS\) and \(SP\) do not intersect except at their endpoints. This figure made up of the four-line segments is called the quadrilateral with vertices \(P, Q, R\) and \(S.\) Below are the images of the quadrilateral.
In a quadrilateral \(PQRS,\) the four-line segments \(PQ, QR, RS\) and \(SP\) are called its sides.
Two sides of a quadrilateral are known as its adjacent sides if they have a common end-point. In the above figure \(PQ, QR; QR, RS; RS, SP\) and \(SP, PQ\) are four adjacent sides of the quadrilateral \(PQRS.\)
Learn About Properties of Quadrilaterals
Two sides of a quadrilateral are called opposite sides if they do not have a common end-point. In the above figures \(PQ, RS\) and \(PS, QR\) are two opposite sides of the quadrilateral \(PQRS.\)
In the quadrilateral \(PQRS,\) the line segments \(PR\) and \(QS\) are called its diagonals.
In the quadrilateral \(PQRS,\) the angles, \(\angle SPQ,{\rm{ }}\angle PQR,{\rm{ }}\angle QRS\) and \(\angle RSP\) are called its angles. These angles are denoted by \(\angle P,{\rm{ }}\angle Q,{\rm{ }}\angle R\) and \(\angle S\) respectively.
Two angles of a quadrilateral are called adjacent angles if they have a common side as an arm. In the above figures \(\angle P,{\rm{ }}\angle Q;{\rm{ }}\angle Q,{\rm{ }}\angle R;{\rm{ }}\angle R,{\rm{ }}\angle S\) and \(\angle S,{\rm{ }}\angle P\) are four pairs of adjacent angles of the quadrilateral \(PQRS.\)
Two angles of a quadrilateral that are not adjacent are known as opposite angles of the quadrilateral. In the given figure, \(∠P, ∠R\) and \(∠Q, ∠S\) are two opposite angles of the quadrilateral \(PQRS.\)
We can see that a quadrilateral has four sides, four angles and two diagonals.
The perimeter of a polygon is equal to the sum of all its sides. Similarly, the perimeter of a quadrilateral is equal to the sum of the lengths of its four sides.
If \(PQRS\) is a quadrilateral, then the perimeter of the quadrilateral \(=PQ+QR+RS+SP\)
A quadrilateral area is a region within four sides of a particular boundary or figure. The region within these four sides of the quadrilateral will typically be defined as an area of the quadrilateral.
We can derive the area of a quadrilateral by dividing it into two triangles. Also, we can obtain the area of the quadrilateral using coordinate geometry. In the below section, we will derive the formulas of the area of the quadrilaterals.
Consider a quadrilateral \(PQRS\) of different lengths. Let us derive a formula for the area of a quadrilateral.
We can view the quadrilateral as a combination of two triangles, with the diagonal \(d\) being the common base.
Let \({h_1}\) and \({h_2}\) be the heights of triangle \(QRS\) and \(SPQ\) respectively.
We know that, area of the triangle \( = \frac{1}{2} \times {\rm{base \times height}}\)
So, the area of triangle \(QRS = \frac{1}{2} \times d \times {h_1}\)
and, area of triangle \(PQS = \frac{1}{2} \times d \times {h_2}\)
From the figure, the area of quadrilateral \(PQRS = {\rm{area\, of\, }}\Delta {\rm{QRS + area\, of\, PQS}}\)
\( = \left( {\frac{1}{2} \times d \times {h_1}} \right) + \left( {\frac{1}{2} \times d \times {h_2}} \right)\)
\( = \frac{1}{2} \times d\left( {{h_1} + {h_2}} \right)\)
Therefore, the area of the quadrilateral \(PQRS = \frac{1}{2} \times d\left( {{h_1} + {h_2}} \right)\)
In other words, the area of the quadrilateral \({\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times\, diagonal\, \times\, sum\, of\, the\, heights}}\)
If \(PQRS\) is a quadrilateral, then considering the diagonal \(PR,\) we can split the quadrilateral \(PQRS\) into two triangles \(PQR\) and \(PRS.\)
Let \(P\left( {{x_1},{y_1}} \right),Q\left( {{x_2},{y_2}} \right),R\left( {{x_3},{y_3}} \right)\) and \(S\left( {{x_4},{y_4}} \right)\) be the vertices of a quadrilateral \(PQRS.\)
Now, Area of quadrilateral \({\rm{PQRS = Area\, of\, the\, \Delta PQR \,+ Area\, of\, \Delta PRS}}\)
We know that, area of the triangle \(= \frac{1}{2}\left[ {{x_1}\left( {{y_2} – {y_3}} \right) + {x_2}\left( {{y_3} – {y_1}} \right) + {x_3}\left( {{y_1} – {y_2}} \right)} \right]\)
Therefore, the area of quadrilateral
\(PQRS = \frac{1}{2}\left\{ {\left( {{x_1}{y_2} + {x_2}{y_4} + {x_4}{y_1}} \right) – \left( {{x_2}{y_1} + {x_4}{y_2} + {x_1}{y_4}} \right)} \right\} +\)
\(\frac{1}{2}\left\{ {\left( {{x_2}{y_3} + {x_3}{y_4} + {x_4}{y_2}} \right) – \left( {{x_3}{y_2} + {x_4}{y_3} + {x_2}{y_4}} \right)} \right\}\)
\( = \frac{1}{2}\left[ {\left( {{x_1}{y_2} + {x_2}{y_3} + {x_3}{y_4} + {x_4}{y_1}} \right) – \left( {{x_2}{y_1} + {x_3}{y_2} + {x_4}{y_3} + {x_1}{y_4}} \right)} \right]\)
\(= \frac{1}{2}\left[ {\left( {{x_1} – {x_3}} \right)\left( {{y_2} – {y_4}} \right) – \left( {{x_2} – {x_4}} \right)\left( {{y_1} – {y_3}} \right)} \right]{\rm{sq}}.\,{\rm{units}}\)
Therefore, the area of quadrilateral \( = \frac{1}{2}\left[ {\left( {{x_1} – {x_3}} \right)\left( {{y_2} – {y_4}} \right) – \left( {{x_2} – {x_4}} \right)\left( {{y_1} – {y_3}} \right)} \right]\)
The following representation helps us to write the above formula very easily. Take the vertices \(P\left( {{x_1},{y_1}} \right),Q\left( {{x_2},{y_2}} \right),R\left( {{x_3},{y_3}} \right)\) and \(S\left( {{x_4},{y_4}} \right)\) in the counter-clockwise direction and write them column-wise as that of the area of a triangle.
Therefore, the area of quadrilateral \(PQRS = \frac{1}{2}\left[ {\left( {{x_1} – {x_3}} \right)\left( {{y_2} – {y_4}} \right) – \left( {{x_2} – {x_4}} \right)\left( {{y_1} – {y_3}} \right)} \right]\)
A straight line from a vertex to a non-adjacent vertex is called a diagonal of a quadrilateral.
Formula to find the areas for different types of quadrilaterals are given below.
Q.1. If \(A( – 5,7),B( – 4, – 5),C( – 1, – 6)\) and \(D(4,5)\) are the vertices of a quadrilateral, find the area of the quadrilateral \(ABCD.\)
Ans: By joining \(B\) to \(D,\) we will get two triangles \(ABD\) and \(BCD.\)
Now, the area of triangle \(ABD = \frac{1}{2}[ – 5( – 5 – 5) + ( – 4)(5 – 7) + 4(7 + 5)]\)
\( = \frac{1}{2}[50 + 8 + 48] = \frac{{106}}{2} = 53\,{\rm{sq}}.\,{\rm{units}}\)
Also, the area of triangle \(BCD = \frac{1}{2}[ – 4( – 6 – 5) – 1(5 + 5) + 4( – 5 + 6)]\)
\( = \frac{1}{2}[44 – 10 + 4] = 19\,{\rm{sq}}.\,{\rm{units}}\)
Therefore, the area of quadrilateral \(ABCD = 53 + 19 = 72\,{\rm{sq}}.\,{\rm{units}}.\)
Q.2. In the given quadrilateral \(ABCD\) the side \(BD = 15\;{\rm{cm}}\) and the heights of the triangles \(ABD\) and \(BCD\) are \(5\,{\rm{cm}}\) and \(7\,{\rm{cm}}\) respectively. Find the area of the quadrilateral \(ABCD.\)
Ans: Given: Diagonal \(d = 15\;{\rm{cm}}\)
Heights \({h_1} = 5\;{\rm{cm}}\) and \({h_2} = 7\;{\rm{cm}}\)
Sum of heights of the triangle \( = 5 + 7 = 12\;{\rm{cm}}\)
We know that, area of quadrilateral \(ABCD = = \frac{1}{2} \times d\left( {{h_1} + {h_2}} \right)\)
\( = \frac{1}{2} \times 15 \times 12 = 90\;{\rm{c}}{{\rm{m}}^2}\)
Therefore, the area of quadrilateral \(ABCD = 90\;{\rm{c}}{{\rm{m}}^2}\)
Q.3. Find the area of the rectangle whose length is \({\rm{15\,cm}}\) and width is \({\rm{10\,cm}}\)
Ans: We know that area of a rectangle with length \(l,\) and breadth \(b\) is given by \(A = l \times b\)
\( \Rightarrow A = 15 \times 10\;{\rm{c}}{{\rm{m}}^2} = 150\;{\rm{c}}{{\rm{m}}^2}\)
Therefore, the area of the given rectangle is \(150\;{\rm{c}}{{\rm{m}}^2}\)
Q.4. Find the area of a kite whose diagonals are \({\rm{15\,m}}\) and \({\rm{18\,m}}\)
Ans: We know that the Area of the kite with diagonals \({d_1}\) and \({d_2}\) is given by
\(A = \frac{1}{2} \times {d_1} \times {d_2}\)
Given: \({d_1} = 15\;{\rm{m}}\) and \({d_2} = 18\;{\rm{m}}\)
So, \(A = \frac{1}{2} \times 15 \times 18 = 15 \times 9 = 135\;{{\rm{m}}^2}\)
Therefore, the area of the given kite is \(135\;{{\rm{m}}^2}\)
Q.5. Find the area of a trapezoid with bases of \({\rm{10\,cm}}\) and \({\rm{14\,cm}}\), and a height of \({\rm{5\,cm}}\)
Ans: We know that area of the trapezoid is \(A = \frac{1}{2}(a + b)h\)
Given: \(a = 10\;{\rm{cm}},b = 14\;{\rm{cm}}\) and \(h = 5\;{\rm{cm}}\)
So, \(A = \frac{1}{2}(10 + 14) \times 5\)
\(A = 60\;{\rm{c}}{{\rm{m}}^2}\)
Therefore, the area of the given trapezoid is \(60\;{\rm{c}}{{\rm{m}}^2}\)
In this article, we learned about the quadrilaterals, different types of quadrilaterals, and formulas for finding the quadrilateral area by dividing it into two parts. We also learned about the quadrilateral area using coordinate geometry and perimeter of the quadrilaterals. Furthermore, we have learnt the formula to find the area of different types of quadrilaterals.
Let’s look at some of the commonly asked questions about quadrilateral formulas:
Q.1. What is the formula of the perimeter of the quadrilateral?
Ans: If \(PQRS\) is a quadrilateral, then the perimeter of the quadrilateral
\(= PQ + QR + RS + SP\)
Q.2. How to find the area of a quadrilateral formula?
Ans: We can find the area of the quadrilateral by dividing a quadrilateral into two triangles. If \(d\) is the diagonal of a quadrilateral and \({h_1}\) and \({h_2}\) are the heights of a triangle.
Therefore, the area of the quadrilateral \(PQRS = \frac{1}{2} \times d\left( {{h_1} + {h_2}} \right)\)
Q.3. How to find the area of quadrilateral using coordinate geometry?
Ans: If \(P\left( {{x_1},{y_1}} \right),Q\left( {{x_2},{y_2}} \right),R\left( {{x_3},{y_3}} \right)\) and \(S\left( {{x_4},{y_4}} \right)\) are the vertices of a quadrilateral then, the area of quadrilateral \(PQRS = \frac{1}{2}\left[ {\left( {{x_1} – {x_3}} \right)\left( {{y_2} – {y_4}} \right) – \left( {{x_2} – {x_4}} \right)\left( {{y_1} – {y_3}} \right)} \right]\)
Q.4. What is the formula to find the area of a trapezoid?
Ans: Area of the trapezoid with side lengths \(a\) and \(b\) and height \(h\) is given by \(A = \frac{1}{2}(a + b)h\)
Q.5. What is the formula to find the area of the kite?
Ans: Area of the kite with diagonals \({d_1}\) and \({d_2}\) is given by \(A = \frac{1}{2} \times {d_1} \times {d_2}\)
Q6. What is the formula to find area of different types of Quadrilateral?
Ans: Formula to find the areas for different types of quadrilaterals are given below.
We hope you find this detailed article on quadrilateral formulas helpful. If you have any queries or doubts regarding this topic, ask them in the comment section. We will assist you at the earliest. Happy learning!