Area Of Triangle Using Points

Expanse of a Triangle and Quadrilateral: Definition, Formula, Examples

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Area of a triangle and quadrilateral: Polygons are simple, closed figures fabricated of only line segments. While triangles are made of three line segments, quadrilaterals take four. A two-dimensional effigy's area is the amount of space it takes up. In other words, it is the number of unit of measurement squares that cover the surface of a airtight figure.

In geometry, in that location are different formulas to calculate the areas and quadrilaterals. Just we apply dissimilar methods to calculate areas of polygons in coordinate geometry. Permit usa acquire to determine the areas of the triangles and quadrilaterals when the coordinates of their vertices are given.

Elementary Polygons in Cartesian Plane

Polygons are two-dimensional geometric figures with a fixed number of sides. A polygon's sides are made upwards of straight lines that are joined end to terminate. As a outcome, the line segments of a polygon are referred to every bit sides or edges. The betoken at which \(2\) or more than line segments come across or intersect is referred to as the vertex or corners, and an bending is formed in that location.

The simplest polygon is a triangle, made up of only three line segments. A triangle can be formed by joining three vertices. The three vertices will be given as three coordinates in a cartesian plane, equally shown below.

Similarly, a quadrilateral can be made in a cartesian plane by joining the coordinates of the iv required vertices, as shown here.

The perimeter of any closed figure is the length of its boundary. The perimeter of the triangle and quadrilateral can be plant easily by calculation the lengths of their sides. Merely, the area cannot be easily constitute like that. The formulas to calculate areas in and out of the cartesian aeroplane varies as the inputs we have to calculate the area difference.

Geometrical Expanse of a Triangle

The area of the triangle is the space occupied by the triangle in a two-dimensional plane. At that place are different formulas to determine the area of the triangle in geometry.

The area of a triangle provided with its base of operations and corresponding meridian (altitude) tin be calculated using the formula

Area of a triangle \(=\frac{ane}{2} \times\) base of operations \(\times\) altitude

We can also use Heron'southward formula to find the expanse of a triangle. It is given past

Expanse \(=\sqrt{(s(s-a)(s-b)(s-c))}\)

Here,

\(s \rightarrow\) semi perimeter \(=\frac{a+b+c}{2}\)

\(a \rightarrow\) length of side \(a\)

\(b \rightarrow\) length of side \(b\)

\(c \rightarrow\) length of side \(c\)

Now, allow us learn to calculate the surface area of a triangle whose coordinates of the vertices are given.

Area of a Triangle Using Coordinates

One fashion to observe the area when coordinates are given is that we tin can find the lengths of the \(iii\) sides using the distance formula. Then we can utilize Heron's formula. But this is a ho-hum process, mainly if the lengths of the sides are irrational numbers.

Let \(A B C\) be a triangle whose vertices are \(A\left(x_{1}, y_{1}\right), B\left(x_{two}, y_{2}\right)\) and \(C\left(x_{iii}, y_{3}\correct)\). Draw \(A P, B Q\) and \(C R\) as perpendiculars from \(A, B\) and \(C\), respectively, to the \(x\)-axis. Clearly \(A B Q P, A P R C\) and \(B Q R C\) are all trapeziums.

Notation that the no coordinates of whatever ii points is the same. In simpler words, the given vertices are non-collinear.

Hence, from the graph to a higher place, we tin can run into that

Area of \(\triangle A B C=\) area of trapezium \(A B Q P+\) area of trapezium \(A P R C-\) area of trapezium \(B Q R C\)

We know that, surface area of a trapezium \(=\frac{1}{2} \times(\)sum of parallel sides\() \times(\)distance between them\()\)

\(\therefore\) Expanse of \(\triangle A B C=\frac{1}{2}(B Q+A P) \times Q P+\frac{1}{2}(A P+C R) \times P R-\frac{ane}{2}(B Q+C R) \times Q R\)

\(=\frac{1}{ii}\left(y_{two}+y_{ane}\right)\left(x_{1}-x_{two}\right)+\frac{ane}{ii}\left(y_{ane}+y_{three}\correct)\left(x_{3}-x_{i}\right)-\frac{1}{2}\left(y_{2}+y_{iii}\right)\left(x_{3}-x_{two}\right)\)

\(=\frac{1}{2}\left[x_{ane}\left(y_{2}-y_{iii}\right)+x_{two}\left(y_{3}-y_{1}\correct)+x_{3}\left(y_{1}-y_{2}\right)\correct.\)

Thus, the area of \(\triangle A B C\) is given by the formula

\(\frac{i}{2}\left[x_{one}\left(y_{2}-y_{iii}\right)+x_{2}\left(y_{3}-y_{1}\correct)+x_{3}\left(y_{1}-y_{ii}\right)\right]\)

Here,

\(\left(x_{i}, y_{1}\right),\left(x_{2}, y_{2}\right)\) and \(\left(x_{3}, y_{3}\right)\) are the coordinates of the vertices of the triangle.

Area by Determinant Method

The area of triangles in coordinate geometry tin can too be calculated using determinants. If vertices of the triangle are given, then we can find the surface area of the triangle using the determinant formula.

We know that the determinant gives a scalar value that can be either positive or negative. Since we are calculating the area of triangles hither, we too know that areas can never be negative. Hence, nosotros consider the absolute value of the determinant as the area of the triangle.

Let \(A B C\) be a triangle whose vertices are \(A\left(x_{1}, y_{one}\correct), B\left(x_{2}, y_{ii}\right)\) and \(C\left(x_{iii}, y_{3}\right)\) in a cartesian airplane, so the surface area of a triangle in determinant form formula is given past,

Area of triangle \(ABC = \frac{ane}{ii}\left| {\begin{assortment}{{c}} {{x_1}} & {{y_1}} & 1 \\ {{x_2}} & {{y_2}} & 1 \\ {{x_3}} & {{y_3}} & 1 \\ \end{array} } \right|\)

After simplification, this formula of the area of a triangle in determinant form can too exist written every bit:

Area of triangle \(A B C=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{ane}\correct)+x_{3}\left(y_{one}-y_{ii}\correct)\right]\)

Collinearity of Points

The expanse of a triangle tin also be used to notice the collinearity of the points. Three or more points are said to be collinear if all of them lie in a straight line. If 3 points are collinear, and then they cannot be the vertices of a triangle. If so, they do not form a triangle, and there volition be no area. This belongings tin exist used to test collinearity. If the points are collinear, and so the surface area of the triangle formed by them will exist \(0\).

That is, \(\frac{1}{ii}\left[x_{1}\left(y_{2}-y_{three}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{ii}\right)\right]=0\), if three points are collinear.

Surface area of the Quadrilateral

A quadrilateral is a four-sided polygon. Quadrilaterals are in various shapes like parallelograms, trapeziums, and kites. Every quadrilateral can be divided into \(ii\) triangles. To calculate the area of a quadrilateral whose coordinates of the vertices are given:

Divide the quadrilateral into two triangles.
Find the area of the individual triangles.
Sum the surface area of the triangles.

Allow \(A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right)\) and \(D\left(x_{four}, y_{4}\right)\) be the vertices of a quadrilateral \(A B C D\).

Now, area of quadrilateral \(A B C D=\) Surface area of the \(\triangle A B D+\) Area of the \(\triangle B C D\)

\( = \frac{1}{2}\left\{ {\left( {{x_1}{y_2} + {x_2}{y_4} + {x_4}{y_1}} \right) – \left( {{x_2}{y_1} + {x_4}{y_2} + {x_1}{y_4}} \right)} \right\}\)
\(+ \frac{1}{ii}\left\{ {\left( {{x_2}{y_3} + {x_3}{y_4} + {x_4}{y_2}} \right) – \left( {{x_3}{y_2} + } \right.} \right.\left. {\left. {{x_4}{y_3} + {x_2}{y_4}} \right)} \right\}\)

\( = \frac{one}{2}\left\{ {\left( {{x_1}{y_2} + {x_2}{y_3} + {x_3}{y_4} + {x_4}{y_1}} \right) – \left( {{x_2}{y_1} + {x_3}{y_2} + {x_4}{y_3} + {x_1}{y_4}} \correct)} \correct\}\)

\( = \frac{1}{two}\left\{ {\left( {{x_1} – {x_3}} \right)\left( {{y_2} – {y_4}} \right) – \left( {{x_2} – {x_4}} \correct)\left( {{y_1} – {y_3}} \right)} \correct\}\)

Hence, the expanse of a quadrilateral in coordinate geometry when all the coordinates of the vertices are given tin be calculated using the formula

Solved Examples – Area of a Triangle and Quadrilateral

Below are a few solved examples that can help in getting a improve idea.

Q.1. Detect the area of a triangle with vertices are \((i,-1),(-iv,6)\) and \((-three,-5)\) .
Ans: Area of the triangle with vertices \(A(1,-i), B(-4,half-dozen)\) and \(C(-three,-5)\), can be calculated using the formula given by

\(\frac{1}{2}\left[x_{one}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{ii}\right)\right]\)

Expanse of the triangle \(=\frac{1}{2}[1(vi+5)+(-)(-5+one)+(-3)(-1-half-dozen)]\)

\(=\frac{1}{2}(11+16+21)\)

\(=24\) foursquare units

Q.2. Find the surface area of the quadrilateral with vertices, taken in guild, are \(A(-3,2), B(5,iv), C(7,-6)\) and \(D(-five,-4)\).
Ans: The area of a quadrilateral is equal to the sum of the area of the two triangles formed past dividing the quadrilateral.

Area of the quadrilateral \(A B C D=\) Area of \(\triangle A B C+\) Area of \(\triangle A C D\)

Area of \(\triangle A B C=\frac{ane}{ii} \mid(-3)(four+6)+5(-6-ii)+7(2-4 \mid\)

\(=\frac{1}{two}|-30-40-14|\)

\(=\frac{1}{ii}|-84|\)

\(=42\) foursquare units

Area of \(\triangle A C D=\frac{1}{two}|-iii(-6+4)+7(-4-2)+(-v)(2+half-dozen)|\)

\(=\frac{1}{2}|+half dozen-42-40|\)

\(=\frac{1}{2}|-76|\)

\(=38\) square units

And then, the area of the quadrilateral \(A B C D=42+38=80\) foursquare units.

Q.three. Determine the expanse of a triangle formed by the points \(A(v,2), B(4,7)\) and \(C(7,-4)\).
Ans: Nosotros can find the expanse of the triangle using the determinants method.

Area of triangle \(ABC = \frac{1}{two}\left| {\brainstorm{array}{{c}} {{x_1}} & {{y_1}} & 1 \\ {{x_2}} & {{y_2}} & i \\ {{x_3}} & {{y_3}} & 1 \\ \end{assortment} } \right|\)
\(= \frac{1}{2}\left| {\begin{array}{{c}} {{5}} & {{2}} & 1 \\ {{4}} & {{7}} & 1 \\ {{7}} & {{- 4}} & ane \\ \cease{array} } \correct|\)

After simplification, this formula of area of a triangle in determinant course can also be written equally:

Area of triangle \(A B C=\frac{i}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{ii}\left(y_{three}-y_{1}\right)+x_{3}\left(y_{1}-y_{two}\correct)\correct]\)

Surface area of the triangle with the vertices \(A(5,2), B(iv,vii)\) and \(C(vii,-4)\)

\(=\frac{1}{ii}|5(7+4)+4(-4-2)+7(2-seven)|\)

\(=\frac{1}{2}|55-24-35|\)

\(=\left|\frac{-4}{2}\right|\)

\(=|-2|\)

\(=2\) square units

Since the area is a measure that cannot be negative, we should accept the numerical value of \(-2\), i.e., \(2\).

Q.iv. Find the value of \(k\) if the points \(A(2,3), B(four, 1000)\), and \(C(6,-iii)\) are collinear.
Ans: The points are said to exist collinear if they prevarication on the same line. When the points are lying on the same line, these points cannot be connected to form a triangle. And hence, the area of the triangle formed past them must exist \(0\).

That is, if the points are collinear, then

Area of the triangle, \(\frac{1}{2}\left[x_{i}\left(y_{2}-y_{three}\right)+x_{2}\left(y_{iii}-y_{1}\right)+x_{iii}\left(y_{ane}-y_{two}\right)\correct]=0\)

\(\Rightarrow \frac{1}{two}|(thousand+3)+four(-3-3)+vi(three-k)=0|\)

\(\Rightarrow \frac{1}{two}(-four k)=0\)

\(\Rightarrow 1000=0\)

Hence, the value of \(m\) for which the given points are collinear, is \(0\).

Q.5. If \(A(-5,7), B(-4,-v), C(-1,-6)\) and \(D(4,five)\) are the vertices of a quadrilateral, observe the area of the quadrilateral \(A B C D\).
Ans: Surface area of a quadrilateral is given by \( = \frac{1}{2}\left\{ {\left( {{x_1} – {x_3}} \right)\left( {{y_2} – {y_4}} \right) – \left( {{x_2} – {x_4}} \right)\left( {{y_1} – {y_3}} \correct)} \right\}\)

\( = \frac{1}{2}\{ ( – 5 – ( – i))( – 5 – 5) – ( – 4 – 4)(vii – ( – 6))\} \)

\( = \frac{1}{two}\{ ( – iv)( – x) – ( – viii)(thirteen)\} \)

\( = \frac{1}{two}\{ 40 + 104)\)

\(=\frac{144}{two}=72\)

Therefore, the area of quadrilateral \(A B C D=72\) square units.

Summary

A polygon is a simple, closed curve made upwardly of line segments. The simplest polygon is the triangle which is a \(3\)-sided polygon. A polygon with \(4\) sides is chosen a quadrilateral. In coordinate geometry, when the coordinates of the three vertices of the triangle are given, and then the area of the triangle tin be determined using the formula, Area \(=\frac{i}{2} x_{1}\left(y_{2}-y_{three}\right)+\) \(x_{ii}\left(y_{3}-y_{ane}\right)+x_{3}\left(y_{i}-y_{ii}\correct)\). When the area of a triangle is zero, then the vertices are collinear. The area of the quadrilateral, if all its coordinates are given, tin can be calculated using the formula \(\frac{1}{2}\left\{ {\left( {{x_1} – {x_3}} \correct)\left( {{y_2} – {y_4}} \right) – \left( {{x_2} – {x_4}} \right)\left( {{y_1} – {y_3}} \correct)} \correct\}\). In simpler words, we tin say that to calculate the expanse of a quadrilateral, divide it into triangular regions with no common area and add their areas together. The aforementioned process can exist used to summate the area of other polygons too.

Frequently Asked Questions (FAQs)

Students might be having many questions with respect to the Area of a Triangle and Quadrilateral. Here are a few usually asked questions and answers.

Q.1. What are a triangle and a quadrilateral?
Ans: A triangle is the simplest polygon with 3 sides, while a quadrilateral has four sides.

Q.2. How are triangles and quadrilaterals related?
Ans: Both triangles and quadrilaterals are polygons. Every quadrilateral tin can be divided into ii triangles.

Q.three. What are the quadrilaterals shapes?
Ans: Quadrilaterals are four-sided polygons. The dissimilar kinds of quadrilateral shapes are parallelograms, squares, rectangles, trapeziums, kites and rhombuses.

Q.4. Can determinants be used to observe the area of a triangle?
Ans: We can find the area of the triangle using the determinants method.

Area of triangle \(ABC = \frac{1}{2}\left| {\begin{assortment}{{c}} {{x_1}} & {{y_1}} & 1 \\ {{x_2}} & {{y_2}} & i \\ {{x_3}} & {{y_3}} & i \\ \end{array} } \right|\)
\(= \frac{1}{2}\left| {\begin{array}{{c}} {{v}} & {{2}} & 1 \\ {{iv}} & {{7}} & 1 \\ {{7}} & {{- iv}} & ane \\ \stop{assortment} } \correct|\)

After simplification, this formula of the area of a triangle in determinant form can also be written every bit:
Area of triangle \(A B C=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{one}-y_{2}\right)\correct]\)

Q.5. How many triangles are nowadays in a quadrilateral?
Ans: A quadrilateral tin be divided into two triangles.