Area of a Quadrilateral Using Coordinates | Class 10 Maths

Quick answer: To find the area of a quadrilateral from the coordinates of its vertices, draw a diagonal to split it into two triangles, find the area of each triangle with $\frac{1}{2}\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right|$, and add the two areas. For the quadrilateral $A(-4, -2)$, $B(-3, -5)$, $C(3, -2)$, $D(2, 3)$, the areas are $10.5$ and $17.5$, so the total area is $28$ square units.

The area of a triangle can be found directly from the coordinates of its three vertices. For a quadrilateral with four vertices, the simplest method is to split it into two triangles with a diagonal, find each triangle's area, and add them.

This guide explains the method and works through a well-known NCERT problem step by step, with a quick way to check the answer.

Syllabus note: The section on the area of a triangle using coordinates has been removed from the current rationalised NCERT Class 10 textbook (Chapter 7, Coordinate Geometry). The method is still useful for older papers, NCERT Exemplar and higher classes, but check your current syllabus before your exam.

What You Will Learn

• How a diagonal splits a quadrilateral into two triangles
• The formula for the area of a triangle from its coordinates
• Why each area must be taken as a positive value
• A full solution to the NCERT problem with vertices $A(-4, -2)$, $B(-3, -5)$, $C(3, -2)$, $D(2, 3)$
• A shortcut check using the shoelace method

1. Splitting a Quadrilateral with a Diagonal

Let the vertices of the quadrilateral, taken in order, be $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$ and $D(x_4, y_4)$.

   D ----------- C
   |  ACD      / |
   |         /   |
   |       /     |
   |     /  ABC  |
   |   /         |
   A ----------- B
   (diagonal AC)

Step 1: Join the diagonal $AC$.

Step 2: This splits $ABCD$ into two triangles, $ABC$ and $ACD$.

Step 3: Add their areas:

$\text{Area}(ABCD) = \text{Area}(\triangle ABC) + \text{Area}(\triangle ACD)$

Area of a Triangle from Coordinates

For a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$:

$\text{Area} = \frac{1}{2}\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right|$

Key point: The expression inside the bars can come out negative, depending on the order of the vertices. Area is never negative, so always take the positive value of each triangle's area before adding.

2. Solved NCERT Problem

Question

Find the area of the quadrilateral whose vertices, taken in order, are $A(-4, -2)$, $B(-3, -5)$, $C(3, -2)$ and $D(2, 3)$.

Solution

Step 1: Split the quadrilateral. Join $AC$ to divide $ABCD$ into $\triangle ABC$ and $\triangle ACD$.

Step 2: Area of $\triangle ABC$, with $A(-4, -2)$, $B(-3, -5)$, $C(3, -2)$.

$\text{Area}(\triangle ABC) = \frac{1}{2}\left|(-4)\left[-5 - (-2)\right] + (-3)\left[-2 - (-2)\right] + 3\left[-2 - (-5)\right]\right|$

$= \frac{1}{2}\left|(-4)(-3) + (-3)(0) + 3(3)\right| = \frac{1}{2}\left|12 + 0 + 9\right| = \frac{21}{2} = 10.5 \text{ sq. units}$

Step 3: Area of $\triangle ACD$, with $A(-4, -2)$, $C(3, -2)$, $D(2, 3)$.

$\text{Area}(\triangle ACD) = \frac{1}{2}\left|(-4)(-2 - 3) + 3\left[3 - (-2)\right] + 2\left[-2 - (-2)\right]\right|$

$= \frac{1}{2}\left|(-4)(-5) + 3(5) + 2(0)\right| = \frac{1}{2}\left|20 + 15 + 0\right| = \frac{35}{2} = 17.5 \text{ sq. units}$

Step 4: Add the two areas.

$\text{Area}(ABCD) = \frac{21}{2} + \frac{35}{2} = \frac{56}{2} = 28 \text{ sq. units}$

Answer: The area of the quadrilateral is 28 square units.

3. Quick Check: The Shoelace Method

List the vertices in order and repeat the first vertex at the end:

Step 1: Multiply each $x$ by the next $y$ and add.

$(-4)(-5) + (-3)(-2) + (3)(3) + (2)(-2) = 20 + 6 + 9 - 4 = 31$

Step 2: Multiply each $y$ by the next $x$ and add.

$(-2)(-3) + (-5)(3) + (-2)(2) + (3)(-4) = 6 - 15 - 4 - 12 = -25$

Step 3: Take half the difference.

$\text{Area} = \frac{1}{2}\left|31 - (-25)\right| = \frac{1}{2} \times 56 = 28 \text{ sq. units}$

This matches the answer above. The shoelace method is a quick check; in the exam, show the triangle method as your main working.

4. Summary Table

5. Exam Tips and Common Mistakes

Exam Tip: Always write the units, square units, in your final answer.

Exam Tip: Be careful with double negatives, for example $-2 - (-5) = -2 + 5 = 3$.

Common Mistake: Taking the vertices out of order. They must go around the quadrilateral ($A \rightarrow B \rightarrow C \rightarrow D$). If you use $A, C, B, D$, the triangles overlap and the answer is wrong.

Common Mistake: Adding a negative triangle area. Take the positive value of each triangle's area before adding.

Frequently Asked Questions

How do you find the area of a quadrilateral using coordinates?

Draw a diagonal to split the quadrilateral into two triangles, find the area of each triangle using the coordinate formula, and add the two areas.

What is the formula for the area of a triangle using coordinates?

Area $= \frac{1}{2}\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right|$.

What is the area of the quadrilateral with vertices (−4, −2), (−3, −5), (3, −2) and (2, 3)?

The area is 28 square units: 10.5 for triangle ABC plus 17.5 for triangle ACD.

Why must the area be taken as positive?

The formula can give a negative value depending on the order of the vertices, but area is a size and cannot be negative.

Is the area of a triangle in the current Class 10 syllabus?

It has been removed from the rationalised NCERT Class 10 textbook, but it is still useful for older papers, NCERT Exemplar and higher classes.