**Area of quadrilateral when four vertices are given :**

Here, we are going to see, how to find area of a quadrilateral when four vertices are given.

Let us consider the quadrilateral given below.

In the above quadrilateral, A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) and D(x₄, y₄) are the vertices.

To find area of the quadrilateral ABCD, now we have take the vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) and D(x₄, y₄) of the quadrilateral ABCD in order (counter clockwise direction) and write them column-wise as shown below.

And the diagonal products x₁y₂, x₂y₃, x₃y₄ and x₄y₁ as shown in the dark arrows.

Also add the diagonal products x₂y₁, x₃y₂, x₄y₃ and x₁y₄ as shown in the dotted arrows.

Now, subtract the latter product from the former product to get area of the quadrilateral ABCD.

Hence, area of the quadrilateral ABCD is

= (1/2) x { (x₁y₂+x₂y₃+x₃y₄+x₄y₁) - (x₂y₁+x₃y₂+x₄y₃+x₁y₄) }

**Problem :**

Find the area of the quadrilateral whose vertices are (-4, -2), (-3, -5), (3, -2) and (2, 3)

**Solution : **

Plot the given points in a rough diagram as given below and take them in order (counter clock wise)

Let the vertices be A(-4, -2), B(-3, -5), C(3, -2) and (2, 3)

Then, we have

(x₁, y₁) = (-4, -2)

(x₂, y₂) = (-3, -5)

(x₃, y₃) = (3, -2)

(x₄, y₄) = (2, 3)

Area of triangle ABC is

= (1/2) x { (x₁y₂+x₂y₃+x₃y₄+x₄y₁) - (x₂y₁+x₃y₂+x₄y₃+x₁y₄) }

= (1/2) x { [20 + 6 +9 - 4] - [6 - 15 -4 -12] }

= (1/2) x { [31] - [-25] }

= (1/2) x { 31 + 25 }

= (1/2) x { 56 }

= 28 square units.

**Hence, are of triangle ABC is 28 square units.**

Find the area of the quadrilateral whose vertices are

(i) (6, 9), (7, 4), (4, 2) and (3, 7).

(ii) (-3, 4), (-5, -6), (4, -1) and (1, 2)

(iii) (-4, 5), (0, 7), (5, -5) and (-4, -2)

**Answers :**

(i) 17 square units.

(ii) 43 square units

(iii) 60.5 square units

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