**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

After having gone through the stuff given above, we hope that the students would have understood "Area of quadrilateral when four vertices are given".

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