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

**Solution:**

Let ABC be any triangle whose vertices are A(x_{1}, y_{1}), B(x_{2}, y_{2}), and C(x_{3}, y_{3}).

Area of a triangle is given by 1/2 (x_{1} (y_{2} - y_{3}) + x_{2} (y_{3}- y_{1}) + x_{3} (y_{1} - y_{2}))

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

Join AC to form two triangles ∆ABC and ∆ACD.

We know that, area of a triangle = 1/2 {x_{1} (y_{2} - y_{3}) + x_{2} (y_{3}- y_{1}) + x_{3}-(y_{1} - y_{2})}

By substituting the values of vertices, A, B, C in the formula.

Area of ΔABC = 1/2 [(- 4){(- 5) - (- 2)} + (- 3){(- 2) - (- 2)}+ 3{(- 2) - (- 5)}]

= 1/2 (12 + 0 + 9)

= 21/2 square units

By substituting the values of vertices, A, C, D in the Equation (1),

Area of ΔACD = 1/2 [(- 4){(- 2) - 3} + 3{(3) - (- 2) + 2{(- 2)} - (- 2)}]

= 1/2 (20 + 15 + 0)

= 35/2 square units

Area of ABCD = Area of ΔABC + Area of ΔACD

= (21/2 + 35/2) square units

= 28 square units

**Video Solution:**

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

### NCERT Class 10 Maths Solutions - Chapter 7 Exercise 7.3 Question 4

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

The area of the quadrilateral whose vertices, taken in order, are (- 4, - 2), (- 3, - 5), (3, - 2) and (2, 3) is 28 square units