Question

area of quadrilateral whose vertices are (1,2) (6,2) (5,3) (3,7)

Answer 1

Make a quadrilateral Abcd divide it in 2 triangles by making diagonal now apply area of triangle formula in both the triangles add the both areas you will get that.
hope it helps

Answer 2

Area of the quadrilateral with vertices is 25 sq units.

Step-by-step explanation:

Given :The vertices are A(1,2) B(6,2) C(5,3) D(3,7)

To find: The area of quadrilateral for the vertices

Formula used: Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD

Area of triangle = $\frac{1}{2} [x_{1} (y_{2} -y_{3} ) + x_{2}( y_{3} - y_{1}) + x_{3}(y_{1} -y_{2})$

Solution

Let us divide the quadrilateral ABCD into ABD and BCD

Now by using the formula,

Area of triangle with coordinates or vertices $(x,y) , (x_{2} ,y_{2} ),(x_{3} ,y_{3} )$ is given by the formula

⇒ $\frac{1}{2} [x_{1} (y_{2} -y_{3} ) + x_{2}( y_{3} - y_{1}) + x_{3}(y_{1} -y_{2})$

Area of a triangle ABD with coordinates A(1,2) B(6,2) and D (3,7)

Area of triangle ABD =  $\frac{1}{2} [ 1(2 - 7) + 6(7 - 2) + 3 (1 - 2)]$

= $\frac{1}{2} [ 1(-5) + 6(5) + 3 (-1)]$

= $\frac{1}{2} [-5+30-3]$

= $\frac{1}{2}[ 22]$

= 11 sq units

Area of triangle BCD with coordinates B(6,2) C(5,3) and D(3,7)

Area of triangle = $\frac{1}{2} [x_{1} (y_{2} -y_{3} ) + x_{2}( y_{3} - y_{1}) + x_{3}(y_{1} -y_{2})$

Area of triangle BCD = $\frac{1}{2} [(6(3-2)+5(7-2) + 3(2 - 3)]$

= $\frac{1}{2}[ (6(1) + 5(5) +3(-1)]$

= $\frac{1}{2} [ 6+25-3]$

= $\frac{1}{2}[28]$

=14 sq units

Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD

= 11 + 14

=25 sq units.

Final answer:

Area of the quadrilateral with vertices is 25 sq units.