Question

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

Answer 1

Answer:

Area of the Quadrilateral is 24 units².

Step-by-step explanation:

Given:

Vertex of Quadrilateral say ABCD

Coordinates of the vertices are A( -4 , -2 ) , B ( 3 , -5 ) , C( 2 , -3 ) and D( 2 , 3 )

To find: Area of the Quadrilateral.

We know that a diagonal break the quadrilateral in two triangles.

Draw diagonal AC. we get ΔABC and ΔADC

Area of Quadrilateral = Area of ΔABC + Area of ΔADC

Area of triangle is given as follows when coordinates are given,

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

Area of ΔABC $=\frac{1}{2}\left|(-4(-5-(-3))+3(-3-(-2))+2(-2-(-5)))\right|$

                       $=\frac{1}{2}\left|(-4(-5+3)+3(-3+2)+2(-2+5))\right|$

                       $=\frac{1}{2}\left|(-4(-2)+3(-1)+2(3))\right|$

                       $=\frac{1}{2}\left|(8-3+6)\right|$

                       $=\frac{1}{2}\left|11\right|$

                       $=\frac{11}{2}\:units^2$

Area of ΔADC $=\frac{1}{2}\left|(-4(3-(-3))+2(-3-(-2))+2(-2-3))\right|$

                       $=\frac{1}{2}\left|(-4(3+3)+2(-3+2)+2(-5))\right|$

                       $=\frac{1}{2}\left|(-4(6)+3(-1)-10)\right|$

                       $=\frac{1}{2}\left|(-24-3-10)\right|$

                       $=\frac{1}{2}\left|-37\right|$

                       $=\frac{37}{2}\:units^2$

Area of the Quadrilateral = 11/2 + 37/2 = 48/2 = 24 units²

Therefore, Area of the Quadrilateral is 24 units².