Question

IF A(-4,8) B(-3,-4) C(0,5) AND D(5,6) ARE THE VERTICES OF A QUADRILATERAL ABCD,find its area .

Answer 1

32 sq. units
By draw a diagonal and thus two triangles in the quadrilateral ABCD. Then calculate the areas of the two triangles ABC and ADC separately by the formula : 1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)] . Then finally add the two areas which gives you the area of the quadrilateral ABCD.

Answer 2

Answer:

The area of ABCD is 32 units².

Step-by-step explanation:

The vertices of quadrilateral ABCD are A(-4,8) B(-3,-4) C(0,5) and D(5,6).

The area of ABCD is the sum of area of ABC and CDA.

The area of a triangle is

$A=\frac{1}{2}(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))$

$Area(ABC)=\frac{1}{2}(-4(-4-5)-3(5-8)+0(8+4))=22.5$

$Area(CDA)=\frac{1}{2}(0(6-8)+5(8-5)-4(5-6))=9.5$

$Area(ABCD)=Area(ABC)+Area(CDA)=22.5+9.5=32$

Therefore the area of ABCD is 32 units².