Question

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

Answer 1

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Answer 2

Answer:

28 square units

Step-by-step explanation:

The given vertices of quadrilateral are A(- 4, - 2), B( - 3 ,- 5), C(3 ,- 2 ) and D(2,3).

Th area of a triangle is

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

Using the above formula, the area of triangle ABC is

$A_1=\frac{1}{2}|-4(-5+2)-3(-2+2)+3(-2+5)|$

$A_1=\frac{1}{2}|-4(-3)-3(0)+3(3)|$

$A_1=\frac{1}{2}|12+9|$

$A_1=\frac{21}{2}$

Using the above formula, the area of triangle ACD is

$A_2=\frac{1}{2}|-4(-2-3)+3(3+2)+2(-2+2)|$

$A_2=\frac{1}{2}|-4(-5)+3(5)+2(0)|$

$A_2=\frac{1}{2}|20+15|$

$A_2=\frac{35}{2}$

The area of quadrilateral is

$A=A_1+A_2$

$A=\frac{21}{2}+\frac{35}{2}$

$A=\frac{21+35}{2}$

$A=\frac{56}{2}$

$A=28$

Therefore the area of quadrilateral is 28 square units.