Question

find area of quadrilateral whose vertices taken in order are A(-5,-3) B(-4,-6) C(2,-1) D(1,2) plez don't give formula tell the whole process

Answer 1

area of quadrilateral = 23 sq. units

Answer 2

Answer:

The area of quadrilateral is 23 square units.

Step-by-step explanation:

We are given a quadrilateral ABCD whose vertices are A(-5,-3) B(-4,-6) C(2,-1) D(1,2). Please find attachment for figure.

Using area of triangle of coordinate,

$\text{Area of triangle}=\dfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2|$

In ΔABD, A(-5,-3) B(-4,-6) D(1,2)

$\text{Area of }\triangle ABD=\dfrac{1}{2}|-5(-6-2)-4(2+3)+1(-3+6)|$

$\text{Area of }\triangle ABD=\dfrac{23}{2}$

In ΔBCD, C(2,-1) B(-4,-6)  D(1,2)

$\text{Area of }\triangle BCD=\dfrac{1}{2}|2(-6-2)-4(2+1)+1(-1+6)|$

$\text{Area of }\triangle BCD=\dfrac{23}{2}$

ar (ABCD) = ar(ABD) + ar(BCD)

ar (ABCD) = 23/2+ 23/2 = 23 square units.

Hence, The area of quadrilateral is 23 square units.