Question

find the area of Quadrilatra ABCD
whose vertices taken in order are
A(-1,-1) B(-1, 4) (5, 4) D(5,-1)​

Answer 1

A(-3,1), B(-2,-4), C(4,-1), D(3,4) are the vertices of quadrilateral ABCD

Let AC be the diagonal of quad ABCD

therefore, ar of quad ABCD= ar triangle ABC+ ar triangle ADC

ar of triangle ABC= 1/2[x1(y2-y3) -x2(y3-y1) -x3(y1-y2)]

= 1/2[ (-3) (-4+1) + (-2) (-1+1) +4(-1+4) ]

=1/2[9+0+12]

=21/2 sq units

ar of triangle ADC= 1/2[ (-3) (4+1) + 3(-1+1) + 4(-1-4)]

=1/2[-15+0-20]

= -35/2 sq units (since area cant be negative so area of triangle ADC=35/2 sq units)

Hence ar of quadrilateral ABCD= 21/2 + ( 35/2)

=56/2= 28 sq units

Answer 2

$\bf\large\underline\blue{Answer:-}$

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.