Question

Find the area of the quadrilateral whose vertices are A(-3,1); B(-2,-2); C(4,1); D(2,3): *​

Answer 1

Answer:

35/2 sq.units

Step-by-step explanation:

Given,

A = (-3 , 1)

B = (-2 , -2)

C = (4 , 1)

D = (2 , 3)

To Find :-

Area of the quadrilateral.

Formula Required :-

$Area=\dfrac{1}{2}\left|\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}+\begin{vmatrix}x_2&y_2\\x_3&y_3\end{vmatrix}+\begin{vmatrix}x_3&y_3\\x_4&y_4\end{vmatrix}+..\begin{vmatrix}x_n&y_n\\x_1&y_1\end{vmatrix}\right|$

Solution :-

• x_1 = -3 , y_1 = 1
• x_2 = -2 , y_2 = -2
• x_3 = 4 , y_3 = 1
• x_4 = 2 , y_4 = 3

$Area=\dfrac{1}{2}\left|\begin{vmatrix}-3&1\\-2&-2\end{vmatrix}+\begin{vmatrix}-2&-2\\4&1\end{vmatrix}+\begin{vmatrix}4&1\\2&3\end{vmatrix}+\begin{vmatrix}2&3\\-3&1\end{vmatrix}\right|$

$=\dfrac{1}{2}\left|(6+2)+(-2+8)+(12-2)+(2+9)\right|$

$=\dfrac{1}{2}\left|8+6+10+11\right|$

$=\dfrac{1}{2}|35|$

$=\dfrac{35}{2}sq.units$