Question

Find the area of the quadrilateral ABCD, the coordinates of whose vertices are A(6,-3),B(-4,-2),C(3,1) and D(5,0)?

• A. 21 unit²
• B. 30 unit²
• C. 25 unit²
• D. 16 unit²

Answer 1

Answer:

A. 21 unit²

Use the shoelace formula for the area of a quadrilateral.

Coordinates:

• (A(6,-3))

• (B(-4,-2))

• (C(3,1))

• (D(5,0))

$\text{Area}=\frac{1}{2}\left|x_1y_2+x_2y_3+x_3y_4+x_4y_1-(y_1x_2+y_2x_3+y_3x_4+y_4x_1)\right|$

Substitute the values:

$= \frac{1}{2}\left|6(-2)+(-4)(1)+3(0)+5(-3)\right.$

$\left.-\left[(-3)(-4)+(-2)(3)+(1)(5)+(0)(6)\right]\right|$

$= \frac{1}{2}|-12-4+0-15-(12-6+5+0)|$

$= \frac{1}{2}|-31-11|$

$= \frac{1}{2}\times 42$

$= 21$

Area of quadrilateral ABCD = 21 square units