Question

find the area of quadrilateral abcd. round the area to the nearest whole number, if r
a(-5, 4)
b(0, 3)
f(-2, 1)
c(4, -1)
e(2, -3)
d(4, -5)
the area is | 28 × square units.

Explanation:

Step1: Enclose the quadrilateral in a rectangle

Enclose ABCD in a rectangle with vertices (-5, - 5), (-5, 4), (4, 4) and (4, - 5). The length of the rectangle's sides are \(4-(-5)=9\) and \(4 - (-5)=9\). So the area of the rectangle \(A_{rect}=9\times9 = 81\).

Step2: Calculate the area of the four right - angled triangles outside the quadrilateral

Triangle 1 with vertices A(-5, 4), B(0, 3), and a point on the rectangle. Its base \(b_1 = 5\) and height \(h_1=1\), so \(A_1=\frac{1}{2}\times5\times1=\frac{5}{2}\).
Triangle 2 with vertices B(0, 3), C(4, - 1), and a point on the rectangle. Its base \(b_2 = 4\) and height \(h_2 = 4\), so \(A_2=\frac{1}{2}\times4\times4 = 8\).
Triangle 3 with vertices C(4, - 1), D(4, - 5), and a point on the rectangle. Its base \(b_3=4\) and height \(h_3 = 4\), so \(A_3=\frac{1}{2}\times4\times4=8\).
Triangle 4 with vertices D(4, - 5), A(-5, 4), and a point on the rectangle. Its base \(b_4 = 9\) and height \(h_4=1\), so \(A_4=\frac{1}{2}\times9\times1=\frac{9}{2}\).

Step3: Calculate the area of the quadrilateral

The area of the quadrilateral \(A = A_{rect}-A_1 - A_2 - A_3 - A_4\).
\[

$$\begin{align*} A&=81-\frac{5}{2}-8 - 8-\frac{9}{2}\\ &=81-( \frac{5 + 9}{2})-16\\ &=81 - 7-16\\ &=58 \end{align*}$$

\]

Answer:

58 square units