# Problem of the Week Problem D and Solution The Area Within

## Problem

Quadrilateral $$ABCD$$ is constructed as follows:

• vertex $$A$$ is located at the origin;

• vertex $$C$$ is at the intersection point of the lines $$\ell_1:~ 5x+2y=30$$ and $$\ell_2:~ x+2y=22$$;

• vertex $$B$$ is at the intersection point of the $$y$$-axis and $$\ell_2$$; and

• vertex $$D$$ is at the intersection point of the $$x$$-axis and $$\ell_1$$.

Determine the area of $$ABCD$$.

## Solution

Let the coordinates of $$C$$ be $$(h,k)$$ where $$h$$ is the horizontal distance from the $$y$$-axis to $$C$$ and $$k$$ is the vertical distance from the $$x$$-axis to $$C$$.

To find the coordinates of $$D$$, let $$y=0$$ in $$5x+2y=30$$. Therefore, the $$x$$-intercept is 6 and the coordinates of $$D$$ are $$(6,0)$$.

To find the coordinates of $$B$$, let $$x=0$$ in $$x+2y=22$$. Therefore, the $$y$$-intercept is 11 and the coordinates of $$B$$ are $$(0,11)$$.

Two methods are provided to find $$C$$, the point of intersection of $$\ell_1$$ and $$\ell_2$$.

• Solving for $$C$$ using the method of substitution:
Rewrite equation $$\ell_2$$ as $$x=22-2y$$.

Substitute for $$x$$ in $$\ell_1$$ so that $$5(22-2y)+2y=30$$. Simplifying, $$110-10y+2y=30$$. This further simplifies to $$-8y=-80$$ and $$y=10$$.

Substituting $$y=10$$ in $$x+2y=22$$ gives $$x+20=22$$ and $$x=2$$.

The coordinates of $$C$$, the point of intersection of $$\ell_1$$ and $$\ell_2$$, are $$(2,10)$$. Therefore, $$h=2$$ and $$k=10$$.

• Solving for $$C$$ using the method of elimination:
\begin{aligned} \ell_1: \quad 5x+2y&=30\\ \ell_2: \quad ~x+2y&=22\\ \text{Subtracting, we obtain,} \quad 4x&=8\\ \mbox{Therefore,} \quad x&=2\\\end{aligned} Substituting $$x=2$$ in $$l_1$$, $$10+2y=30$$ and $$y=10$$. The coordinates of $$C$$, the point of intersection of $$\ell_1$$ and $$\ell_2$$, are $$(2,10)$$. Therefore, $$h=2$$ and $$k=10$$.

Quadrilateral $$ABCD$$ can be divided into two triangles, $$\triangle ABC$$ and $$\triangle ACD$$.

Therefore, \begin{aligned} \text{Area } ABCD&=\text{Area }\triangle ABC + \text{Area }\triangle ACD\\ &=\frac{1}{2}(h \times AB) + \frac{1}{2}(k \times AD)\\ &=\frac{1}{2}(2)(11) + \frac{1}{2}(10)(6)\\ &=11+30\\ &=41\end{aligned} Therefore, the area of $$ABCD$$ is $$41 \text{ units}^2$$.