Question

In Exercises 27-32 find and then compare lengths of segments.

Find the area of the rectangle with vertices B(8, 0). T(2, -9), R(-1, -7), and C(5, 2).

Short answer

The area of a rectangle is 39.

Step-by-step solution

Step-1 – Given

The given coordinates are $B\left(8,0\right)\text{,}T\left(2,-9\right)\text{,}R\left(-1,-7\right)\text{,and}C\left(5,2\right)$.

Step-2 – To determine

We have to find the area of the rectangle.

Step-3 – Calculation

Using the distance formula, we will find the length and breadth of the rectangle.

BT and RC are the lengths of the rectangle.

TR and CB are the breadths of the rectangle.

So,

$\begin{array}{l}BT=\sqrt{{\left(2-8\right)}^2+{\left(-9-0\right)}^2}\left[\text{since}B\left(8,0\right)\text{and}T\left(2,-9\right)\right]\\ BT=\sqrt{{\left(-6\right)}^2+{\left(-9\right)}^2}\\ BT=\sqrt{36+81}\\ BT=\sqrt{117}\end{array}$

$\begin{array}{l}TR=\sqrt{{\left(-1-2\right)}^2+{\left(-7-\left(-9\right)\right)}^2}\left[\text{since}T\left(2,-9\right)\text{and}R\left(-1,-7\right)\right]\\ TR=\sqrt{{\left(-1-2\right)}^2+{\left(-7+9\right)}^2}\\ TR=\sqrt{{\left(-3\right)}^2+{\left(2\right)}^2}\\ TR=\sqrt{9+4}\\ TR=\sqrt{13}\end{array}$

$\begin{array}{l}RC=\sqrt{{\left(5-\left(-1\right)\right)}^2+{\left(2-\left(-7\right)\right)}^2}\left[\text{since}R\left(-1,-7\right)\text{and}C\left(5,2\right)\right]\\ RC=\sqrt{{\left(5+1\right)}^2+{\left(2+7\right)}^2}\\ RC=\sqrt{{\left(6\right)}^2+{\left(9\right)}^2}\\ RC=\sqrt{36+81}\\ RC=\sqrt{117}\end{array}$

$\begin{array}{l}CB=\sqrt{{\left(8-5\right)}^2+{\left(0-2\right)}^2}\left[\text{since}C\left(5,2\right)\text{and}B\left(8,0\right)\right]\\ CB=\sqrt{{\left(3\right)}^2+{\left(-2\right)}^2}\\ CB=\sqrt{9+4}\\ CB=\sqrt{13}\end{array}$

We know that in a rectangle the pair of opposite sides are equal. So, BTRC is a rectangle.

Area of the rectangle is:

$\begin{array}{l}a=l\times b\left[\begin{array}{l}a=\text{area}\\ l=\text{length}\\ b=\text{breadth}\end{array}\right]\\ a=\left(BT\right)\left(BC\right)\\ a=\left(\sqrt{117}\right)\left(\sqrt{13}\right)\\ a=\sqrt{1521}\\ a=39\end{array}$

So, the area of the rectangle is 39.