How do you find the area of the rectangle with vertices A(-5,1), B(-3,-1), C(3,5), D(1,7)?

Jul 8, 2016

Area $\square A B C D = 24$ sq. units

Explanation:

The area of $\square A B C D$ could be calculated in several ways.
One would be to divide the rectangle into two triangles, then use the Pythagorean Theorem to find the distances between the points, and finally use Heron's formula to find the areas of the two triangles.

However, I think the method below is easier.

Enclose the rectangle $A B C D$ in another rectangle $\square P Q R S$ with sides parallel to the x and y axis:

$A r e {a}_{\square A B C D}$
$= A r e {a}_{\square P Q R S} - \left(A r e {a}_{\triangle D P A} + A r e {a}_{\triangle A Q B} + A r e {a}_{{\triangle}_{B} R C} + A r e {a}_{\triangle C S D}\right)$

$A r e {a}_{\square P Q R S} = 8 \times 8 = 64$
$A r e {a}_{\triangle D P A} = A r e {a}_{\triangle B R C} = \frac{6 \times 6}{2} = 18$
$A r e {a}_{\triangle A Q B} = A r e {a}_{\triangle C S D} = \frac{2 \times 2}{2} = 2$

Therefore
$A r e {a}_{\square P Q R S} = 64 - \left(18 + 2 + 18 + 2\right) = 24$