Area of a Rectangle: Calculation using the diagonal

Look at the following rectangle:

BC = 8

BD = 17

Calculate the area of the rectangle ABCD.

We will find side DC by using the Pythagorean theorem in triangle DBC:

$BC^2+CD^2=BD^2$

Let's substitute the known data:

$8^2+CD^2=17^2$

$CD^2=289-64=225$

Let's take the square root:

$CD=15$

Now we have the length and width of rectangle ABCD and we'll calculate the area:

$15\times8=120$

120

The rectangle ABCD is shown below.

$BD=25,BC=7$

Calculate the area of the rectangle.

We will use the Pythagorean theorem in order to find the side DC:

$(BC)^2+(DC)^2=(DB)^2$

We begin by inserting the existing data into the theorem:

$7^2+(DC)^2=25^2$

$49+DC^2=625$

$DC^2=625-49=576$

Finally we extract the root:

$DC=\sqrt{576}=24$

168

Below is the rectangle ABCD.

O is the intersection point of the diagonals of the rectangle.

AD = 8

BO = 8.5

Calculate the area of the triangle ABD.

According to the given information, we can claim that:

$BD=2BO=8.5\times2=17$

Now let's look at triangle ABD to calculate side AB

$AB^2+AD^2=BD^2$

Let's input the known data:

$AB^2+8^2=17^2$

$AB^2=289-64=225$

We'll take the square root

$AB=15$

Now let's calculate the area of triangle ABD:

$\frac{15\times8}{2}=\frac{120}{2}=60$

60