Area of a Rectangle: Using Pythagoras' theorem

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:

$15\times8=120$

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:

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

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:

$DC=\sqrt{576}=24$

Let's calculate side BG using the Pythagorean theorem:

$BG^2+GC^2=BC^2$

We'll substitute the known data:

$BG^2+3^2=5^2$

$BG^2+9=25$

$BG^2=16$

$BG=\sqrt{16}=4$

Now we can calculate the area of rectangle ABGE since we have the length and width:

$5\times4=20$

Let's look at triangle ADK in order to calculate side AD:

$AD^2+DK^2=AK^2$

Now let's substitute in our values:

$AD^2+4^2=5^2$

$AD^2+16=25$

We'll then move 16 to the other side and change the sign to the appropriate one:

$AD^2=25-16$

$AD^2=9$

Next, we'll take the square root and get:

$AD=3$

Since AD is a side of rectangle ABCD, we can calculate side AB as follows:

$S=AB\times AD$

Let's substitute in our values:

$24=3\times AB$

Finally, we'll divide both sides by 3:

$AB=8$

To find the missing side, we use the Pythagorean theorem in the upper triangle.

Since the triangle is isosceles, we know that the length of both sides is 7.

Therefore, we apply Pythagoras$A^2+B^2=C^2$$7^2+7^2=49+49=98$

Therefore, the area of the missing side is:$\sqrt{98}$

The area of a rectangle is the multiplication of the sides, therefore:

$\sqrt{98}\times10=98.99\approx99$