Area of a Rectangle: Using Pythagoras' theorem

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

The trapezoid ABCD and the rectangle ABGE are shown in the figure below.

Given in cm:

AB = 5

BC = 5

GC = 3

Calculate the area of the rectangle ABGE.

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:

$5\times4=20$

20

Shown below is the rectangle ABCD.

Given in cm:

AK = 5

DK = 4

The area of the rectangle is 24 cm².

Calculate the side AB.

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

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

Let's input the given data:

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

$AD^2+16=25$

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

$AD^2=25-16$

$AD^2=9$

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 input the given data:

$24=3\times AB$

We'll divide both sides by 3:

$AB=8$

8

Shown below is a rectangle and an isosceles right triangle.

What is the area of the rectangle?

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:

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

$\approx99$

Given the rectangle ABCD

AB=X

The ratio between AB and BC is $\sqrt{\frac{x}{2}}$

We mark the length of the diagonal A the rectangle in m

Check the correct argument:

Given that:

$\frac{AB}{BC}=\sqrt{\frac{x}{2}}$

Given that AB equals X

We will substitute accordingly in the formula:

$\frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}$

$x\sqrt{2}=BC\sqrt{x}$

$\frac{x\sqrt{2}}{\sqrt{x}}=BC$

$\frac{\sqrt{x}\times\sqrt{x}\times\sqrt{2}}{\sqrt{x}}=BC$

$\sqrt{x}\times\sqrt{2}=BC$

Now let's focus on triangle ABC and use the Pythagorean theorem:

$AB^2+BC^2=AC^2$

Let's substitute the known values:

$x^2+(\sqrt{x}\times\sqrt{2})^2=m^2$

$x^2+x\times2=m^2$

We'll add 1 to both sides:

$x^2+2x+1=m^2+1$

$(x+1)^2=m^2+1$

$m^2+1=(x+1)^2$

Given the rectangle ABCD

AB=X the ratio between AB and BC is equal to$\sqrt{\frac{x}{2}}$

We mark the length of the diagonal $A$ with $m$

Check the correct argument:

Let's find side BC

Based on what we're given:

$\frac{AB}{BC}=\frac{x}{BC}=\sqrt{\frac{x}{2}}$

$\frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}$

$\sqrt{2}x=\sqrt{x}BC$

Let's divide by square root x:

$\frac{\sqrt{2}\times x}{\sqrt{x}}=BC$

$\frac{\sqrt{2}\times\sqrt{x}\times\sqrt{x}}{\sqrt{x}}=BC$

Let's reduce the numerator and denominator by square root x:

$\sqrt{2}\sqrt{x}=BC$

We'll use the Pythagorean theorem to calculate the area of triangle ABC:

$AB^2+BC^2=AC^2$

Let's substitute what we're given:

$x^2+(\sqrt{2}\sqrt{x})^2=m^2$

$x^2+2x=m^2$

$x^2+2x=m^2$