Diagonals: Using Pythagoras' theorem

Let's focus on triangle BCD in order to find side DC.

We'll use the Pythagorean theorem and input the known data:

$BC^2+DC^2=BD^2$

$6^2+DC^2=10^2$

$DC^2=100-36=64$

Let's now remove the square root:

$DC=8$

Since in a rectangle each pair of opposite sides are equal to each other, we know that:

$DC=AB=8$

$BC=AD=6$

Now we can calculate the perimeter of the rectangle by adding all sides together:

$8+6+8+6=16+12=28$

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:

$DC=\sqrt{576}=24$

We will use the Pythagorean theorem in order to find BD:

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

Let's input the known data:

$BD^2=3^2+4^2$

$BD^2=9+16$

$BD^2=25$

We'll take the square root:

$BD=\sqrt{25}=5$

In a rectangle, each pair of opposite sides are equal to each other, therefore:

$AB=DC=4$

We will use the Pythagorean theorem to find AC:

$AC^2=BC^2+DC^2$

Let's substitute the known data:

$AC^2=3^2+4^2$

$AC^2=9+16$

$AC^2=25$

Let's take the square root:

$AC=\sqrt{25}=5$

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:

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

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

We know that:

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

We also know that AB equals X.

First, we will substitute the given data into the formula accordingly:

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

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

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

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

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

Now let's look at triangle ABC and use the Pythagorean theorem:

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

We substitute in our known values:

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

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

Finally, we will add 1 to both sides:

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

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