Square of sum: Using Pythagoras' theorem

To find $x$ in the given triangle, let's apply the Pythagorean Theorem. The squared lengths of the triangle's legs and hypotenuse are related by this equation:

First, expand each term:

• $(x+1)^2 = x^2 + 2x + 1$
• $x^2 = x^2$
• $(x+2)^2 = x^2 + 4x + 4$

Plug these into the Pythagorean Theorem equation:

Combine like terms:

Rearrange the equation to isolate terms on one side:

Simplify to get a quadratic equation:

Now, solve for $x$ using factoring. Look for two numbers that multiply to $-3$ and add to $-2$. These numbers are $-3$ and $1$:

Set each factor equal to zero:

• $x - 3 = 0 \Rightarrow x = 3$
• $x + 1 = 0 \Rightarrow x = -1$

Given the condition $x > 0$, the valid solution is:

$x = 3$

To solve this problem, we'll follow these steps:

• Identify the given information.
• Use the geometric properties of a circle and a right triangle to find the radius.
• Express the area of the circle in terms of the given expression.

Now, let's work through each step:

Step 1: Given a circle with center $O$ and radii $AO$ and $OB$ such that $AO\perp OB$, each is a radius $r$, and $AB = \text{and}+2$.

Step 2: By the Pythagorean theorem, we know:

Step 3: Solving for the area of the circle:

The radius $r$ can be expressed by rearranging:

The area of the circle using this radius is:

Therefore, the expression for the area of the circle is $\frac{\pi}{2}[y^2+4y+4]$.

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:

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

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