Solving the Isosceles Right Angle Triangle Problem

We are given an isosceles right angle triangle ABC, where the angle at B is 90 degrees, and the side AB is 12 cm. In an isosceles right angle triangle, the two sides forming the right angle are equal in length. Therefore, BC must also be equal to AB.

• Given: $\angle ABC = 90^\circ$
• Given: $AB = 12$ cm
• Since triangle ABC is isosceles and right-angled at B, $BC = AB = 12$ cm.

Calculating the Hypotenuse Length

To find the radii of the inscribed and circumscribing circles, we first need to find the length of the hypotenuse AC. We can use the Pythagorean theorem since it's a right-angled triangle.

According to the Pythagorean theorem:

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

Substituting the values of AB and BC:

$\quad AC^2 = 12^2 + 12^2$

$\quad AC^2 = 144 + 144$

$\quad AC^2 = 288$

Now, we find the square root of 288:

$\quad AC = \sqrt{288} = \sqrt{144 \times 2} = 12\sqrt{2}$ cm.

So, the length of the hypotenuse AC is $12\sqrt{2}$ cm.

Determining the Radius of the Inscribed Circle (Inradius)

For a right-angled triangle with legs of length 'a' and 'b' and hypotenuse 'c', the radius of the inscribed circle (inradius), denoted by 'r', is given by the formula:

$\quad r = \frac{a + b - c}{2}$

In our triangle ABC, the legs are AB = 12 cm and BC = 12 cm, and the hypotenuse is AC = $12\sqrt{2}$ cm.

Substituting these values into the formula:

$\quad r = \frac{12 + 12 - 12\sqrt{2}}{2}$

$\quad r = \frac{24 - 12\sqrt{2}}{2}$

Now, divide both terms in the numerator by 2:

$\quad r = 12 - 6\sqrt{2}$ cm.

Determining the Radius of the Circumscribing Circle (Circumradius)

For a right-angled triangle, the center of the circumscribing circle (circumcenter) is located exactly at the midpoint of the hypotenuse. The radius of the circumscribing circle (circumradius), denoted by 'R', is half the length of the hypotenuse.

The hypotenuse AC is $12\sqrt{2}$ cm.

The formula for the circumradius of a right-angled triangle is:

$\quad R = \frac{\text{Hypotenuse}}{2}$

Substituting the length of AC:

$\quad R = \frac{12\sqrt{2}}{2}$

$\quad R = 6\sqrt{2}$ cm.

Calculating the Ratio of Inradius to Circumradius

We need to find the ratio of the radius of the inscribed circle (r) to the radius of the circumscribing circle (R), which is r : R.

We found that $r = 12 - 6\sqrt{2}$ and $R = 6\sqrt{2}$.

The ratio is: $(12 - 6\sqrt{2}) : (6\sqrt{2})$

To simplify this ratio, we can divide both parts by the greatest common divisor of the coefficients, which is 6.

Divide the first part by 6: $\frac{12 - 6\sqrt{2}}{6} = \frac{12}{6} - \frac{6\sqrt{2}}{6} = 2 - \sqrt{2}$.

Divide the second part by 6: $\frac{6\sqrt{2}}{6} = \sqrt{2}$.

So, the simplified ratio is $(2 - \sqrt{2}) : \sqrt{2}$.

Let's check the given options:

Option 1: $6 - \sqrt{2} : 3\sqrt{2}$

Option 2: $2 - \sqrt{2} : \sqrt{2}$

Option 3: $6 - 3\sqrt{2} : 1\sqrt{2}$ (Can be simplified by dividing by 3: $2 - \sqrt{2} : \frac{\sqrt{2}}{3}$) - Incorrect

Option 4: $6 - 3\sqrt{2} : 6\sqrt{2}$ (Can be simplified by dividing by 3: $2 - \sqrt{2} : 2\sqrt{2}$) - Incorrect

Our calculated ratio $(2 - \sqrt{2}) : \sqrt{2}$ matches Option 2.

Revision Table: Formulas for Right-Angled Triangles

Additional Information: Incenter and Circumcenter

• The incenter of any triangle is the point where the angle bisectors meet. It is the center of the inscribed circle.
• The circumcenter of any triangle is the point where the perpendicular bisectors of the sides meet. It is the center of the circumscribing circle.
• For a right-angled triangle:
  • The incenter is inside the triangle.
  • The circumcenter is always at the midpoint of the hypotenuse.
• For an isosceles right-angled triangle, the incenter lies on the altitude from the right angle to the hypotenuse (which is also the median and angle bisector).