The best strategy to solve this problem is to just dive in.

Largest circle inscribed within a right triangle means that this is a isosceles right triangle. For an isosceles right triangle there are two equal sides a. Using the Pythagoream theorem you can determine that the hypotenuse side:

\(a^2\) + \(a^2\) = \(c^2\)

Thus, c = \(\sqrt{2}a\)

With that in mind, the perimeter of the triangle is 20 units.

\(a + a + \sqrt{2}a\) = 20

a ~ 6

Now, look at the diagram shown below. For a right triangle with sides a, b, and hypotenuse c, the area is \(A = \frac{1}{2}ab\)

The inradius can be found by equating the area of the triangle ABC with sum of the areas of the triangles ACI, BCI and ABI.

The altitudes for these triangles serve as the inradii for the circle. Let's call this radius (shown in red), r. This gives us

\(\frac{1}{2}ab = \frac{1}{2}ra + \frac{1}{2}rb + \frac{1}{2}rc = \frac{1}{2}r(a + b + c)\)

Solving for r gives \(r = \frac{ab}{a + b + c}\)

In this case we already established above that \(a = b\) and \(c = \sqrt{2}a\) and that \(a = 6\)

Thus, using these values we establish that the radius of the circle, \(r = 1.8\)

Plugging this into the formula for area of a circle, \(A = pi (1.8)^2\)

This gives us approx. \(4pi\)

Thus, the answer is C

For more Right Triangle properties, check out:

http://mathworld.wolfram.com/RightTriangle.htmlHope this helps! Good luck studying!

Attachments

right triangle.jpg [ 26.23 KiB | Viewed 2188 times ]