We are given a diagram of a right triangle. The hypotenuse has length 6. The triangle is divided into two smaller triangles by the altitude to the hypotenuse. The two smaller triangles are similar to each other and to the original triangle. The length of one of the sides is given as $x$ and the length of the hypotenuse is 6. We need to find the length of $x$ to the nearest tenth. The tick marks on the original triangle indicate that the altitude to the hypotenuse divides the hypotenuse into two equal segments. This means the original triangle is an isosceles right triangle, which means the two legs have equal length.

1. Problem Description

We are given a diagram of a right triangle. The hypotenuse has length

6. The triangle is divided into two smaller triangles by the altitude to the hypotenuse. The two smaller triangles are similar to each other and to the original triangle. The length of one of the sides is given as $x$ and the length of the hypotenuse is

6. We need to find the length of $x$ to the nearest tenth. The tick marks on the original triangle indicate that the altitude to the hypotenuse divides the hypotenuse into two equal segments. This means the original triangle is an isosceles right triangle, which means the two legs have equal length.

2. Solution Steps

Let the length of each leg be $l$.

Using the Pythagorean theorem on the original right triangle, we have:

$l^2 + l^2 = 6^2$

$2l^2 = 36$

$l^2 = 18$

$l = \sqrt{18} = 3\sqrt{2}$

Since the two smaller triangles are similar to the original triangle and the altitude to the hypotenuse bisects the hypotenuse, then the two legs of the original right triangle are equal in length. The altitude to the hypotenuse is also $l/2$

Let $x$ be the length of the side of one of the smaller triangles. We know that the hypotenuse of the small triangle has length equal to one half of the hypotenuse of the original triangle. So the length of the hypotenuse of the small triangle is $6/2 = 3$.

Also, the small triangle is an isosceles triangle.

Let $x$ represent the length of the leg of the small isosceles right triangle. Then

$x^2 + x^2 = 3^2$

$2x^2 = 9$

$x^2 = \frac{9}{2}$

$x = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$

$x = \frac{3(1.414)}{2} = \frac{4.242}{2} = 2.121$

Rounding to the nearest tenth, we get $x \approx 2.1$.

3. Final Answer

2.1