Question

solve for ( x ) to the nearest tenth.
answer
( x=)

Explanation:

Step1: Use the Pythagorean theorem for similar right - angled triangles

By the geometric mean theorem (a property of right - angled triangles), we have the relationship \(x=\frac{\sqrt{(9^{2}-6^{2})(5^{2})}}{5}\). First, find the length of the common hypotenuse - related segments.
For the larger right - angled triangle with sides \(6\), \(x\) and hypotenuse \(9\), and the smaller right - angled triangle with sides \(5\), \(x\) and the other part of the hypotenuse. Using the fact that in right - angled triangles, if two right - angled triangles are similar (by AA similarity, as they both have a right angle and share an acute angle), we can also use the formula \(x=\sqrt{9^{2}-6^{2}-5^{2}+ \frac{(9^{2}-6^{2})\times5^{2}}{9^{2}}}\) (derived from Pythagorean theorem and similarity). A more straightforward way is using the formula \(x = \sqrt{6^{2}-(9 - 5)^{2}}\) (another approach based on Pythagorean theorem in the sub - triangles).
Another way:
Let's use the property of right - angled triangles. If we consider the two right - angled triangles (the smaller one with legs \(5\) and \(x\) and the larger one with hypotenuse \(9\) and one leg \(6\)).
We know that by the Pythagorean theorem, for the triangle with sides \(6\), \(x\) and hypotenuse \(9\): the other non - hypotenuse side (let's call it \(a\)) of the larger triangle (excluding \(x\)) is \(a=\sqrt{9^{2}-6^{2}}=\sqrt{81 - 36}=\sqrt{45}=3\sqrt{5}\).
For the smaller triangle with sides \(5\) and \(x\), using the similarity of triangles (the two right - angled triangles are similar).
We can also use the formula \(x=\sqrt{6^{2}-(9 - 5)^{2}}\) (based on the fact that if we consider the projection and Pythagorean theorem in sub - triangles).

$$x=\sqrt{6^{2}-(9 - 5)^{2}}=\sqrt{36-16}=\sqrt{20}$$

Step2: Calculate the value of \(x\)

$$x=\sqrt{20}\approx4.5$$

Answer:

\(4.5\)