Chapter 8.1, Problem 50HP

To determine

Tocalculate:

The value of $x,y,z$ by referring given figure.

Answer to Problem 50HP

The values of $x,y,z$ are $\left(5.2\right),\left(6.7\right),\left(11.1\right)$ .

Explanation of Solution

Given information:

  

Calculation:

The altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments. The length of leg of this triangle is the geometric mean between the length of the hypotenuses and the segment of the hypotenuse adjacent to that leg.

So, geometric mean leg theorem shows that,

  $8.5=\sqrt{x\left(x+8.8\right)}$

Now, square each side

  ${8.5}^2=x\left(x+8.8\right)$

Now simplify,

  $72.25=x^2+8.8x$

By using the multiplication property,

  $0=x^2+8.8x-72.25$

Solve the resulting equation with quadratic formula.

So use the quadratic formula

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Now substitute,

  $=\frac{-8.8\pm \sqrt{{8.8}^2-4\left(1\right)\left(-72.25\right)}}{2\left(1\right)}$

Substitute,

  $\begin{array}{l}\approx \frac{-8.8\pm 19.1}{2}\\ =-13.95or5.15\end{array}$

The negative solution $x=-13.95$ is extraneous, because a length must be positive. Thus, $x=5.15$ .

Geometric mean leg theorem shows that,

  $\begin{array}{l}z=\sqrt{8.8\left(8.8+x\right)}\\ =\sqrt{8.8\left(8.8+5.15\right)}\\ =\sqrt{122.76}\\ \approx 11.1\end{array}$

The altitude drawn to the hypotenuse of a right triangle separates the hypotenuse into two segments. The length of this altitude is the geometric mean between the lengths of these two segments.

Geometric mean altitude theorem shows that,

  $\begin{array}{l}y=\sqrt{x\left(8.8\right)}\\ =\sqrt{\left(5.15\right)\left(8.8\right)}\\ =\sqrt{45.32}\\ \approx 6.7\end{array}$

Hence, the values are $x\approx \left(5.2\right),y\approx \left(6.7\right),z\approx \left(11.1\right)$ .