Chapter 8.2, Problem 18WE

To determine

To find: length of second diagonal of rhombus.

Answer to Problem 18WE

  $16\text{ cm}$

Explanation of Solution

Given:

The perimeter of rhombus is 40 cm and one diagonal of a rhombus of length is 12 cm.

Concept Used:

Theorem 5-13: Rhombus’s diagonals are perpendicular bisector.

Theorem 8-2: In right triangle, square of hypogenous is equal to the square of base and perpendicular.

  $\text{Perimeter of rhombus}=4\times \text{side}$

It is given that perimeter of rhombus is 40 cm.

Then,

  $\begin{array}{c}\text{Perimeter of square}=4\times \text{side}\\ 40=4\times \text{side}\\ \frac{40}{4}=\text{side}\\ 10=\text{side}\end{array}$

Draw a rhombus with sides of length 10 cm and one diagonal of a rhombus of length 12.

Now, from theorem 5-13, the diagonals $\overline{AC}$ and $\overline{BD}$ bisect each other and are also perpendicular. That is,

  $AO=OC$

  $DO=OB$

  $m\angle AOD=90$

Now, it is given that $DB=12$ , so

  $\begin{array}{c}DB=12\\ DO+OB=12\\ DO+DO=12\end{array}$

  $\begin{array}{c}2DO=12\\ DO=6\end{array}$

Now, let $AO=x$

Consider the right triangle

  $△ADO$ .

Now, $\overline{AD}$ is the hypotenuse and has length 10 and the lengths of the legs are $AO=x$ and

  $DO=6$ .

So, by theorem 8-2,

  $\begin{array}{c}A{D}^2=A{O}^2+O{D}^2\\ {10}^2=x^2+6^2\\ 100=x^2+36\end{array}$

  $\begin{array}{c}100-36=x^2\\ 64=x^2\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{64}\\ x=8\end{array}$

That is,

  $AO=OC=8$

And thus,

  $\begin{array}{c}AC=AO+OC\\ =8+8\\ =16\end{array}$

The length of the other diagonal of rhombus is 16 cm.