Chapter 8.2, Problem 17WE

To determine

To find: perimeter of rhombus.

Answer to Problem 17WE

  $68$

Explanation of Solution

Given:

The diagonals of a rhombus of lengths are 16 and 30.

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.

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

Draw a rhombus with diagonals of a rhombus of lengths 16 and 30.

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 $AC=30$ , so

  $\begin{array}{c}AC=30\\ AO+OC=30\\ AO+AO=30\end{array}$

  $\begin{array}{c}2AO=30\\ AO=15\end{array}$

Similarly, it is given that $DB=16$ , so

  $\begin{array}{c}DB=16\\ DO+OB=16\\ DO+DO=16\end{array}$

  $\begin{array}{c}2DO=16\\ DO=8\end{array}$

Now, let the length of the side of rhombus be x. Then,

Consider the right triangle $△ADO$ .

Now, $\overline{AD}$ is the hypotenuse and has length x and the lengths of the legs are 8 and 15.

So, by theorem 8-2,

  $\begin{array}{c}A{D}^2=A{O}^2+O{D}^2\\ x^2={15}^2+8^2\\ x^2=225+64\\ x^2=289\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{289}\\ x=17\end{array}$

The length of side of rhombus is 17.

Now, perimeter of rhombus is,

  $\begin{array}{c}\text{Perimeter of rhombus}=4\times \text{side}\\ =4\times 17\\ =68\end{array}$