Chapter 12.8, Problem 11P

To determine

Find the length of other diagonal.

Answer to Problem 11P

5.5

Explanation of Solution

Formula Used:

Law of Cosine :

  $a^2=b^2+c^2-2bc\cos A$

Law of Sine:

  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

  

Given:

In quadrilateral ABCD, AB = 3, BC = 4, CD = 5, and DA = 6.The length of diagonal $\overline{BD}$ is 7.

Calculation:

Represent the situation and solve :

  

Use Law of Cosine:

  $\begin{array}{l}D{C}^2=B{D}^2+B{C}^2-2(BD)(BC)\cos \angle DBC\\ \Rightarrow 5^2=7^2+4^2-2(7)(4)\cos \angle DBC\\ \Rightarrow \cos \angle DBC=0.7143\\ \Rightarrow \angle DBC=44°\end{array}$

  $\begin{array}{l}A{D}^2=B{D}^2+A{B}^2-2(BD)(AB)\cos \angle ABD\\ \Rightarrow 6^2=7^2+3^2-2(7)(3)\cos \angle ABD\\ \Rightarrow \cos \angle ABD=0.5238\\ \Rightarrow \angle ABD=58°\end{array}$

So, $\angle ABC=58°+44°=102°$

Now, Use Law of Cosine to find AC:

  $\begin{array}{l}A{C}^2=A{B}^2+B{C}^2-2(AB)(BC)\cos \angle ABC\\ \Rightarrow A{C}^2=3^2+4^2-2(3)(4)\cos 102°\\ \Rightarrow AC=\sqrt{30}\\ \Rightarrow AC\approx 5.5\end{array}$

So, the length of the other diagonal is 5.5