Chapter 6, Problem 23RE

To determine

To Find: The lengths of the sides of the parallelogram

Answer to Problem 23RE

The lengths of the sides of the parallelogram are $4.3\text{ feet and 12}\text{.6 feet}$

Explanation of Solution

Given information: The length of the diagonals of a parallelogram are 10 feet and 16 feet

Concept used: ABC is a triangle with sides $a,b,c$ then Law of cosine

  $c^2=a^2+b^2-2ab\cos \theta$

Calculation:

The figure of the parallelogram is shown below:

  

From, the above figure, lengths of the sides of the parallelogram are obtained as:

  $\begin{array}{l}{c}_1{}^2=a^2+b^2-2ab\cos \theta \\ c_1{}^2=8^2+5^2-2\times 8\times 5\cos {28}^{\circ }\end{array}$

Solve,

  $c_1=4.3\text{ feet}$

And,

  $\begin{array}{l}{c}_2{}^2=a^2+b^2-2ab\cos \theta \\ c_2{}^2=5^2+8^2-2\times 5\times 8\cos {152}^{°}\\ c_1\approx 12.6 \text{ feet}\end{array}$ [Solve]

Hence,

The lengths of the sides of the parallelogram are $4.3\text{ feet and 12}\text{.6 feet}$