Chapter 11.1, Problem 34WE

To determine

To Find: The area of rectangle.

Answer to Problem 34WE

  $81\sqrt{3}c{m}^2$

Explanation of Solution

Given: The diagonals of a rectangle is $18cm$ long and intersect at ${60}^{°}$ angle.

The diagonals of a rectangle is $18cm$ long

i.e.

  $AC=BD=18cm$

  

We know that the diagonals of a rectangle bisect each other.

Thus

  $\begin{array}{l}OC=\frac{AC}{2}=\frac{18}{2}=9cm\\ \angle COE=\frac{\angle COD}{2}=\frac{60}{2}={30}^{°}\\ OE=\frac{AD}{2}=\frac{l}{2}\\ CE=\frac{CD}{2}=\frac{b}{2}\end{array}$

Now in triangle $OCE$

  $\begin{array}{l}\sin {30}^{°}=\frac{1}{2}=\frac{CE}{OC}\\ b=9cm\\ \cos {30}^{°}=\frac{\sqrt{3}}{2}=\frac{OE}{OC}\\ l=9\sqrt{3}cm\end{array}$

Thus

  $l=9\sqrt{3}cm,b=9cm$

Then area of rectangle

  $\begin{array}{l}A=lb\\ A=(9\sqrt{3})(9)\\ A=81\sqrt{3}c{m}^2\end{array}$

Hence, $81\sqrt{3}c{m}^2$