Chapter 11, Problem 2CR

To determine

To Find: The area of rectangle.

Answer to Problem 2CR

  $8\sqrt{5}sq.unit$

Explanation of Solution

Given: $length=4,diagonal=6$

Given: $length=4,diagonal=6$

Using Pythagoras theorem to find the width of the rectangle.

  $\begin{array}{l}{6}^2=4^2+w^2\\ 36-16=w^2\\ w^2=20\\ w=2\sqrt{5}\end{array}$

Area of rectangle.

  $\begin{array}{l}A=lb\\ A=4\times 2\sqrt{5}\\ A=8\sqrt{5}sq.unit\end{array}$