Chapter 6, Problem 58SGR

To determine

The perimeter and area of the rectangle having length $8+\sqrt{3}$ and width $6-\sqrt{2}$ .

Answer to Problem 58SGR

The perimeter of rectangle is $28+2\sqrt{3}-2\sqrt{2}$ units and area of rectangle is $48-8\sqrt{2}+6\sqrt{3}-\sqrt{6}$ square units.

Explanation of Solution

Given information: It is given that length and width of rectangle are $8+\sqrt{3}$ and $6-\sqrt{2}$ respectively.

Formula used:

  $\begin{array}{c}\text{Perimeter of rectangle}=2\left(\text{length+width}\right)\\ \text{Area of rectangle}=\text{length}\times \text{width}\end{array}$

Calculation: With the help of formulas the perimeter and area of rectangle are evaluated as,

  $\begin{array}{c}\text{Perimeter of rectangle}=2\left(\left(8+\sqrt{3}\right)\text{+}\left(6-\sqrt{2}\right)\right)\\ =2\left(14+\sqrt{3}-\sqrt{2}\right)\\ =28+2\sqrt{3}-2\sqrt{2}\text{ units}\end{array}$

Similarly for area of rectangle,

  $\begin{array}{c}\text{Area of rectangle}=\left(8+\sqrt{3}\right)\times \left(6-\sqrt{2}\right)\\ =8\left(6-\sqrt{2}\right)+\sqrt{3}\left(6-\sqrt{2}\right)\\ =48-8\sqrt{2}+6\sqrt{3}-\sqrt{6}\text{ square units}\end{array}$