Chapter 7.4, Problem 61PPS

To determine

To determine the perimeter and the area of the given rectangle

Answer to Problem 61PPS

Perimeter = $10\left(\sqrt{7}+\sqrt{5}\right)$

Area = $72+12\sqrt{35}$

Explanation of Solution

Given:

  

Formula used:

Let, L and B be the length and breadth of the rectangle.

Perimeter = P = 2( L + B )

Area = A = LB

Calculation:

From the given figure, it is clear that,

  $\begin{array}{l}L=3\sqrt{7}+3\sqrt{5}\\ B=2\sqrt{7}+2\sqrt{5}\end{array}$

Hence,

Area of the rectangle:

  $\begin{array}{l}\Rightarrow A=L\times B\\ \Rightarrow A=(3\sqrt{7}+3\sqrt{5})(2\sqrt{7}+2\sqrt{5})\\ $ \Rightarrow A=6{(\sqrt{7}+\sqrt{5})}^2$\Rightarrow A=6({\sqrt{7}}^2+{\sqrt{5}}^2+2\times \sqrt{7}\times \sqrt{5})\\ \Rightarrow A=6(7+5+2\sqrt{35})\\ \Rightarrow A=72+12\sqrt{35}\end{array}$

Perimeter of rectangle:

  $\begin{array}{l}\Rightarrow P=2(L+B)\\ \Rightarrow P=2(3\sqrt{7}+3\sqrt{5}+2\sqrt{7}+2\sqrt{5})\\ \Rightarrow P=2(5\sqrt{7}+5\sqrt{5})\\ \Rightarrow P=2\times 5\times (\sqrt{7}+\sqrt{5})\\ \Rightarrow P=10(\sqrt{7}+\sqrt{5})\end{array}$

Conclusion:

Hence, the perimeter is $10\left(\sqrt{7}+\sqrt{5}\right)$ and area of the rectangle is $72+12\sqrt{35}$ .